Question

A 1.0-kg ball is attached to the end of a 2.5-m string to form a pendulum. This pendulum is released from rest with the string horizontal. At the lowest point in its swing when it is moving horizontally, the ball collides elastically with a 2.0-kg block initially at rest on a horizontal frictionless surface. What is the speed of the block just after the collision

Answer 1

The speed of the block just after the collision is equal to the speed of the ball divided by the square root of 2.

Step-by-step explanation:

When the pendulum reaches the lowest point of its arc, it has the maximum kinetic energy and minimal potential energy. The energy is shared between the ball and the block after the collision. Since the collision is elastic, the total kinetic energy is conserved. Therefore, we can equate the initial kinetic energy of the ball with the final kinetic energy of the block.

Using the equation for kinetic energy, KE = 1/2 * mass * velocity^2, we find that the initial kinetic energy of the ball is (0.5 * 1.0 kg * v^2), where v is the speed of the ball. The final kinetic energy of the block is (0.5 * 2.0 kg * v_b^2), where v_b is the speed of the block after the collision.

Equating the initial kinetic energy of the ball to the final kinetic energy of the block gives us:

1. (0.5 * 1.0 kg * v^2) = (0.5 * 2.0 kg * v_b^2)
2. (v^2) = 2 * (v_b^2)
3. v_b = v / sqrt(2)

Therefore, the speed of the block just after the collision is equal to the speed of the ball divided by the square root of 2.