Collins Cambridge International AS & A Level Physics

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Dynamics

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prior understanding

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Who crash-tests aeroplanes? This empty Cessna 172 aeroplane has just been dropped from a height of 25 m at NASA’s Landing and Impact Research Facility.

The purpose of this crash test is to test whether an emergency locator beacon will still work after the impact. If the emergency beacon still works, the aeroplane will be found more quickly after an accident. Damage in impacts like this happens because of the large forces that act during sudden deceleration.

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You may remember adding and resolving vectors from Chapter 1. You will be finding resultant forces and components of momentum vectors in in this Chapter.

You may recall the definitions of velocity and acceleration from Chapter 2 and the use of velocity-time graphs. You may have considered acceleration in free fall. It will also be useful if you recall the concepts of weight, mass, kinetic energy, forces and equilibrium that you may have covered previously.

Learning objectives

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In this chapter you will learn the meaning of and the relationship between mass and weight. You will learn how force, mass and acceleration are related and will discover the concept of momentum. You will learn Newton’s laws of motion and how to apply them. You will also discover how friction and drag forces affect the motion of objects. You will explore whether momentum and kinetic energy are conserved in elastic and inelastic interactions and discover the principle of conservation of momentum.

3.1 Mass and weight (syllabus 3.1.1, 3.1.6) 3.2 Newton’s first and second laws of motion; momentum (syllabus 3.1.2–3.1.5) 3.3 forces in interactions: Newton’s third law (syllabus 3.1.5) 3.4 Motion with resistive forces (syllabus 3.2.1–3.2.3) 3.5 conservation of momentum (syllabus 3.3.1, 3.3.2) 3.6 Momentum and kinetic energy (syllabus 3.3.3, 3.3.4)

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3 Dynamics

3.1 Mass and weight What is mass?

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Weight and mass are two terms that are often confused. This is because they are not always used correctly in everyday speech.

Weight is the force of gravity on an object. Its unit is newtons, N, and it is a vector quantity. Weight acts in the same direction as gravity. On Earth, this is towards the centre of the Earth. Weight of an object is not constant as it varies according to the strength of the gravitational field.

The term gravitational field is used to describe any place where the effects of gravity can be detected. Hence, all objects will have weight when they are in a gravitational field.

The gravitational force on an object of mass m is F = mg

where g is the gravitational field strength in N kg–1, defined as the gravitational force (in N) on a 1 kg mass. The term g is also the acceleration of free fall (or acceleration due to gravity) in m s–2.

In Topic 3.2 you will learn that the equation F = ma is a form of Newton’s second law.

Comparing F = mg with F = ma shows you that g is the acceleration of the mass m.

Since g is constant in a uniform gravitational field, all masses have the same acceleration when they fall vertically in this field, if gravity is the only force acting.

Near the Earth’s surface g = 9.81 N kg–1 = 9.81 m s–2.

The gravitational force F is the object’s weight, and so can be assigned the term W, and we have W = mg

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The concept of free fall was introduced in Topic 2.3. All objects in free fall have the same constant acceleration near the Earth’s surface, ignoring the effect of air resistance.

What is weight?

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You will learn more about the forces acting on objects due to gravitational fields in Chapter 13. You will discover how the force between two masses depends on the masses and the separation between them.

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Mass can be defined as the quantity of matter in an object. Mass is a scalar quantity which has no direction associated with it. Mass is constant for any object that is at rest or travelling much slower than the speed of light. The unit of mass used in equations is the SI unit kilogrammes, kg.

The mass of an object is also a measure of its resistance to change in motion, that is a measure of the object’s inertia.

Think of a table tennis ball and a golf ball. Both of these have a diameter of about 40 mm, but their masses are very different. The mass of the table tennis ball is only 2.7 g while the mass of the golf ball is 46 g. The difference in mass is because the table tennis ball only has air inside, while the golf ball is solid rubber inside.

Both these balls are placed on a flat surface such as a table top. Now imagine blowing on them. Which one is easier to move?

The table tennis ball is easier to move because it has less mass.

Link In Topic 3.2 we will show that, by definition, 1 N = 1 kg m s–2. From this we can see that 1 N kg–1 = 1 m s–1.

weight in N = mass in kg × acceleration of free fall in m s–2 Consider a 1 kg mass and a 2 kg mass being dropped together. Their weights are the only forces acting on them. The weight of the 1 kg mass is 1 kg × 9.81 m s–2 = 9.81 N and the weight of the 2 kg mass is 2 kg × 9.81 m s–2 = 19.62 N. 9.81 kg ms −2 The acceleration of the 1 kg mass is given by a = F = = 9.81 m s–2 1 kg m 19.62 kg ms −2 The acceleration of the 2 kg mass is given by a = F = = 9.81 m s–2 2 kg m

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3.2 Newton’s first and second laws of motion; momentum

Key ideas ➜➜Mass is a property of an object that resists a change in its motion. This resistance to change in motion is called inertia. ➜➜Mass is a scalar quantity with unit kg. ➜➜The gravitational force on an object is its weight. Weight is a vector with unit N. ➜➜Weight is the product of an object’s mass its acceleration of free fall.

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1. A block of metal has a mass of 6.7 kg. (a) Calculate the weight of the block on Earth where g = 9.81 m s–2. (b) Calculate the weight of the block on the Moon where g = 1.63 m s–1.

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3.2 Newton’s first and second laws of motion; momentum Newton’s first law

Bus slows down

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Imagine you are sitting on a moving bus when the driver suddenly applies the brakes.

What happens to you?

Link Inertia is the resistance to motion that an object has because of its mass.

See Topic 3.1.

Figure 3.1 You continue forwards when the bus decelerates. Your tendency to continue forwards is due to your inertia.

You keep moving forward when the bus brakes because, while there is a force on the bus to make it stop, no force acts on you.

In 1687 Isaac Newton published Mathematical Principles of Natural Philosophy in which he described the motion of objects. His laws of motion were formulated in the book, and they can be used to describe the movement of all objects (although some corrections need to be applied for very small particles, such as electrons, and for objects travelling close to the speed of light).

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3 Dynamics Newton’s first law states that:

The word normal means perpendicular.

The normal contact/ reaction force is the component of the contact/reaction force that acts perpendicular to the surface.

An object remains at rest or continues to move with uniform velocity unless acted upon by a resultant force.

Think of a person skating in a straight line on horizontal smooth ice. Imagine there is no friction between the ice and the skates, and no air resistance. The skater will continue at the same speed and in the same straight line.

In all real scenarios there are several forces acting. When a car is travelling at a constant velocity in a straight line on a horizontal road, the forces on the car can be represented by the four vectors in Figure 3.2. normal contact forces

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resistive forces

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Figure 3.2 The forces acting on a car when it is in uniform motion.

• First, the car is not moving up or down because the total of the normal contact forces are equal and opposite to its weight. The normal contact forces here come from the ground pushing up on all four tyres. This type of force is often called a ‘normal reaction force’.

• The car is not getting faster or slower because the driving force is equal and opposite to the resistive forces.

• The car is travelling in a straight line because there are no forces acting to its right or left (into or out of the page).

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sum of normal contact forces

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The resultant force is found by the vector addition of all the forces acting. Recall vector addition from Chapter 1. You will meet resultant forces again in the context of equilibrium in Chapter 4, topic 00. You will also learn more about other conditions for equilibrium in Chapter 4.

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The car is said to be in uniform motion because the forces on it are balanced. The resultant force on it is zero. The resultant force is the sum of all forces acting driving resistive on an object. As forces are vector quantities, those of equal magnitude acting in force forces opposite directions will have a sum of zero. When all the forces on an object are balanced, an object is said to be in translational equilibrium.

It is useful to draw a simple diagram that represents all the forces acting on an object. A diagram that shows only the forces acting on one object, which is weight represented by a dot, is called a free body diagram.

Figure 3.3 A free body diagram for the Newton’s first law describes why the car, if travelling at a constant speed in a car in Figure 3.2 straight line, will not speed up, slow down or change direction unless one of the forces becomes larger than the one opposing it.

Tip Newton’s first law also describes why the car, if stationary, will not begin to move until We cannot tell the one of the forces becomes larger than the one opposing it. difference between an object being at rest (stationary) and it having uniform velocity by looking at a vector diagram. All of the forces will be equal and opposite, giving a resultant force of zero in both cases.

2. (a) (i) Name the force or forces acting on an object at rest on a flat surface. (ii) Draw a free body diagram to illustrate your answer to part (i). (b) Draw two other free body diagrams showing the forces: (i) when the object is being pushed against resistive forces to start movement (ii) when the object is moving at constant speed with resistive forces.

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3.2 Newton’s first and second laws of motion; momentum

Newton’s second law

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Look again at the free body diagram for the car in equilibrium in Figure 3.4. At equilibrium, the driving force and the resistive forces are equal and opposite, so there is no resultant force forward or backward.

If the driving force starts to become larger than the resistive forces then the car will accelerate forwards (to the left in the diagram). If the driving force again becomes equal to the resistive forces, the car will again move at a constant speed. If the resistive forces become larger than the driving force, then the car will decelerate.

Newton’s second law describes the relationship between resultant force and the changes in motion that are produced by the resultant force. You may recall from previous courses the relationship between resultant force F, mass m and acceleration a, that is

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force (in N) = mass (in kg) × acceleration (in m s–2) F = ma

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This equation is a useful summary of Newton’s second law for an object of constant mass m. The equation is used to define the SI unit of force, the newton, N. One newton is the resultant force that gives a mass of one kilogram an acceleration of one metre per second.

