LINEAR MOMENTUM

CHAPTER 6 LINEAR MOMENTUM 6.1 Definition of linear momentum 6.2 Conservation of linear momentum and its applications

6.2.1 Elastic collision 6.2.2 Inelastic collision 6.3 Impulse

6.1 Definition of linear momentum The linear momentum of an object with mass m travelling with velocity v is defined as: “The product of mass and velocity of an object.�

• It is a vector quantity. • The formula of momentum

Momentum is “MASS in MOTION�

đ?‘? = đ?‘šđ?‘Žđ?‘ đ?‘ Ă— đ?‘Łđ?‘’đ?‘™đ?‘œđ?‘?đ?‘–đ?‘Ąđ?‘Ś đ?‘? = đ?‘šđ?‘Ł SI unit: kg.m/s @ N.s The direction of the momentum is the same as the direction of the velocity.

‌‌‌(1)

6.2 Conservation of Linear Momentum and Its Applications For closed system (isolated system), the law of conservation of linear momentum states that “If the net force acting on the system is zero, the total linear momentum of a system is constantâ€? đ??šđ?‘›đ?‘’đ?‘Ą = 0 ∆đ?‘? =0 ∆đ?‘Ą ∆đ?‘? = 0 đ?‘?đ?‘“ − đ?‘?đ?‘– = 0

đ?‘?đ?‘– =

đ?‘?đ?‘“

‌‌‌(2)

�1 �1 + �2 �2 = �1 �1 + �2 �2

‌‌‌(3)

The total momentum of an isolated system of objects remains constant.

Collision When two objects collide, the total linear momentum is conserved. However, the kinetic energy after the collision is generally less than that before collision in two ways: i. ii.

It can be converted into heat because of friction.

It is spent in creating permanent distortion or damage, as in automobile collision.

Collision can be classified into:

Elastic collision – one in which the total kinetic energy of the system after the collision is equal to the total kinetic energy before the collision.

Inelastic collision – one in which the total kinetic energy of the system is not the same before and after the collision; if the objects stick together after colliding, the collision is said to be completely inelastic.

Collision Collisions are common occurrence in everyday life.

Before collision

Collision Elastic

Inelastic

Collision in which kinetic energy & momentum is conserved m1v1

1

2

m2 v2

Collision in which kinetic energy is not conserved, momentum conserved

Must Remember !!!

Collision

Momentum

Elastic

ÎŁđ?‘?đ?‘– = ÎŁđ?‘?đ?‘“ đ?‘š1 đ?‘˘1 + đ?‘š2 đ?‘˘2 = đ?‘š1 đ?‘Ł1 + đ?‘š2 đ?‘Ł2

Kinetic energy

đ??žđ??¸đ?‘– = đ??žđ??¸đ?‘“ 1 1 (đ?‘š1 đ?‘˘12 + đ?‘š2 đ?‘˘22 ) = (đ?‘š1 đ?‘Ł12 + đ?‘š2 đ?‘Ł22 ) 2 2

Completely collision

− đ?‘˘2 − đ?‘˘1 = đ?‘Ł2 − đ?‘Ł1

Inelastic

ÎŁđ?‘?đ?‘– = ÎŁđ?‘?đ?‘“

đ?‘š1 đ?‘˘1 + đ?‘š2 đ?‘˘2 = (đ?‘š1 + đ?‘š2 )đ?‘Ł2 đ??žđ??¸đ?‘– = đ??žđ??¸đ?‘“ + (−đ??žđ??¸đ?‘™đ?‘œđ?‘ đ?‘ ) 1 1 đ?‘š1đ?‘˘12 + đ?‘š2 đ?‘˘22 + đ??žđ??¸đ?‘™đ?‘œđ?‘ đ?‘ = (đ?‘š1 + đ?‘š2 )đ?‘Ł 2 2 2

đ??žđ??¸đ?‘– > đ??žđ??¸đ?‘“

Example elastic collision : http://www.physicsclassroom.com/mmedia/momentum/trece.cfm http://www.physicsclassroom.com/mmedia/momentum/crete.cfm http://www.physicsclassroom.com/mmedia/momentum/cthoe.cfm Example inelastic collision : http://www.physicsclassroom.com/mmedia/momentum/creti.cfm http://www.physicsclassroom.com/mmedia/momentum/cthoi.cfm http://www.physicsclassroom.com/mmedia/momentum/treci.cfm

Example 1:

uB  3 m s

1

u A  6 m s 1 A

B

Figure above shows an object A of mass 200 g collides head-on with object B of mass 100 g. After the collision, B moves at a speed of 2 m s-1 to the left. Determine the velocity of A after collision.

Solution :

mA  0.200 kg; mB  0.100 kg; u A  6 m s 1 u B  3 m s 1; vB  2 m s 1

   pi   p f m Au A  mB u B  mAv A  mB vB

0.200  6  0.100 3  0.200 v A  0.100  2 v A  3.5 m s

1

to the left

6.3 Impulse Impulse is the change in momentum The formula of impulse:

J = Final momentum – Initial momentum = P f – Pi ‌‌‌(4) = mv – mu Impulse is also defined as the average of force, F times the length of time, t. ∆đ?‘? đ??š=

∆đ?‘Ą ‌‌‌(5)

J = FΔt

Substitute equation (4) into (5)

Ft = mv – mu Impulse is a vector and SI unit for impulse is kg m/s @ N.s

Example 2: A volley ball player changes the velocity of the ball from 5.0 m/s to -30 m/s along a certain direction. If the impulse delivered to the ball by the player is -15.0 kg.m/s, calculate the mass of the valley ball.

Solution :

J  mv  mu J  m( v  u) J m v u  15   30  5  0.43 kg

Example 3: A baseball with a mass of 0.2 kg is moving towards a player with a speed of 50 m/s. The player then hits the ball with his bat. The bat exerts an average force of 8.0 × 103 N on the ball for 1.6 ms. The average force is directed in the opposite direction to the initial velocity of the ball. Calculate the final speed of the ball.

[Assume that u is to the left and F is to the right] Solution :

Ft  mv  mu Ft  m( v  u) (8  103 )(1.6  10 3 )  0.2( v  ( 50)) v  14.0 m / s

Youtube link for reference https://www.youtube.com/watch?v=Xe2r6wey26E

https://www.youtube.com/watch?v=8ko3qy9vgLQ

Example on youtube https://www.youtube.com/watch?v=bOK3NyQNkl0

https://www.youtube.com/watch?v=pZtbcvodfmI