Question

A 3.2-kg ball is attached to the end of a 0.80-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 1.6-kg block initially at rest on a horizontal frictionless surface. the speed of the block just after the collision is 6.0 m/s. what is the speed of the ball just after the collision? (elastic collision)

Answer 1

To find the speed of the 3.2-kg ball after an elastic collision with a 1.6-kg block, use the conservation of momentum and conservation of kinetic energy to solve a system of equations with the given variables.

Step-by-step explanation:

In the scenario where a 3.2-kg ball attached to a pendulum collides with a 1.6-kg block initially at rest, we can determine the speed of the ball just after the collision using the principles of conservation of momentum and conservation of kinetic energy that apply to elastic collisions.

Since the initial speed of the block after collision is given as 6.0 m/s, we can set up two equations based on these principles: one for conservation of momentum (m1v1i + m2v2i = m1v1f + m2v2f) and one for conservation of kinetic energy (0.5m1v1i2 + 0.5m2v2i2 = 0.5m1v1f2 + 0.5m2v2f2). Solving the system of equations with the known variables will yield the speed of the ball after collision.