1 N = 1 kg m s–2.

At the start of topic 3.1 we thought about why a table tennis ball is easier to blow across a table than a golf ball, and introduced the concept of inertia. Newton’s second law allows us to quantify the effect of inertia. It tells us why mass makes a difference when you try to accelerate an object. A larger mass will require a larger force to accelerate it by the same amount than a smaller mass.

Figure 3.4 In order to get the same acceleration, you need to push the car with a much greater force than the shopping trolley.

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3. The person in Figure 3.4 pushes the car and the shopping trolley separately on level ground with a force of 520 N on each object. Ignore friction. (a) Calculate the initial acceleration of the car, whose mass is 950 kg. (b) Calculate the initial acceleration of the shopping trolley, whose mass is 34 kg.

Momentum

A fast moving object with a large mass, such as the truck in Figure 3.5, is difficult to stop. Newton’s second law tells us why.

Say the truck has a mass of 44 000 kg and a velocity of 25 m s–1.

What force is needed to bring the lorry to a stop in a time of 10 s?

The average acceleration is change in velocity divided by the time taken, a= 0−v t

Figure 3.5 A large truck moving at speed needs a large to stop it.

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3 Dynamics Using F = ma, the magnitude of F is 4.4 × 104 kg × 25m s-1 = 1.1 × 105 N F = mv = 10s t This is a very large force.

You can see from the equation F = mv/t above that an object with a smaller mass but a very high speed will also require a very large force to stop it. The product mv is a property of a moving object that we call momentum.

Momentum is given the symbol p and is defined as the product of mass and velocity. It is a vector quantity. Its units are the product of the units of mass and velocity, that is kg m s–1.

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momentum (kg m s−1) = mass (kg) × velocity (m s–1) p = mv

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The units of momentum can also be written as N s. You saw earlier that, from F = ma, the newton N is equivalent to the base SI units kg m s–2. Hence N s = kg m s–2 × s = kg m s–1 Worked example

The truck in Figure 3.5 has a mass of 44 000 kg and a velocity of 25 m s–1.

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Calculate the momentum of the truck.

Answer p = mv = 44 000 kg × 25 m s–1 = 1.1 × 106 kg m s–1

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4. (a) A bullet has a mass of 22 g and a velocity of 490 m s–1. Calculate its momentum. (b) The New Horizon space probe, mass 470 kg, left Earth’s orbit with a speed of 16 000 m s–1. Calculate its momentum. Give your answer in standard form.

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The force required to bring the lorry to a stop depends not only on its momentum, but also on how quickly we want to stop it. To stop it in a shorter time would need a larger force than to stop it in a longer time.

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Worked example

What force would be required to stop the lorry in Figure 3.5 in 5 s rather than in 10 s?

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4.4 × 104 kg × 25m s-1 F = ma = mv = = 2.2 × 105 N 5.0s t

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You can refer to the Experimental Skills section in Topic 2.3 for a reminder of direct proportionality.

To stop the lorry in half the time requires double the force. We describe this relationship between force and time as inversely proportional.

Inversely proportional means that as one quantity increases the other quantity decreases by the same factor. For example, if y is inversely proportional to x, then as y doubles, x halves. We can write this inverse proportional relationship as y ∝ 1 . In this x case, F ∝ 1 . t The force needed to stop the lorry is directly proportional to its momentum and inversely proportional to the time taken.

5. A tennis ball of mass 65 g is thrown with a velocity of 22 m s–1. A person catches the ball and brings it to rest in a time of 0.25 s.

Calculate the force required to bring the ball to rest.

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3.2 Newton’s first and second laws of motion; momentum

Experimental skills: Force, mass and acceleration

remaining masses

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Apparatus The apparatus is shown in Figure 3.6.

Position the pulley (or smooth cylinder) so that the mass hanger can fall vertically through a distance of 1.00 m or more.

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The task In Figure 3.6, a falling mass is attached to the trolley by a thread that has very little stretch. The trolley is held in place then accelerates from rest when released.

If we ignore friction the accelerating force is the same size as the weight of the falling masses. The size of the force can be altered by varying these masses.

The average acceleration of the trolley can be measured by timing how long it takes the trolley to move a distance of 1.00 m between two marks, assuming the trolley accelerates uniformly.

Questions P1. (a) Sketch a velocityLink time graph for a You may want to uniformly refer back to the accelerating trolley. equations of motion (b) Show how to in Topic 2.2. calculate the acceleration of the trolley from measurements of the time taken to move a distance of 1.00 m from rest.

P2. Describe what precautions you should take to prevent injury to yourself and damage to the floor.

P3. (a) Describe how you would ensure that the acceleration is as uniform as possible between the start and the 1.00 m mark. (b) Explain the purpose of the trial runs in selecting the starting mass on the hanger. (c) Explain why masses are removed from the trolley and added to the hanger, rather than keeping the mass of the trolley constant and only increasing the mass on the hanger. (d) Explain how to calculate the force acting on the trolley from the mass on the hanger.

P4. Draw a suitable table to record the results for this experiment, allowing for repeat measurements and for calculated quantities derived from the raw data.

P5. Below are the results recorded by a student in this investigation. The values for acceleration are mean values from repeat measurements. (a) Use these results to plot a graph to determine whether they show that acceleration is directly proportional to force. Draw a straight line of best fit.

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Newton’s second law describes how a resultant force affects the motion of an object. It is summarised in the form F = ma for an object with constant mass m. The acceleration a is directly proportional to the resultant force F causing the acceleration and so a graph of a against F should be a straight line through the origin.

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Figure 3.6

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Techniques Record the mass of the trolley.

Mark a distance of 1.00 m in the direction of the string from the starting point of the trolley.

Carry out some trials to see how long it takes the trolley to cover 1.00 m. Use no extra mass on the trolley for the trial and vary the masses on the hanger.

In the main investigation, start with the maximum mass, chosen from the trial results. Place all but one of the masses on the trolley, and only one of the masses on the hanger. In each new run a mass is removed from the trolley and placed on the hanger. (Or if a mass is added to the trolley, it must be taken from the hanger.)

F = 0.981 N, a = 0.531 m s–2; F = 1.96 N, a = 1.14 m s–2; F = 2.94 N, a = 1.79 m s–2; F = 3.92 N, a = 2.41 m s–2; F = 4.91 N, a = 3.10 m s–2 (b) Explain whether or not the graph shows that acceleration is directly proportional to force. (c) Suggest why the graph does not pass through the origin.

A level Analysis P6. Describe how to adapt the method and the analysis of results, to show that the acceleration of an object is inversely proportional to the mass of the object.

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3 Dynamics

General form of Newton’s second law In the case of the truck (Figure 3.5), we looked at reducing its velocity from v to zero, that is, changing its momentum from mv to zero. In general, for a change in momentum Δ(mv) in a time Δt, the equation for the resultant force is ∆p F = ∆(mv ) or F = ∆t ∆t The right-hand side of this equation is rate of change of momentum.

The general form of Newton’s second law is stated as:

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The rate of change of momentum of an object is directly proportional to the resultant force acting on it. The change in momentum occurs in the direction of the resultant force.

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The equals sign in the equation above comes about from the definition of the newton as an SI unit: 1 N = 1 kg m s–2.

6. A football player kicks a football of mass 0.43 kg. The player’s foot exerts an average force of 130 N on the ball, which leaves the player’s foot with a speed of 31 m s–1.

Calculate the time that the ball was in contact with the player’s foot.

7. A box is being transported in a van as shown in Figure 3.8. The box has a mass of 100 kg and the van is travelling at 20 m s–1.

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Remember that Newton’s second law in the form F = ma is valid only for an object of constant mass m. Then Δ(mv) = m Δv, so ∆(mv ) = ∆t ∆v = ma. m ∆t

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Force in N = rate of change of momentum in kg m s–2

20 m s–1

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Figure 3.7

(a) Calculate the momentum of the box. (b) The driver applies the brakes and the van takes 5 s to stop. Calculate the frictional force required to prevent the box from sliding forward.

The general form of Newton’s second law can be used in problems where the mass is not constant, in particular when there is a steady flow of mass.

Worked example A small jet engine releases 5.0 kg of exhaust per second. The exhaust gas comes out at 65 m s–1. Calculate the force with which the exhaust gases are emitted.

Consider the mass of gas that leaves the engine in 1 second.

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3.2 Newton’s first and second laws of motion; momentum The momentum given to this gas is p = mv = 5.0 kg × 65 m s–1 = 325 kg m s–1 The force on the gases to give this momentum change is 325 kg m s −1 ∆p = = 325 N F= 1s ∆t 8. A hosepipe releases water at a rate of 3.0 kg s–1. The water jet comes out of the hose at 4.7 m s–1. Calculate the force with which the water is expelled from the end of the hose.

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Many injuries in accidents are caused by the rapid change in momentum that happens in a collision. For example, in some road accidents, vehicles can be brought to rest from speeds of around 55 km h–1 in around 0.1 s.

Devices such as airbags can increase the time taken for a person in a vehicle to come to a stop. Increasing the time the person takes to stop will decrease the rate at which their momentum changes, and therefore reduce the force on the person.

Models of the human body, called crash test dummies, like the one in Figure 3.8, are used to investigate what happens in a collision. The dummies are made to be as similar to a human body as possible. They are used to test the effect an impact has on the body without harming a real person and allow repeated testing to investigate the effect of changing a design variable.

They are fitted with sensors that record the changes in force, acceleration and displacement with time.

Figure 3.9 is a graph of the acceleration of a crash test dummy’s head during a collision.

