# 2 Bullets, same mass, same v, fired at wood and steel block, which block has > v?

by BC2210
Tags: block, bullets, fired, mass, steel, wood
 P: 17 1. The problem statement, all variables and given/known data 2 bullets of the same mass are fired at the same velocities. One hits a wood block, and embeds itself. One hits a steel block, and bounces off. Both blocks are on a frictionless table. Compare the two blocks speed after. 2. Relevant equations 3. The attempt at a solution The wood block is a elastic collision because it embeds itself correct? The steel block is a perfectly inelastic collison which causes the bullet to bounce of with the same v, and the block moves away with the same v. But what about the v of the wooden block? All the energy is transfered into the block so it moves with the same v as well?
 P: 67 ahh we just did this in physics. what you would have to do is the mass of the bullet times the velocity of the bullet plus the mass of the block times its initial velocity (which would come out to zero) and set it equal to the mass of the bullet and block added together and your variable would be v. so ill put that all in an equation for you. m1v1 + m2v2 = (m1+m2)vf where m1 and v1 are the mass and velocity of the bullet. m2 and v2 are the mass and velocity of the block initially (if its at rest then that equals 0) and vf is the final velocity your trying to find. plug it all in and you will get your answer! let me know if this helps.
 P: 67 do you happen to have all the information?
P: 17

## 2 Bullets, same mass, same v, fired at wood and steel block, which block has > v?

All the information for the two scenarios are exactly the same. The mass of the block and bullet, initial v, and all is the same. Only difference is one block is steel and one is wood. One is elastic and one is inelastic collision. But do the V's of the blocks after the collision equal each other?
 P: 67 Hmmm, I would want to say no. For an inelastic equation, where the bullet would go into the wooden block, the equation is m1v1 + m2v2 = (m1 + m2)vf. which is different from an elastic equation where everything remains constant. m1v1 + m2v2 = m1fv1f + m2fv2f (sorry my notation is a bit off) where the left side is the initial and the right side is the final. So inorder for the bullet to reboud at the same velocity, the mass of the metal block would have to remain zero, i.e. no kinetic energy is lost. does this make sense? originally you had your elastic and inelastic's mixed.
 Mentor P: 40,278 Hint: Compare the momentum transferred to the block in each case. (When the bullet bounces off and reverses direction, how has its momentum changed?)
 P: 17 I know momentum is conserved in each case. In the case of the wood block, the two objects stick together and move with a common velocity. But is that velocity the initial v of the bullet? However, kinetic energy is not conserved when the bullet hits the wood block correct? The kinetic energy in the elastic collision with the steel block IS conserved...correct? SO wouldnt the steel block move with a faster velocity after impact over the wood block?
 P: 67 no because it repelled the bullet. the steel block has no velocity. the wooden block does. If the steel block did move, and it also repeled the bullet at the same velocity, then it would have spontaneously gained energy, which is impossible.
Mentor
P: 40,278
 Quote by derek.basler no because it repelled the bullet. the steel block has no velocity. the wooden block does. If the steel block did move, and it also repeled the bullet at the same velocity, then it would have spontaneously gained energy, which is impossible.
While it's true that the bullet cannot bounce back with exactly the same speed, the problem never stated that it did. The steel block will definitely move.

Again I suggest that you forget about energy and just consider the momentum transferred to the block in each case. Momentum conservation is all you need to answer this one.
 P: 67 "One hits a steel block, and bounces off. Both blocks are on a frictionless table. The steel block is a perfectly inelastic collison which causes the bullet to bounce of with the same v, and the block moves away with the same v." if it is an elastic collision, then kinetic energy must be conserved. either way, the velocity of the steel block would be much less, if anything.
Mentor
P: 40,278
 Quote by derek.basler "One hits a steel block, and bounces off. Both blocks are on a frictionless table. The steel block is a perfectly inelastic collison which causes the bullet to bounce of with the same v, and the block moves away with the same v."
That was the OP's attempt at a solution, not part of the problem statement.

 if it is an elastic collision, then kinetic energy must be conserved.
That's certainly true.
 either way, the velocity of the steel block would be much less, if anything.
Why do you say that?
 P: 67 because in an elastic collision that repels a bullet, thereby causing it to go back in the oppisite direction with an almost equivalent speed, the kinetic energy must be conserved. if the bullet was stopped and all the kinetic energy was transfered into the steel block, then maybe it could have a greater speed. but to me it doesnt seem very likely. As for an inelastic collision, where the bullet goes into the box, all the kinetic energy of the bullet is being transfered over to the final mass and velocity, the mass of the box plus the mass of the bullet times the final velocity. Therefore, since more kinetic energy is being put into the wooden block, it moves faster, unlike the steel box.
Mentor
P: 40,278
 Quote by derek.basler because in an elastic collision that repels a bullet, thereby causing it to go back in the oppisite direction with an almost equivalent speed, the kinetic energy must be conserved.
The problem doesn't state that the collision is perfectly elastic, but let's assume it is. (The actual speed of the rebounding bullet will depend on the relative masses of bullet and block.)

Yes, kinetic energy is conserved. But so is momentum, and that's the key here.
 As for an inelastic collision, where the bullet goes into the box, all the kinetic energy of the bullet is being transfered over to the final mass and velocity, the mass of the box plus the mass of the bullet times the final velocity. Therefore, since more kinetic energy is being put into the wooden block, it moves faster, unlike the steel box.
The problem with this thinking is that since KE is not conserved in the inelastic collision, all of the bullet's KE does not go into increasing the speed of the wood block--much of the energy is lost as thermal energy and deformation.

One more time: Which block--wood or steel--ends up with the most momentum?
 P: 67 The wooden block obviously. One, it has more final mass than the steel one, because it has the bullet plus the mass of the wooden block. And did we agree that the wooden block would have more velocity? because that would further corroborate my thinking. it would be much easier if we had actually number masses and velocities.
Mentor
P: 40,278
 Quote by Doc Al One more time: Which block--wood or steel--ends up with the most momentum?
 Quote by derek.basler The wooden block obviously.
Well... no.
 One, it has more final mass than the steel one, because it has the bullet plus the mass of the wooden block.
That's true.
 And did we agree that the wooden block would have more velocity?
Nope. That's the very point we are discussing.
 because that would further corroborate my thinking. it would be much easier if we had actually number masses and velocities.
You don't need actual numbers. Call the mass of the bullet m and the block M; call the initial speed of the bullet v and the final speed of the block V. Thus the total momentum of the system in both cases is mv.

Bullet vs wood: In this case the momentum of bullet and block equals the initial momentum of the bullet (mv), so the final speed of the block is V = mv/(m + M).

Bullet vs steel: Let's call the speed of the bullet after it bounces off v', thus the final momentum of the bullet is -mv'. From momentum conservation: mv = -mv' + MV, thus V = m(v + v')/M.

Now which situation do you think gives the block the greatest final speed?

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