Question

A 1.3-kg ball is attached to the end of a 0.8-m string to form a pendulum. This pendulum is released from rest with the string horizontal. At the lowest point of its swing, when it is moving horizontally, the ball collides with a 0.6-kg block initially at rest on a horizontal frictionless surface. The speed of the block just after the collision is 2.2 m/s. What is the speed of the ball just after the collision

Answer 1

Answer:

2.9 m/s

Explanation:

Momentum will be conserved

Speed of the ball just before collision is

v = √2gh = √(2(9.8)(0.8)) = 3.96 m/s

The initial momentum is 1.3(3.96) = 5.15 kg•m/s

The block takes away momentum of 0.6(2.2) = 1.32 kg•m/s

Leaving the ball with momentum of 5.15 - 1.32 = 3.83 kg•m/s

vf(ball) = 3.83 / 1.3 = 2.946... ≈ 2.9 m/s