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Rapid changes in momentum

Figure 3.8 Crash test dummies have a similar mass to a person and are jointed to move like a human body would during a crash. Here the airbag increases the time taken for the dummy to come to a stop during the collision.

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25 50 75 100 125 150 175 200 225 250 275 300 Time / ms

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Figure 3.9 Output from an acceleration sensor placed in the head of a crash test dummy. The car hit a concrete block at 56 km h–1.

Worked example

Use the information in Figure 3.9 to answer these questions. (a) The car makes contact with the concrete block at time = 0 on the graph. Explain why the head of the dummy does not begin to decelerate until about 12 ms after this time. (b) State the maximum acceleration of the head of the dummy in m s–1.

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3 Dynamics (c) The head of the dummy has a mass of 5 kg. Calculate the maximum magnitude of the force acting on the head of the dummy during the collision. (d) The head of the dummy has a velocity of zero at 180 ms. (i) Calculate the maximum change in momentum of the head of the dummy during the first 180 ms of the collision. (ii) Calculate the average force acting on the head of the dummy to bring it to a stop.

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Answer (a) The head of the dummy has inertia. It will continue with uniform velocity until acted upon by a force. This will be the restraining force from the airbag. (b) From the graph, the maximum deceleration is about −36 g. 1 g is 9.81 m s–2. −36 × 9.81 m s–2 = −350 m s–2 (c) magnitude of maximum acceleration = 350 m s–2 F = ma = 5 kg × 350 m s–2 = 1800 N or 1.8 kN (d) (i) initial velocity = 56 km h–1 56 km h–1 = 56 000 m h–1 56 000 m h–1 = 56 000 m s–1 = 16 m s–1 3600 initial momentum = mass × velocity = 5 kg × 16 m s–1 = 78 kg m s–1 As final velocity = 0, then 78 kg m s–1 is the maximum change in momentum (ii) force = change in momentum ÷ time taken time taken = 180 ms − 12 ms = 170 ms F = 78 kg m s–1 ÷ 0.17 s = 460 N

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9. A car of mass 1350 kg is travelling at 15.4 m s–1 when it collides with a wall.

The car comes to a stop in 115 ms. (a) Calculate the change in momentum of the car. (b) Calculate the average force on the car. (c) Cars are designed to have front and rear sections that crumple quite easily on impact. Explain how this can help reduce forces on people in the car during a collision.

10. Fragile items like laptops are often delivered in boxes that contain polystyrene packing. Polystyrene is lightweight and easily deformed.

Explain, using the terms momentum and force, how polystyrene packing protects fragile items during delivery.

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Key ideas

➜➜Newton’s first law: an object remains at rest or continues to move with uniform velocity unless acted upon by a resultant force. This means that an object with zero or constant velocity either has no forces acting on it, or all of the forces acting on it are balanced; this is called translational equilibrium. ➜➜The product of an object’s mass and its velocity is called its momentum. ➜➜Newton’s second law: The rate of change of momentum of an object is proportional to the resultant force acting on it. The change in momentum occurs in the direction of the resultant force. ➜➜The SI unit of force the newton is defined so that the resultant force on an object is equal to its rate of change of momentum. ∆p ∆(mv ) which for constant m becomes F = ma. ➜➜F = = ∆t ∆t

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3.3 Forces in interactions: Newton’s third law

3.3 Forces in interactions: Newton’s third law

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When an object exerts a force on a second object, the second object simultaneously exerts a force of equal magnitude and opposite direction on the first object.

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How does a jet engine work?

A jet engine takes air in at the front, burns fuel in this air, and then forces the exhaust gas out at the back at high speed. When the engine forces the exhaust gases out at the back, there is an equal and opposite force of the gases on the engine, pushing it forward.

This is an example of Newton’s third law, which states:

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Imagine sitting on a chair at a very heavy desk. Both the desk and the chair are fitted with small wheels. You try to Figure 3.10 Jet engines like this can propel aeroplanes of 400 000 kg to push the heavy desk away from you. What happens? altitudes of 12 000 m and velocities of 920 km h–1.

You will move backwards. The desk will only move a little. Although you are exerting a force on the desk, Newton’s third law states that the Link desk will also be pushing back on you with an equal and opposite force. Your inertia is Refer back to much less than that of the massive desk, so your acceleration is much greater.

Newton’s second law Newton’s third law holds for all types of forces: not only for contact forces such as in Topic 3.2. pushes and pulls, but also for non-contact forces such as gravitational, magnetic and electrostatic forces.

All forces are the result of interactions between objects. Any interaction involves a Tip pair of forces; each object experiences a force of the same magnitude but in opposite Remember that one directions. The opposing forcesº of the pair are always the same type, for example force of the pair acts gravitational. on one object and B the other, opposite A force acts on the other free body diagram A B object. force from B on A

force from A on B

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Figure 3.11 Equal and opposite forces in a simple interaction

11. Use Newton’s third law to describe the force which is equal in magnitude and opposite in direction to each of these. (a) A heavy book exerting a contact force, equal to its weight, on a table. (b) A sprinter at the start of a race pushing back with a contact force on a starting block. (c) A horse pulling forwards on a heavy cart. (d) A bird pushing air downward with its wings when flying. (e) A ball being pulled down to the Earth’s surface by gravitational force. (f) A bar magnet being used to repel an identical bar magnet.

12. Two ice skaters, A and B, of equal mass are standing opposite each other.

Skater A pushes on skater B. Skater B does not push. Neither skater loses their balance. (a) Explain what happens.

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3 Dynamics

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13. A person is about to step off a small boat and onto a river bank, as shown in Figure 3.12.

Figure 3.12

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(a) Draw a free body diagram to show the forces on the person and on the boat, due to the person stepping off the boat. (b) Explain, in terms of forces, why it would be safer for the person if the boat was tied to the river bank.

14. A jet engine releases 17 kg of exhaust gas every second. The exhaust gas comes out at 85 m s–1. Calculate the force that propels the engine.

Key ideas

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➜➜Newton’s third law: When an object exerts a force on a second object, the second object simultaneously exerts a force of equal magnitude and opposite direction on the first object. ➜➜This holds for all types of force including non-contact forces such as gravitational, magnetic and electrostatic forces.

3.4 Motion with resistive forces Friction Friction is a resistive force that acts where surfaces contact each other. When one surface moves over another, friction acts in the opposite direction to the motion. It resists the motion. The magnitude of the friction force depends on the materials that are in contact and on the magnitude of the contact force.

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3.4 Motion with resistive forces According to Newton’s first law of motion, if an object is moving and in translational equilibrium, it will continue to move until some non-zero resultant force acts on it. In practice, for an object moving on a solid surface, this force will usually be friction.

Figure 3.13a shows the force of friction on a moving object where friction is the only force acting. The object will decelerate because the friction force is in the opposite direction to the motion. direction of motion

friction

(b)

See Topic 3.2 for Newton’s first law of motion.

direction of motion driving force

friction

Figure 3.13 (a) Friction opposes motion (b) The friction force reduces the resultant forward force

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Now, consider the same object with a driving force and a force of friction acting, as shown in Figure 3.13b. The driving force on the object is greater than the friction force so the resultant force forwards (to the right) causes the object to accelerate. The friction reduces the acceleration that would be produced by the driving force alone.

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(a)

Link

Drag

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Imagine standing in a swimming pool where more than half of your body is beneath the surface of the water. It is difficult to walk as fast as when you are on land. This is due to the resistive force of the water being greater than that of the air.

Gases and liquids are types of fluid. All fluids resist the motion of an object that is moving through the fluid. The resistive force is called drag.

Drag forces arise partly because of the density of the fluid − movement of an object through the fluid requires the fluid to be pushed out of the way − and partly due to the viscosity of the fluid. Viscosity is the resistance of a fluid to flow. Sometimes drag is referred to as ‘viscous drag’.

The size of the drag force in a particular fluid depends on the speed of the object through the fluid, and on its size and shape:

direction of motion

(b)

direction of motion

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(a)

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• The greater the speed of the object, the greater the drag force.

• The greater the area of the object’s surface that is presented to the fluid as it moves through the fluid, the greater the drag force. This is because more fluid needs to be pushed out of the way.

• The less streamlined an object’s shape, the greater the drag force. A streamlined shape is one that allows steady, non-turbulent flow of the fluid past it.

Figure 3.14 (a) A streamlined object with a smaller cross-sectional area experiences less drag when moving through a fluid than (b) a larger object that is not streamlined.

Tip Two objects of the same size and shape, moving at the same speed through the same fluid, experience the same drag force.

The drag force does not depend on the mass of the object.

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3 Dynamics

Motion under gravity with air resistance

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Air resistance is the drag force that acts on all objects moving through air.

In Topic 2.3 we looked at free fall in the absence of air resistance. In practice, air resistance is often not negligible and will significantly affect the motion of a falling object. Consider an object that has just been dropped. At the instant it is released, its velocity will be zero, so there will be no air resistance.

The object’s downward velocity increases due to the gravitational force which is equal to mg. This gravitational force is constant provided the gravitational field strength g is constant.

As the velocity increases, air resistance will begin to increase.

The increasing air resistance means that the resultant downward force, and hence the acceleration, decrease. These eventually become zero. Then the air resistance becomes constant, equal to the gravitational force (Figure 3.15).

From this time onward, the forces on the object remain balanced and, according to Newton’s first law, there will be no further acceleration. We call this velocity terminal velocity. velocity = 0 air resistance = 0

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velocity increasing air resistance increasing

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The term free fall is only used for falling objects when air resistance is not present or is ignored, that is when the only force acting on an object is its weight.

velocity increasing air resistance increasing

mg

velocity constant air resistance = mg mg

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Figure 3.15 As the velocity of a falling object increases, air resistance also increases up to a maximum equal to the gravitational force mg at which point the object has reached terminal velocity.

