Momentum and Impulse Equations!!!!?

Question

A 5.00 g bullet is fired with a velocity of 1.00 x 10^2 m/s toward a stationary solid block resting on a frictionless surface.

a) What is the change in momentum of the bullet if it embeds in the block?

b)What is the change in momentum of the bullet if it ricochets in the opposite direction with a speed of 99 m/s - almost the same speed as it had originally?

c) In which case does the block acheive the greatest velocity?

Answer 1

For a), let the mass of the block be B. Before the impact, the momentum of the two is 5 * 10^2, which it must be after the impact. The velocity of the combined block and bullet is

v = 5 * 10^2 / (B + 5)

The momentum of the bullet alone is 5v, or

25 * 10^2 / (B + 5)

If the block is extremely light, the momentum of the bullet will be close to the original 5 * 10^2. If the block is extremely heavy, the momentum of the bullet will be close to zero.

For b), the change in momentum is 9.99 * 10^2. The bullet starts with momentum of 5 * 10^2 and ends with momentum -4.99 * 10^2.

For c), the ricochet gives the block greater velocity, because that is the circumstance in which the bullet loses the most momentum.

We do not compute any impulse. To do so, we would need to know the time taken by certain events, and we do not have that information.

Answer 2

You need to know the mass of the block to solve this.