Question

Angie, Brad, and Carlos are discussing a physics problem in which two identical bullets are fired with equal speeds at equalmass wood and steel blocks resting on a frictionless table. One bullet bounces off the steel block while the second becomes embedded in the wood block. “All the masses and speeds are the same,” says Angie, “so I think the blocks will have equal speeds after the collisions.” “But what about momentum?” asks Brad. “The bullet hitting the wood block transfers all its momentum and energy to the block, so the wood block should end up going faster than the steel block.” “I think the bounce is an important factor,” replies Carlos. “The steel block will be faster because the bullet bounces off it and goes back the other direction.” Which of these three do you agree with, and why?

Short answer

The reply of Carlos is right and it is explained below.

Step-by-step solution

Given information

We have given that One bullet bounces off the steel block while the second becomes embedded in the wood block. “All the masses and speeds are the same,” says Angie, “so I think the blocks will have equal speeds after the collisions.” “But what about momentum?” asks Brad. “The bullet hitting the wood block transfers all its momentum and energy to the block, so the wood block should end up going faster than the steel block.” “I think the bounce is an important factor,” replies Carlos. “The steel block will be faster because the bullet bounces off it and goes back the other direction.”

We need to find that which of these three do you agree with, and why.

Simplify

For the steel block collision:

The initial velocity of the block is zero. We can use the conversation of the momentum for this collision to be in the form.

$m{v}_{b,i}=m(-v_{b,f})+M{v}_{s,f}$ $\left(1\right)$

Where $m$is the mass of the bullet, $M$is the mass of the steel block, $v_{b,i}$is the initial velocity of the bullet,$v_{b,f}$is the final velocity of the bullet, and $v_{s,f}$is the final velocity of the steel block. Solve equation $\left(1\right)$for $v_{s,f}$

$v_{s,f}=\frac{m(v_{b,i}+v_{b,f})}{M}$ $\left(2\right)$

For the wood block collision:

The initial velocity of the block is zero and the final velocity of the bullet is same for the wood as they move together.

Apply the conservation law of momentum by

$m{v}_{b,i}=(m+M)v_{\omega ,f}$ $\left(3\right)$

Where $v_{\omega ,f}$is the final velocity of the wood block. Solve equation $\left(3\right)$for $v_{\omega ,f}$

$v_{\omega ,f}=\frac{m{v}_{b,i}}{m+M}$ $\left(4\right)$

Comparing these two equations $\left(2\right)$and $\left(4\right)$, We will find the velocity of the steel block is larger than the velocity of the wood block.

$v_{steel}>v_{wood}$

So, we agree with Carlos.