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terminal velocity

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Figure 3.16 shows a velocity-time graph for a falling object in the presence of air resistance.

t Figure 3.16 Velocity-time graph for an object falling in air.

15. A table tennis ball of mass 0.027 kg is dropped vertically off a high bridge. (a) Calculate the air resistance when the table tennis ball reaches terminal velocity. (b) Draw a free body diagram of the table tennis ball at terminal velocity.

16. A brick of mass 3.5 kg is dropped off a cliff.

At time t after being dropped, the air resistance on the brick is 0.2 N. (a) Draw a vector diagram to show the magnitude and direction of the forces acting on the brick at time t. (b) Calculate the acceleration of the brick at time t.

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What affects the shape of the velocity-time graph?

Consider two balls of equal diameter but with very different masses, like a table tennis ball and a golf ball. The mass of the golf ball is approximately 20 times greater than that of the table tennis ball.

If both balls are dropped from a high position, the table tennis ball will reach terminal velocity first. The golf ball will continue accelerating for longer, so reach a larger terminal velocity. Therefore, the golf ball will hit the ground first. It may seem from this that air resistance is greater for a smaller mass, but this is not so.

The force of air resistance depends on the area and shape of the moving object, and on the velocity of the object, but not on the mass.

When both balls are falling at the same velocity, the forces of air resistance on them are equal. The slower drop of the table tennis ball can be explained using the relationship between force, mass and acceleration, F = ma. As the table tennis ball has a much smaller mass than the golf ball, the same resistive force will have a greater effect on its motion, causing a larger deceleration and so reducing its velocity first.

If these two balls were dropped in a vacuum, they would both hit the ground at exactly the same time.

An object falling through a more viscous fluid would reach terminal velocity in a shorter time because the resistive force is greater and becomes equal to the gravitational force sooner.

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3.4 Motion with resistive forces

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Worked example

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Three identical steel balls, A, B and C are dropped from a height of 1.0 m.

Ball A is dropped in a vacuum; ball B is dropped in air; ball C is dropped in water.

Sketch graphs for the motion of A, B and C on the same axes to show the variation of: (a) velocity with time for each ball (b) acceleration with time for each ball.

(a)

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Answer (a) No resistive force acts on ball A, and so its velocity will increase uniformly with time.

Ball B is subject to air resistance, which will cause the velocity to increase less rapidly with time. It will not reach terminal velocity from a height of 1.0 m.

Ball C has a greater resistive force on it than B as the drag force from water is greater than that of air, so its velocity will increase even less rapidly. It will probably not reach terminal velocity from a height of 1.0 m. (b)

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a

A

B

B

C

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Figure 3.17 (a) Sketch graph of velocity with time, (b) sketch graph of acceleration with time

(b) The acceleration of ball A will not change with time.

The acceleration of ball B will decrease slightly with time.

The acceleration of ball C will decrease significantly with time.

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3 Dynamics 17. Sketch a velocity-time graph for a small object falling through two different viscous fluids. Use a solid line for the more viscous fluid and a dashed line for the less viscous fluid.

18. A crate of relief supplies is attached to a parachute and dropped from a plane. Sketch an acceleration-time graph for the crate falling to the ground.

Experimental skills: Investigating air resistance

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Apparatus You will need • to make a scale that is 2.00 m tall to be fixed to a wall. You can do this with a measuring tape or pieces of paper that are accurately marked. The intervals should be 0.10 m apart. The scale needs to be vertical in both planes (front to back and side to side) • a table tennis (ping pong) • a balance to measure the mass of the table tennis ball • a timer or stopwatch • a video camera that will allow frame-by-frame replay.

Questions P1. (a) Explain whether the displacements should be measured from the top, middle or bottom of the ball. (b) Explain how a suitable position for the camera should be decided. (c) Explain why repeats are carried out and an average calculated.

P2. The average results from a pair of students is shown in Table 3.1. (a) Use the students’ results to plot a displacement-time graph of the falling table tennis ball. (b) Draw a smooth best-fit curve through the points. (c) Describe the relationship shown on the graph. (d) Use your graph to estimate the terminal velocity of the table tennis ball.

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In this investigation you will investigate air resistance by attempting to estimate the terminal velocity of a table tennis ball falling vertically. A table tennis ball is has a small mass, so when the ball is dropped the increasing air resistance quickly becomes significant compared with the ball’s weight.

Techniques Place the scale vertically with 0 at the top, as the scale will be used to measure displacement of the ball as it falls and not it height.

Position the timer or stopwatch close to the scale so that it can be seen when a video is made of the drop.

Place the camera in a suitable position so that the entire 2.00 m drop can be recorded.

Start the timer, start the video camera, drop the table tennis ball from a height of 2.00 m in front of the scale. Do this three times so that an average displacement can be calculated for each time interval.

Replay the video and record the displacement of the table tennis ball at regular time intervals, such as 0.10 s.

average displacement / m

0.00

0.00

0.13

0.08

0.20

0.19

0.24

0.28

0.28

0.38

0.31

0.47

0.34

0.54

0.38

0.66

0.40

0.73

0.44

0.86

0.50

1.10

0.55

1.35

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0.60

1.62

You may recall the shapes of displacement-time graphs for objects that are accelerating and for objects that are at constant velocity from Chapter 2.

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2.00

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time / s

Analysis

Table 3.1

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3.4 Motion with resistive forces P3. Use the mass of the table tennis ball to calculate the force of air resistance when the ball is at terminal velocity.

P4. Sketch a graph to show how air resistance varies with time for a falling table tennis ball. Label any key values on your sketch graph.

P5. (a) State two main sources of uncertainty in this experiment. (b) Suggest two improvements to the procedure.

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Air resistance acts on all objects moving in air. It contributes to the resultant resistive force on a moving vehicle.

Air resistance increases with the speed of the vehicle relative to the surrounding air; it is proportional to the square of the relative speed. This is why it requires more driving force to ride a bicycle in a direction opposite to the wind than it does to ride in conditions with no wind.

Assuming the wind is constant, air resistance increases with the speed of the vehicle, so the resultant resistive force on the vehicle will also increase with speed.

For a given driving force, for example from an engine, the acceleration will become zero when the resistive forces increase to become equal to the driving force. The vehicle then has constant velocity. It is in translational equilibrium because the resultant force will be zero, as shown in Figure 3.3.

Any moving vehicle will therefore have a maximum theoretical speed that will be reached when the resistive forces become equal to the maximum driving force.

Air resistance is reduced by streamlining an object so that its area perpendicular to motion is minimised. The car shown in Figure 3.18 was designed to achieve the lowest possible resistive force and, at the same time, the highest possible driving force. This enabled it to become the fastest vehicle to travel on land at 1228 km h–1, which is faster than the speed of sound in air.

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Horizontal motion with air resistance

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You will learn about relative speed in Topic 3.6.

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19. A large cargo ship, initially at rest, produces a constant driving force.

The driving force causes the ship to accelerate. Explain why the ship reaches a constant velocity although the driving force remains the same.

20. The Thrust SSC vehicle shown in Figure 3.18 broke the world record for speed on land.

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The rules for the world record state that a vehicle must travel at the record-breaking speed for a distance of 1.6 km. This is called a pass.

The same vehicle must make another pass in the opposite direction within 1 hour of the first pass.

Both passes are on horizontal ground.

Explain why the vehicle must make two passes in opposite directions to qualify for a world speed record.

Figure 3.18 Thrust SSC became the fastest vehicle to travel on land in 1997 partly due to its design to reduce air resistance.

Projectile motion with air resistance A projectile is an object moving in air with no driving force. In Chapter 2, we stated that the horizontal component of the velocity of a projectile would remain constant, but this is only when air resistance is neglected.

A free body diagram of a projectile that is subject to air resistance will show only two forces: its weight that acts vertically downwards and the air resistance that acts in the opposite direction to the velocity of the projectile at that instant.

Link It may be helpful to look back at Chapter 2, Topic 2.4, for descriptions of projectile motion.

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3 Dynamics velocity of projectile at this instant

air resistance

weight, mg

Figure 3.19 In practice, a projectile is acted upon by its weight and by air resistance. Air resistance acts in a direction opposite to the velocity.

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Vertical height

No air resistance

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You saw in Topic 2.4 that the path taken by a projectile is a curve. This curve will be a parabola when there is no air resistance.

As air resistance always acts in a direction opposite to the velocity of the projectile, this force will reduce both the height and the range of the projectile.

The path of a projectile with and without air resistance is shown in Figure 3.20.

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Significant air resistance

Horizontal distance

Figure 3.20 The path of projectile motion with and without air resistance.

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In Figure 3.20, notice how the path of the projectile is symmetrical about the highest point without air resistance. This is because the only acceleration is due to gravity and is in the vertical direction. The horizontal component of the velocity remains constant.

In contrast, the path of the projectile with significant air resistance is not symmetrical.

This is because the horizontal component of the velocity is decreasing with time in the air.

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21. The diagram in Figure 3.21 shows how a projectile would move in the absence of air resistance.

Figure 3.21

Copy the diagram and sketch the path of the same projectile in the presence of air resistance.

22. A projectile is launched horizontally from the edge of a cliff.

Air resistance is not negligible.

When the projectile hits the ground it still has a horizontal component to its velocity.

Sketch a graph to show the variation of (a) the horizontal component of the velocity of the projectile with time (b) the vertical component of the velocity of the projectile with time.

