Question

A bullet with mass m = 5.21 g is moving horizontally with a speed v = 443 m/s when it strikes a block of hardened steel with mass M = 14.8 kg (initially at rest). The bullet bounces off the block in a perfectly elastic collision. (a) What is the speed (m/s) of the block immediately after the collision?

Answer 1

Answer:

0.312 m/s

Explanation:

Elastic collisions conserve momentum and kinetic energy

The velocity of the center of mass will not change. It continues at

0.00521(443) / 14.80521 = 0.155893... ≈ 0.156 m/s

To conserve kinetic energy we can think of the center of mass (CoM) as an ideal spring returning to each mass that strikes it an identical speed of collision in the opposite direction.

The CoM sees the target approach at - 0.156 and will see it depart at 0.156 m/s

A ground based observer sees the target depart at the velocity of the CoM plus the relative velocity .

v = 0.156 + 0.156  = 0.312 m/s