1. A river flows at a rate of 5 m / min from west to east. A man on the river’s south bank, competent at swimming at 10 m / min in calm water, wants to swim throughout the river in the shortest time. In which specific direction should the man swim to reach his destination.
Ans. Time to cross river = t t
For time t to be minimum, cos should be at its maximum.
∴ θ should be 0.
Hence, the swimmer should swim due north
2. A 0.2-kilogramme ball is suspended from a 5 m vertical. A bullet with a mass of 0.01 kg and a horizontal velocity of V m/s collides with the centre of the ball. Following the contact, the bullet and ball move separately. They strike the ground at a distance of 20 metres and the bullet at a distance of 100 metres from the post’s foot. The bullet’s initial velocity V is
Ans. Assume time taken by bullet to fall = t
½ gt = 5
T = 1 sec
Vball x 1= 20; Vball = 20 m/s
Vbullet x 1= 100; Vbullet = 100 m/s
M bullet V = M bullet V bullet + M ball V ball
0.01v = 0.01 x 100 + 0.2 x 20
0.01v = 1 + 4 = 5
∴ V = 500 m/s
A 50 kg bucket is being lifted by a rope. The rope will not break if the tension is 525 N or less. The bucket started at rest and after being lifted 3 meters, it's moving at 5 m/s. If the acceleration is constant, is the rope in danger of breaking?
Answer : The work done on an object is equal to the change in its kinetic energy. In this case, the work done by the tension in the rope lifts the bucket against gravity, increasing its gravitational potential energy.
The initial kinetic energy of the bucket is zero since it starts at rest, and the final kinetic energy is given by:
KE_final = (1/2) * m * v^2
where m is the mass of the bucket (50 kg) and v is the final velocity (5 m/s).
The work done by the tension in the rope is equal to the change in gravitational potential energy:
Work = m * g * h
where g is the acceleration due to gravity (approximately 9.8 m/s^2) and h is the height the bucket is lifted (3 meters).
Equating the work done by the tension to the change in kinetic energy, we have:
m * g * h = (1/2) * m * v^2
Now, we can calculate the tension in the rope using Newton's second law:
Tension = m * (g + a) where a is the acceleration of the bucket. Given that the acceleration is constant, we have:
Tension = m * (g + a)
Substituting the values:
Tension = 50 kg * (4.17 m/s^2 + a)
To determine if the rope is in danger of breaking, we need to check if the tension is within the safe limit of 525 N or less:
Tension ≤ 525 N
Substituting the expression for tension:
Tension = 50 kg * (4.17 m/s^2 + a)
To determine if the rope is in danger of breaking, we need to check if the tension is within the safe limit of 525 N or less:
Tension ≤ 525 N
Substituting the expression for tension:
50 kg * (4.17 m/s^2 + a) ≤ 525 N
Simplifying: 208.5 N + 50 kg * a ≤ 525 N 50 kg * a ≤ 525 N - 208.5 N
50 kg * a ≤ 316.5 N
Dividing both sides by 50 kg: a ≤ 6.33 m/s^2
Since the acceleration a must be less than or equal to 6.33 m/s^2 for the tension to be within the safe limit, we can conclude that the rope is not in danger of breaking since the calculated acceleration is approximately 4.17 m/s^2, which is less than 6.33 m/s^2.