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3.5 Conservation of momentum

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➜➜Solid objects moving on solid surfaces experience friction that opposes their motion, reducing their acceleration. ➜➜Solid objects moving through gases or liquids experience a resistive force called the drag force. ➜➜Drag force is due partly to the viscosity of the fluid. ➜➜Drag force increases with the speed of the object and the area of the object and also depends on the shape of the object. ➜➜Air resistance is the drag force for an object moving through air. ➜➜Air resistance reduces the resultant downward force on a falling object and hence its acceleration. ➜➜The air resistance increases as the falling object’s speed increases. When the air resistance becomes equal to the object’s weight there is no resultant force so the object falls at a constant speed called its terminal velocity. ➜➜Air resistance is one of the resistive forces on a vehicle. It acts against the driving force and increases with the vehicle’s speed. ➜➜If the total resistive forces on a vehicle become equal to the driving force, the velocity becomes constant. ➜➜For a projectile, air resistance causes the height and range to be reduced.

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Key ideas

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3.5 Conservation of momentum

Link Refer back to Topic 3.2 for how to calculate an object’s momentum. You may recall from your previous courses that momentum is always conserved in interactions such as collisions.

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In Topic 3.3 we looked at interactions between objects in terms of forces. When objects interact, we can also predict what will happen after the interaction using the concept of momentum.

In a game of snooker, a wooden cue is pushed towards a ball to strike it rapidly. This ball then rolls away from the cue and collides with another ball.

Figure 3.22 A cue is being used to strike the white ball. The white ball will roll away from the cue and collide with the red ball. What will happen next?

The white ball in Figure 3.22 has the same mass as the red ball. If the white ball hits the red ball directly in a straight line, then the white ball will stop and the red ball will start to move at the same speed that the white one was moving before the collision.

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3 Dynamics This demonstrates the conservation of momentum.

The principle of conservation of momentum states that, if no external force acts on a system of objects, then the momentum of the system will remain unchanged.

Link Refer back to Topic 3.2 to remind yourself of Newton’s second law.

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Refer back to Topic 3.3 to remind yourself of Newton’s third law.

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‘No external force acts’ means that there is no resultant force on the objects immediately before their interaction. This means that they are both moving in a straight line with uniform velocity immediately before the interaction.

Conservation of momentum follows directly from Newton’s third law. Imagine two balls A and B colliding head-on. From Newton’s third law, the force FA on A by B is equal and opposite to the force FB on B by A. FA = –FB Refer back to Topic 3.3 to remind yourself of Newton’s third law.

From Newton’s second law, the force on an object is equal to its rate of change of momentum. So ∆pA ∆p =− B ∆t ∆t The collision duration Δt is the same for both, so

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ΔpA = − ΔpB (pafter − pbefore)A = −(pafter − pbefore)B (pafter)A + (pafter)B = (pbefore)A + (pbefore)B In words,

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total momentum before collision = total momentum after collision

Conservation of momentum in one dimension

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Motion or interactions ‘in one dimension’ means that the objects move and interact only in one straight line.

Momentum is always conserved when objects collide or separate from each other, provided no force acts except the forces of these objects on one another.

In the example of snooker balls colliding, we can represent the interaction with the simple diagram shown in Figure 3.23.

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v = 0.5 m s–1

v=0

before

v=0

v = 0.5 m s–1

after

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Figure 3.23 Before and after a collision between two snooker balls

In Figure 3.234, the white ball initially has a momentum, p, of mv = 0.16 kg × 0.5 m s–1 = 0.08 kg m s–1.

As momentum is conserved, then the momentum before the collision will be equal to the momentum after the collision.

The momentum of the red ball after the collision is therefore also 0.08 kg m s–1, so its p = 0.08 = 0.5 m s–1. velocity, v, is m 0.16 Worked example A ball of mass 0.9 kg rolls with a speed of 3.5 m s–1 towards another ball of mass 2.2 kg which is initially at rest.

After they collide, the 2.2 kg ball rolls with a speed of 1.2 m s–1 in the same direction as the 0.9 kg ball was rolling.

Calculate the speed and direction of the 0.9 kg ball after they collide.

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3.5 Conservation of momentum Answer v1b = 3.5

m s–1

v2b = 0

v1a

1

2

1

0.9 kg

2.2 kg

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v2a

Drawing a simple diagram to show before and after the interaction will help you to structure your calculation and help to avoid confusion between positive and negative directions.

2

Figure 3.24

total momentum before collision = total momentum after collision We can call the balls 1 and 2, and use b for before and a for after.

0.51kg m s −1 = 0.57 m s–1 0.9 kg

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v=

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m1v1b + m2v2b = m1v1a + m2v2a (0.9 kg × 3.5 m s–1) + (2.2 kg × 0) = (0.9 kg × v) + (2.2 kg × 1.2 m s–1) 3.15 kg m s–1 = 0.9v + 2.64 kg m s–1

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As this value is positive, we know the 0.9 kg ball moves in the same direction as the 2.2 kg ball.

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23. A trolley, P, of mass 0.87 kg travelling at 0.56 m s–1 collides with a stationary trolley, R, of mass 0.78 kg. After the collision, both trolleys move in the same straight line.

Trolley P travels at 0.22 m s–1 after the collision.

Calculate the speed of trolley R after the collision.

Worked example

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A railway wagon of mass 8.1 × 104 kg travelling at a speed of 2.4 m s–1 collides with a railway wagon of mass 4.5 × 104 kg travelling at 0.75 m s–1 in the opposite direction. They are on a level track. The two wagons join together.

Calculate the speed of the two wagons after the collision, and state their direction.

Take care with signs. Remember that velocity and momentum are vector quantities, so opposite signs mean opposite directions in one-dimensional problems.

0.75 m s–1

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Answer Draw a diagram like the one in Figure 3.25.

before

Figure 3.25

v

after

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total momentum before collision = total momentum after collision We can call the wagons 1 and 2, and use b and a for before and after. m1v1b + m2v2b = m1v1a + m2v2a Since the two wagons join together in the collision, v1a = v2a. We will call this v.

(8.1 × 104 kg × 2.4 m s–1) + (4.5 × 104 kg × (−0.75 m s–1) ) = (8.1 × 104 kg + 4.5 × 104 kg) × v 1.61 × 105 kg m s–1 = 1.26 × 105 kg × v

v=

1.61 × 105 kg m s −1 = 1.28 m s–1 1.26 × 105 kg

As this value is positive, we can say that the combined wagons move in the direction that the 8.1 × 104 kg wagon was initially moving.

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3 Dynamics 24. Two cars collide as shown in Figure 3.26.

before

A

mass, 1100 kg velocity, 6.4 ms–1

A

mass, 1200 kg velocity 0 B

v

v

B

after 1100 kg

1200 kg

Figure 3.26

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Car A of mass 1100 kg is travelling at 6.4 m s–1 when it collides with car B. Car B of mass 1200 kg is initially at rest. The two cars become locked together as a result of the collision and move together.

Calculate the speed of the two cars together immediately after the collision.

25. A bullet of mass 5 g is fired into a bag of sand. The bag of sand has a mass of 10 kg and is hanging by a rope, free to swing. The bullet enters the bag at a velocity of 300 m s–1 and stops in the sand.

Calculate the initial velocity with which the bag is made to move by the bullet.

26. A car of mass 890 kg travelling at 18 m s–1 collides with a stationary car. The two cars become locked together during the impact and move forward with a velocity of 9.7 m s–1.

Calculate the mass of the stationary car.

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When objects with a large difference in mass collide with each other, the effect of the conservation of momentum can be difficult to detect.

Consider a common event. A car of mass 1100 kg is travelling at 35 m s–1 when it collides with a fly of mass 15 mg travelling in the opposite direction at 1.0 m s–1. As a result of the collision, the fly sticks to the front of the car.

The momentum of the car before the collision is 1100 kg × 35 m s–1 = 3.85 × 104 kg m s–1 The momentum of the fly before the collision is 1.5 × 10 –5 kg × −1.0 m s–1 = −1.5 × –5 10 kg m s–1 So the total momentum after the collision is 3.85 × 104 kg m s–1 + (−1.5 × 10 –5 kg m s–1) The change in momentum of the car is not detectable working at 3 or even 4 significant figures. Effectively the car continues with the same speed.

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27. A concrete block falls from a height of several metres onto soft soil. The block becomes stuck in the soil and does not bounce or break.

Explain, with reference to Newton’s laws, whether momentum is conserved in the collision between the block and the Earth.

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The word explosion is sometimes used in dynamics for objects that are separating. It does not necessarily mean a chemical or nuclear explosion.

Explosions Momentum is conserved not only when objects collide, but also when a single or composite object separates into parts that move away from each other.

This is why a gun recoils (moves backward) when a bullet is fired and why a hosepipe moves backward when water comes out through a nozzle.

Worked example A cannon of mass 100 kg is used to fire a cannonball of mass 5 kg. The cannon is at rest and the cannon ball is fired at a velocity of 80 m s–1.

Calculate the recoil velocity of the cannon.

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3.5 Conservation of momentum

400 kg m s −1 so v = −4 m s–1 100 kg

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−v m s–1 =

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Answer total momentum before explosion = total momentum after explosion Before the event, the total momentum of the system (cannon plus cannonball) is zero.

Remember to add the masses in a combined system like this.

We can call the cannon 1 and the cannonball 2, and use b and a for before and after. m1v1b + m2v2b = m1v1a + m2v2a Here v1b = v2b = 0. total momentum before explosion = (100 kg + 5 kg) × 0 m s–1 = 0 kg m s–1 This means that after the event the total momentum must also be zero. Remember that momentum and velocity are vector quantities. This means the forward momentum of the cannonball will be equal to the backward momentum of the cannon. Let the velocity of the cannon after the explosion be −v. The negative velocity indicates an opposite direction to the positive value (the cannonball).

0 = (5 kg × 80 m s–1) + (100 kg × −v m s–1) 0 = 400 kg m s–1 + (100 kg × −v m s–1) 400 kg m s–1 = 100 kg × –v m s–1

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28. A spacecraft is made of two parts, A and B. A has mass 800 kg and B has mass 150 kg. Part B is released with a velocity of 10 m s–1 relative to the original spacecraft.

Calculate the velocity of part A after the release.

29. A large gun fires an artillery shell of mass 15 kg at a velocity of 680 m s–1. The initial recoil of the gun is 23 m s1.

Calculate the mass of the gun.

30. A large fire hose releases water at a rate of 30 kg s–1 and with a speed of 15 m s–1.

Calculate the force needed to keep the end of the hose from moving backwards.

Conservation of momentum in two dimensions

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Momentum is a vector quantity, so for situations involving more than one dimension, momentum must be conserved in any direction (provided there is no external force in that direction).

It is easiest to consider the momentum in the initial direction of motion of one of the objects, and the momentum in a direction perpendicular to that.

For example, think of a moving snooker ball colliding with a stationary one in a line that does not pass through the centre of both balls as shown in Figure 3.27.

Motion or interactions in two dimensions means that the objects may move and interact in different directions not in the same straight line.

Link

p2 p1

p3 before

Tip

after

Figure 3.27 A two dimensional collision between snooker balls viewed from above. The vectors represent the momentum of each ball but are not to scale.

You may want to look back at resolving vectors in Chapter 1.

In this section, you will be working out the components of momentum in two perpendicular directions.

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3 Dynamics The momentum of the white ball before the collision only has a forward component.

There is no component in the direction perpendicular to its initial motion. (Remember, Figure 3.27 is viewed from above.) This means that the perpendicular momentum before the collision is zero.

After the collision, the white ball and the red ball each have forward and perpendicular components to their momentum, as shown in Figure 3.28.

The vector sum of the forward components must equal the initial forward momentum of the white ball. The vector sum of the perpendicular components must be zero.

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p2

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p3

Figure 3.28 Resolving the momentum of each ball after the collision into components.

Worked example

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In a game of bowls, one player rolls a ball towards their opponent’s ball. The aim is for the balls to collide and so move the opponent’s ball further away from a target.

One player rolls a ball, A, at a velocity of 3.2 m s–1 to collide with an opponent’s ball, B, which is initially at rest. Both balls have equal mass of 2.0 kg each.

After the collision, ball A travels at an angle of 25o to the left of its initial direction. Its speed is 1.7 m s–1. Immediately after the collision, ball B moves at 40° from the direction of A’s initial motion (Figure 3.29).

1.7 m s–1

3.2 m s–1

A

B before

A 25° 40° B after

Figure 3.29

Calculate the speed of ball B immediately after the collision.

Answer Total momentum before collision parallel to initial motion = total momentum after collision parallel to initial motion Let the direction of the initial motion of A be along the x-axis. mAvAx + 0 = mAv’Ax + mBv Bx where v’Ax denotes the speed of A in the x-direction after the collision.

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3.5 Conservation of momentum 2.0 kg × 3.2 m s–1 = (2.0 kg × (1.7 cos 25°) m s–1) + (2.0 kg × (v Bx cos 40°) m s–1) 6.4 kg m s–1 = 3.08 kg m s–1 + (2.0 kg × (v Bx cos 40°) m s–1) 6.4 kg m s–1 − 3.08 kg m s–1 = 2.0 kg × (v Bx cos 40°) m s–1 3.32 kg m s–1 = 2.0 kg × (v Bx cos 40°) m s–1 3.32 kg m s −1 = 2.16 m s–1 2.0 kg × cos40

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In the y-direction: 0 = mAv’Ay + mBv By where v’Ay denotes the speed of A in the y-direction after the collision.

0 = (2.0 kg × (1.7 sin 25°) m s–1) + (2.0 kg × (−v By sin 40°) m s–1) Notice that v is negative as the y-component of its velocity is opposite to that of ball A. 0 = 1.44 kg m s–1 + (2.0 kg × (−v By sin 40°) m s–1) −1.44 kg m s–1 = 2.0 kg × (−v By sin 40°) m s–1

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v Bx =

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−1.44 kg m s −1 = −v By, so v By = 1.12 m s–1 2.0 kg × sin 40

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The speed of the ball, v B , can be found using Pythagoras’ theorem: v B2 = (v Bx2 + v By2) = (2.16 m s–1)2 + (1.12 m s–1)2 = 5.92 m2 s–2 v B = 2.4 m s–1

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31. A red ball of mass 0.5 kg travelling at 1.0 m s–1 collides with a stationary green ball of mass 0.5 kg.

After the collision, the red ball travels at 0.27 m s–1 in a direction 30° left of its original direction. The green ball travels in a direction 35° to the right of where it was hit.

Calculate the speed of the green ball.

Worked example

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Two ice hockey pucks, A and B, are sliding on ice. Puck A, of mass 2.5 kg is travelling at 1.3 m s–1 then it collides with puck B, of mass 2.0 kg travelling at 1.1 m s–1. A and B are travelling in perpendicular directions when they collide and they become locked together as a result of the collision.

Calculate the speed and direction of the pucks after collision.

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Answer Draw a diagram.

vA = 1.3

m s–1

mA = 2.5 kg

B

va = 1.1 m s–1 ma = 2.0 kg

v

q

B

A

A

before

after

m = 4.5 kg

Figure 3.30

total momentum before collision = total momentum after collision

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3 Dynamics pa = 2.0 kg × 1.1 m s–1 = 2.2 kg m s–1

pa = 2.5 kg × 1.3 m s–1 = 3.25 kg m s–1 q Figure 3.31 the momentum vectors before collision are at right angles

3.92 kg m s −1 p = = 0.87 m s–1 m 4.5 kg 2.2 kg m s −1 , so θ = tan–1 3.25 kg m s −1

 2.2 kgms−1     3.25 kgms−1  = 34°

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tan θ =

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v=

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(total momentum after collision)2 = (2.2 kg m s–1)2 + (3.25 kg m s–1)2 total momentum after collision = 3.92 kg m s–1

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Worked example

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32. A car of mass 950 kg is travelling on a straight road at a uniform speed of 15 m s–1.

The car is approaching a road junction.

A truck of mass 2.4 × 103 kg travelling at 12 m s–1 has a direction perpendicular to that of the car when the two vehicles collide.

The car and the truck become locked together as a result of the collision.

Calculate the speed and direction of the two vehicles immediately after the collision.

Give the direction relative to the direction of the car before collision.

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Two tennis balls have equal mass. Ball A, travelling at 2.4 m s–1 collides with ball B, initially at rest. After the collision, ball A moves in a direction 30° away from its original direction and ball B moves in a direction 40° from the initial direction of ball A as shown in Figure 3.32.

A

35° 40°

original direction of ball A

B Figure 3.33 Motion of balls A and B after the collision

Calculate the speeds of ball A and ball B after the collision.

Answer total momentum before collision in original direction of ball A = total momentum after collision in original direction of ball A m × 2.4 m s–1 = mvA cos 30° + mv B cos 40° where vA and v B denote the speeds of A and B after the collision.

2.4 m s–1 = vA cos 30° + v B cos 40° 2.4 m s–1 = 0.866vA + 0.766v B [1]

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3.6 Momentum and kinetic energy in interactions

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total momentum before collision in direction perpendicular to that of ball A = total momentum after collision in direction perpendicular to that of ball A 0 = mvA sin 30° − mv B sin 40° vA sin 30° = v B sin 40° 0.5vA = 0.643v B [2] We have two variables, vA and v B and two equations, so we can solve them using simultaneous equations. solving for vA we need to eliminate v B from equation 2: v B = 0.5v A 0.643 substituting this for v B in equation 1: 2.4 m s–1 = 0.866vA + 0.766 × 0.5v A 0.643 2.4 m s–1 = 0.866vA + 0.596vA 2.4 m s–1 = 1.462vA vA = 1.64 m s–1 This value of vA can now be substituted into either equation 1 or equation 2 to find v B. substituting in equation 2: 0.5 × 1.64 m s–1 = 0.643v B v B = 1.28 m s–1

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33. Two balls, P and R, of equal mass collide. Ball P is initially travelling at 3.9 m s–1 and ball R is initially stationary. After the collision, ball P travels at an angle of 25° to its initial direction, and ball R travels at an angle of 30° to the initial direction of P.

Calculate the speeds of both balls P and R after the collision.

Key ideas

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➜➜The principle of conservation of momentum: if no external force acts on a system of interacting objects, then the momentum of the system will remain unchanged. ➜➜Momentum is always conserved when objects collide or separate from each other, provided no external force acts. ➜➜In two dimensions, momentum is conserved in any one direction. ➜➜Momentum problems in two directions can be solved by resolving known values of momentum into two perpendicular components. Momentum in each of these directions must be conserved.

3.6 Momentum and kinetic energy in interactions In Topic 3.5 you saw that momentum is always conserved when objects collide or separate. Will kinetic energy always be conserved in such interactions?

Kinetic energy, Ek is energy of movement and is calculated using Ek = kinetic energy (J) =

1 2

1 2

mv2

× mass (kg) × speed2 (m s–1)2 81

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3 Dynamics

Elastic and inelastic interactions Consider the two balls in the game of snooker from Topic 3.5.

Snooker balls have a mass of 0.16 kg. The rolling white ball has a velocity of 1.0 m s–1 and collides in a straight line with a stationary red ball of the same mass.

We can use the subscript letters w and r for the white and red ball, and b and a for before and after.

Total momentum before the collision = mwvwb + mrvrb Total momentum before the collision = (0.16 kg × 1.0 m s–1) + (0.16 kg × 0) = 0.16 kg m s–1.

After the collision, the first rolling ball comes to rest and the other ball rolls away in a straight line. Momentum is always conserved, so the ball must roll away with a velocity of 1.0 m s–1 to give a total momentum after the collision of 0.16 kg m s–1.

The total kinetic energy before the collision = ( 12 mv2)wb + ( 12 mv2)rb = ( 12 × 0.16 kg × (1.0 m s–1)2) + ( 12 × 0.16 kg × 02) = 0.08 J. The total kinetic energy after the collision = ( 12 mv2)wa + ( 12 mv2)ra = ( 12 × 0.16 kg × 02) + ( 12 × 0.16 kg × (1.0 m s–1)2) = 0.08 J. Here, kinetic energy has been conserved. When kinetic energy is conserved in an interaction, the event is described as an elastic interaction.

In practice, interactions are rarely perfectly elastic, as some of the initial kinetic energy will be transferred to another form.

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In Chapter 5 you will learn how to derive the equation that relates kinetic energy to mass and velocity. You will also learn more about the principle of conservation of energy.

You may also recall from previous courses that the total quantity of energy in a system is always conserved in any event (the principle of conservation of energy).

The total kinetic energy of objects that interact may or may not be conserved. If kinetic energy is not conserved in an interaction, then some of this energy must be transferred to other forms such as thermal or elastic potential.

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Consider two railway wagons. One wagon of mass 1900 kg is travelling at 3 m s–1 on a level track and collides with another stationary wagon of mass 2200 kg. They join and continue to move together. Is this collision elastic?

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Answer First, use the principle of conservation of momentum to calculate the velocity with which the conjoined wagons move after the collision. total momentum before collision = mv = (1900 kg × 3 m s–1) + (2200 kg × 0) = 5700 kg m s–1 total momentum after collision = 5700 kg m s–1

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velocity after collision, v =

Tip The momentum will always be conserved in any interaction, even when kinetic energy is not conserved.

5700 kg ms −1 p = = 1.4 m s–1 m (1900 + 2200 ) kg

Next, compare the total kinetic energy of the system before and after the collision. total kinetic energy before collision = 12 mv2 = ( 12 × 1900 kg × (3 m s–1)2) + ( 12 × 2200 kg × 02) = 8550 J. total kinetic energy after collision = 12 mv2 = 12 × (1900 + 2200) kg × (1.4 m s–1)2) = 4020 J. Kinetic energy has not been conserved, so the collision is not elastic.

When kinetic energy is not conserved in an interaction, the event is described as an inelastic interaction.

34. A trolley of mass 0.95 kg travelling at 1.4 m s–1 collides with an identical trolley which is initially at rest. The two trolleys become locked together as a result of the collision.

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3.6 Momentum and kinetic energy in interactions (a) Calculate the speed of the trolleys after collision. (b) Show by calculation whether the collision is elastic or inelastic.

35. An air track is a piece of equipment used to produce an almost frictionless surface.

Vehicles called gliders slide over the surface of the air track and can be used to study collisions.

Two gliders are placed on an air track. Each has a bar magnet attached. The north poles of the magnets are facing each other as shown in Figure 3.34. light gates magnet glider A

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glider B

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air track

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Figure 3.33

path of alpha particle

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Glider A has a total mass of 200 g including the magnet.

Glider B has a total mass of 100 g including the magnet.

Glider A is gently pushed towards glider B. The velocity of glider A is 5 cm s–1 immediately before reaching glider B, which is initially at rest. The two gliders do not make contact, as the magnets repel each other.

After the interaction, glider A has a velocity of 1.7 cm s–1. (a) Calculate the velocity of glider B after the interaction. (b) Show by calculation whether this interaction is elastic or inelastic.

36. An alpha particle of mass 6.68 × 10 –27 kg has a velocity of 1.42 × 107 m s–1. The alpha particle approaches a stationary helium atom. The alpha particle and the nucleus of the helium atom repel each other with electrostatic forces, because they are both positively charged, but they do not make contact. The mass of the helium atom is equal to that of the alpha particle.

The alpha particle and the helium atom move in different directions after the interaction as shown in Figure 3.34.

path of helium atom

Figure 3.34

The speed of the alpha particle after the interaction is 1.01 × 107 m s–1. The speed of the helium atom after the interaction is 9.98 × 106 m s–1.

Show that this interaction is elastic.

37. A plastic disc, A, of mass 165 g slides across a frictionless surface with a speed of 1.30 m s–1.

The disc collides elastically with another disc, B, of equal mass which is initially at rest.

The two discs then move in different directions as shown in Figure 3.35.

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3 Dynamics 0.760 m s–1 A 1.30 m s–1 A

B B

before collision

VB

after collision

Figure 3.35

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Calculate the speed of disc B after the collision.

38. A proton of mass 1.67 × 10 –27 kg with a uniform velocity of 2.0 × 104 m s–1 approaches a massive positively charged stationary object, O. The proton is repelled in the opposite direction from P without contacting O. The interaction is elastic. (a) Show that the speed of O after the interaction is negligible. (b) Calculate the change in momentum of the proton.

39. Particles in a gas collide with each other and with the walls of the container. These collisions are almost perfectly elastic.

Explain what would happen to a gas in a sealed container if these collisions were inelastic.

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Relative speed of approach and separation

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Relative speed means the speed of one object measured in comparison to another.

Imagine you are in a vehicle like a car or train. You are travelling in a straight line and another vehicle passes you in the opposite direction. It appears to be going much faster than your vehicle. This is because of relative speed.

For two objects travelling in opposite directions, their relative speed is the sum of their speeds.

Now imagine your vehicle passes another vehicle travelling in the same direction. You appear to pass it quite slowly. This is again due to relative speed.

For two objects travelling in the same direction, their relative speed is the difference between their speeds. The numerical value of relative speed is always positive, so when finding the difference, take the smaller speed away from the larger one.

For two objects, one of which is not moving and the other is moving straight towards it, their relative speed is equal to the speed of the moving one.

These situations are shown by the diagrams in Figure 3.36.

VA A

VB

VB

VA A

VB = 0 relative speed = VA

B

VB

VA = 0 A

relative speed = VA – VB

B

VA A

relative speed = VA + VB

B

B

relative speed = VB

Figure 3.36 Determining relative speed of two objects

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3.6 Momentum and kinetic energy in interactions 40. A train is travelling at a constant velocity of 52 m s–1 going north. It passes another train travelling at a constant velocity of 35 m s–1 going south.

What is the relative speed of the two trains?

B 35 m s–1 C 52 m s–1 D 87 m s–1 A 17 m s–1 41. A car is travelling at a constant speed of 76 km h–1 when it overtakes a truck travelling at 51 km h–1 in the same direction.

What is the relative speed of the two vehicles?

B 51 km h–1 C 76 km h–1 D 127 km h–1 A 25 km h–1

A

3.0 m s–1

3.0 m s–1

B

before collision

3.0 m s–1

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Consider two balls of equal mass rolling towards one another, each with a speed of 3.0 m s–1 (Figure 3.37). Before collision the balls have a relative speed of 3.0 m s–1 + 3.0 m s–1 = 6.0 m s–1. We call this the relative speed of approach.

Before collision the total momentum is zero. For zero momentum afterwards, the velocity of A and B must be equal and opposite − they bounce apart. If the collision is elastic, the total kinetic energy afterwards is equal to that before, so the balls must both have speed 3.0 m s–1.

After collision their speed relative to each other is also 3.0 m s–1 + 3.0 m s–1 = 6.0 m s–1.

We call this the relative speed of separation.

A

B

after collision

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Figure 3.37

3.0 m s–1

In general, for masses of any size, in a perfectly elastic collision:

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relative speed of approach = relative speed of separation

42. Two bowling balls, P and Q, of equal mass are rolling in the same direction in the same straight line. P is behind Q and going 1.2 m s–1 faster than Q.

P collides with Q from behind. Q now goes 1.2 m s–1 faster than P. No external forces act during the collision (Figure 3.38).

Figure 3.38

after collision

Q

v m s–1

P

v m s–1

Q

(v + 1.2) m s–1

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P

(v + 1.2) m s–1

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before collision

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Explain whether: (a) momentum has been conserved in the collision (b) t he collision is elastic or inelastic.

43. Use calculations to show whether these collisions are elastic or inelastic. (a) A ball of mass 5 kg travelling at 10 m s–1 collides with a stationary ball of mass 3 kg. After the collision, the 5 kg ball has a speed of 4 m s–1. Both balls move in the same straight line. (b) A truck of mass 3500 kg travelling at a speed of 20 m s–1 collides into the back of a car of mass 900 kg. The car is travelling at 15 m s–1 in the same direction as the truck at the time of collision. Both vehicles become joined together and move forward in the same straight line.

Explosions Will the separation of a composite object into parts be elastic or inelastic?

The relative speed of the parts of a composite object before separation will always be zero, and their relative speed of separation will always be greater than zero. Therefore, the separation of objects must always be inelastic.

We can show this by calculating the kinetic energy.

Tip Comparing relative speed before and after an interaction in order to show whether the interaction is elastic or inelastic can be more straightforward than calculating kinetic energy before and after.

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3 Dynamics Worked example A gun of mass 3 kg fires a 5 g bullet with a velocity of 400 m s–1.

Show using kinetic energy that the interaction between the gun and the bullet is inelastic.

Answer momentum before firing = mgvgb + mbv bb = (3 kg × 0 m s–1) + (0.005 kg × 0 m s–1) = 0 momentum of bullet leaving the gun = mbv ba = 5 × 10 –3 kg × 400 m s–1 = 2 kg m s–1 momentum of gun immediately after firing = −2 kg m s–1

total kinetic energy before firing = 0

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In Chapter 5, you will learn more about the different forms of energy and how they can be transferred.

total kinetic energy after firing = ( 12 mv2)b + ( 12 mv2)g

= ( 12 × 5 × 10 –3 kg × (400 m s–1)2) + ( 12 × 3 kg × (0.7 m s–1)2) = 400 J. So this separation is inelastic.

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2 kg m s −1 momentum of gun after = = 0.7 m s–1. mg 3 kg

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In a separation, or explosion, the energy which is transferred to the moving parts comes from the energy that was stored to cause the separation or explosion. For example, if two stationary trolleys are pushed apart by a compressed spring, then the kinetic energy transferred to the trolleys has come from the elastic potential energy in the compressed spring.

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44. A cannon of mass 650 kg fires a ball of mass 25 kg at 32 m s–1. (a) Calculate the recoil speed of the cannon. (b) Show, using kinetic energy, whether the interaction between the cannon and the ball is elastic or inelastic.

Key ideas

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➜➜In an elastic collision, momentum and kinetic energy are both conserved. ➜➜In an inelastic collision, momentum is conserved but kinetic energy is not conserved. Some energy is transferred to other forms. ➜➜For a perfectly elastic collision, the relative speed of approach is equal to the relative speed of separation. ➜➜Momentum is also conserved in the separation of a composite object into parts.

Such a separation is always inelastic.

ASSIGNMENT 000: Collisions in space Background Scientists have known for many years that collisions occur in space. There are large craters visible on the Moon and also on Earth, some of which must have been caused by collisions.

Relatively small objects in space, like asteroids and comets, can become caught in the gravitational field of a much larger object, like a planet. This can result in the smaller object starting to orbit the larger one. Or this can result in a collision.

Until 1994, scientists had never been able to watch a collision between a space object and a planet. In 1993 they had seen that a comet was travelling on a path that would take it very close to the planet Jupiter. They thought there was a very high chance of a collision, and observed the events that followed very closely.

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Figure 3.39 One of the fragments of comet Shoemaker-Levy 9 collides with Jupiter.

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In 1993, astronomers Carolyn and Eugene Shoemaker working with David Levy, discovered a comet. This comet was named after them as Shoemaker-Levy 9 or just SL9. Most of the known comets in the Solar System orbit the Sun, but this comet was in orbit around the planet Jupiter.

A1. Comet SL9 was in orbit around Jupiter long before 1993. (a) Describe the nature of the forces which were acting to cause this orbit. (b) Describe the magnitude and direction of these forces. (c) Explain using Newton’s first law, how a comet can follow a curved path around a larger object like a planet or the Sun.

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It was estimated that SL9 had been in orbit around Jupiter for 20−30 years before its discovery.

In the year before it was discovered, calculations showed that SL9 had passed around 40 000 km from the upper atmosphere of Jupiter.

When a smaller object, like a comet, approaches another larger one, like a planet, the force due to gravity from the planet can cause the comet to break apart (Figure 3.40).

planet

p F

f

comet

Figure 3.40 Gravitational forces from a planet can cause a comet to break apart. Not to scale.

Consider a particle P which is part of the comet in Figure 3.40. If P is close to the surface of the comet, then the force, F, due to gravity from the planet will be much greater than the force, f, due gravity from the comet itself.

If he difference between these forces is large enough, the comet will break apart.

A2. Newton’s third law predicts that there will be a pair of equal and opposite forces acting when one object exerts a force on another. This can be called an N3 pair.

Describe the force that makes up the N3 pair in Figure 3.40 for: (a) force F (b) force f.

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3 Dynamics This close approach of SL9 to Jupiter caused the comet to break up into 23 fragments that could be seen from Earth. Scientists did not see this happen, but calculated that it happened in 1992.

A3. Assume that SL9 broke apart into all 23 parts in one instant. (a) Explain whether the event of the comet breaking apart was elastic or inelastic. (b) Explain why all 23 fragments continued to travel in approximately the same direction as each other.

A4. Which property of the SL9 fragments could scientists on Earth observe directly?

A their momentum B their velocity C their kinetic energy D their mass

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One of the fragments had a mass of 1013 kg and was travelling at 60 km s–1 on impact with Jupiter.

A5 (a) Calculate the velocity of this fragment in m s–1. (b) Calculate:

(i) the momentum of this fragment (ii) the kinetic energy of this fragment.

This fragment collided with the atmosphere of Jupiter at an angle of 45° to the top of the atmosphere.

A6 (a) Draw a vector diagram to represent the momentum just before impact of this fragment. (b) Calculate the component of this fragment’s momentum towards the centre of Jupiter.

When the fragments of SL9 impacted with Jupiter, each of them broke apart into many smaller particles.

Careful observations from Earth indicated that some of these particles that moved across the top of Jupiter’s atmosphere had a constant deceleration.

A7. Suggest what can be concluded about the force on these particles that caused them to have constant deceleration.

Jupiter has the strongest gravitational field of all the planets in the Solar System. Some scientists state that Jupiter reduces the occurrence of impacts of objects like comets with Earth.

A8. Suggest an explanation for this statement.

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Chapter overview

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Newton’s first law

Forces and fields

inertia

mass resultant force

resistive forces

acceleration of free fall acceleration

Newton’s second law rate of change of momentum

Testing predictions against evidence momentum

principle of conservation of momentum

Newton’s third law Models of physical systems

weight

elastic and inelastic interactions

kinetic energy

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Chapter review

Learning outcomes

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• understand that mass is a property of an object that resists movement or change in motion • recall and be able to use the equation F = ma and understand that acceleration and the resultant force causing it are always in the same direction • be able to define linear momentum as the product of mass and velocity, p = mv, and use this equation to perform calculations on momentum • be able to define force as the rate of change of an object’s momentum • be able to state Newton’s first, second and third laws of motion and apply them to given situations • describe and understand the concept of weight as the effect of a gravitational field on a mass and recall that the weight of a body is equal to the product of its mass and the acceleration of free fall, W = mg • know about frictional forces and viscous/drag forces including air resistance • know that drag force increases with the speed of an object • describe the motion of objects in a uniform gravitational field with air resistance • understand that objects moving against a resistive force may reach a terminal (constant) velocity • state the principle of conservation of momentum • be able to apply the principle of conservation of momentum to solve problems involving objects in both one and two dimensions • know the difference between elastic and inelastic interactions • understand that, while momentum of a system is always conserved in interactions between bodies, a change in kinetic energy takes place if the collision is inelastic • recall that, for a perfectly elastic collision, the relative speed of approach of the interacting objects is equal to the relative speed of separation of the objects

Chapter review

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1. (a) State Newton’s first law of motion. (b) State whether a resultant force acts on a car when it is: (i) parked and not moving (ii) accelerating rapidly (iii) travelling at a constant high speed in a straight line (iv) travelling at a constant slow speed around a corner.

2. A bowling ball has a mass of 6.8 kg. A soccer ball has a mass of 0.55 kg. Both balls have approximately the same diameters. When both balls are rolling together at the same speed, explain which ball is easier to stop.

3. (a) State Newton’s second law of motion. (b) Write the two equations that can be used to summarise Newton’s second law. (c) The SI derived unit of force is the newton, N. Use Newton’s second law to express the newton in terms of SI base units. (d) Calculate the acceleration produced when a resultant force of 650 N acts on a 2.5 kg mass.

4. (a) Define momentum. (b) Calculate the momentum of each of these objects. Give your answers in standard form. (i) a 12 mg housefly travelling at 3 m s–1 (ii) a 5.5 × 107 kg cruise ship travelling at 12 m s–1 5. An artificial satellite passes through a cloud of cosmic dust particles. Each cosmic dust particle has a mass of 1 pg. The particles collide with the satellite at 70 km s–1 at a rate of 106 per second and are each brought to a stop in 1 µs.

Calculate the average force exerted by the dust particles on the satellite.

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3 Dynamics

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6. A model railway wagon of mass 120 g is travelling at 0.11 m s–1 when it collides with another stationary model wagon of mass 150 g. The two carriages become joined together. (a) Calculate the speed of the two carriages immediately after collision. (b) Determine whether the collision is elastic or inelastic.

7. Air track rider P, travelling at 4.0 cm s–1 collides with rider Q, travelling at 2.5 cm s–1 in the same forwards direction. Both riders have the same mass.

Which of these could describe their velocities after an elastic collision?

A A forwards at 2.0 cm s–1 and B forwards at 6.5 cm s–1.

B A backwards at 1.0 cm s–1 and B forwards at 5.5 cm s–1.

C A forwards at 2.5 cm s–1 and B forwards at 4.0 cm s–1.

D A backwards at 0.5 cm s–1 and B forwards at 6.0 cm s–1.

8. A skydiver of mass 70.0 kg jumps out of a stationary hot air balloon and reaches a terminal velocity of 55.0 m s–1. The skydiver then opens their parachute and slows to a new terminal velocity of 6.00 m s–1.

Describe how the force of air resistance changes during the fall. Calculate the force of air resistance where possible.

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