# Speed of bullet and kinetic energy lost in the impact

1. May 11, 2015

### moenste

1. The problem statement, all variables and given/known data
A bullet of mass 2.0 * 10-3 kg is fired horizontally into a free-standing block of wood of mass 4.98 * 10-1 kg, which it knocks forward with an initial speed of 1.2 m s-1.
(a) Estimate the speed of the bullet.
(b) How much kinetic energy is lost in the impact?
(c) What becomes of the lost kinetic energy?

(a) 300 m s-1;
(b) 89.64 J.

2. Relevant equations
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3. The attempt at a solution
(a) 2.0 * 10-3 kg * X m s-1 = 4.98 * 10-1 kg * 1.2 m s-1
X = 298.8 m s-1
(b) Kinetic energy = 1/2 * m * v2 = 0.5 * (2.0 * 10-3) * 298.8 m s-1 = 89.28 J.
(c) Kinetic energy lost is potential energy gained.

However, as the answers don't fit the book answers 100% and I have no answer for the (c) part, I am not sure whether I am right or wrong. Any help please? Thanks in advance for any tips.

2. May 11, 2015

### PeroK

Can you see what you've forgotten? There's a hint in the numbers they gave you for the mass of the block and the bullet.

3. May 11, 2015

### moenste

For (b) I think I got it: KE of bullet - KE of block = 90 J - .35856 J = 89.64 J.

But for parts (a) and (c)? I did take the book answer of 300 m s-1 in the (b) calculations.

4. May 11, 2015

### jbriggs444

Go back to part a and see if you can spot your error. Conservation of momentum says that the momentum before the collision is equal to the momentum after the collision. You've almost written that down in equation form. But you missed a term. PeroK gave you a good hint.

5. May 12, 2015

### ehild

A bullet is fired horizontally into a free-standing block of wood.

What do you think "into" means? What happens to the bullet when it hits the target? Your solution means that it loses speed, and falls down in front of the block.

6. May 12, 2015

### moenste

Thanks guys, I got from ehild's words that the bullet is still moving but only with a new velocity. So:
mbullet*vbullet = mblock*vblock + mbullet*vnew speed of bullet

2*10-3*Vb = 4.98*10-1*1.2 + 2*10-3*1.2
Vb = 300 ms-1

And any tips on (c) What becomes of the lost kinetic energy? My guess is kinetic energy lost is potential energy gained, so lost KE transforms into potential energy... But I'm not sure.

7. May 12, 2015

### PeroK

You need to try to relate the physics you are learning to the real world. What two things would you expect to happen if you fired a bullet into a block of wood?

8. May 12, 2015

### AlephNumbers

Hmmm. The problem statement is a bit ambiguous. It does not explicitly say what happens when the bullet is fired into the block of wood. It seems to me that there may be two reasonable solutions to this problem.

9. May 12, 2015

### moenste

Hm, the bullet should deform, decrease its' speed and continue moving until rest. The block should deform and start moving until rest. So form change and change in speed for both objects?

10. May 12, 2015

### AlephNumbers

Perhaps. But there is a much more likely scenario, which I am sure is what the questioner intended. What do you think would happen if you fired a bullet into a really thick, dense block of wood?

The block in the problem statement is pretty small for a ballistics test, but it is free standing...

11. May 12, 2015

### moenste

I think the bullet would deform, lose speed and fall on the ground if the block would be very thick or it would do what I said in the prevous post, depending on the block. But how does that relate to the question of "What becomes of the lost kinetic energy?"?

12. May 12, 2015

### AlephNumbers

It has everything to do with the lost kinetic energy. The kinetic energy lost in the collision is dependent upon the elasticity of the collision.

Look, this isn't heavy armor plating we are talking about, this is wood! Have you ever thrown a dart at a dartboard? A dart, which has significantly less penetrating power than a bullet, can easily become embedded in a wooden dartboard. How can you extrapolate this situation to that of firing a bullet into a free standing block of wood?

13. May 12, 2015

### moenste

The bullet should deform, decrease its' speed and continue moving until rest. The block should deform and start moving until rest. Lost kinetic energy is transformed from the bullet to the block?

14. May 12, 2015

### ehild

The question is, with what velocity the bullet would move further? The block is in the way, the bullet can not go round.
It can happen that it bounces back from the block, but but the block is from wood, a soft material...
What would happen if the block was made of butter?

15. May 14, 2015

### moenste

If the block would made of butter the velocity would stay the same and the bullet would go through it.

16. May 14, 2015

### ehild

That can happen, too. I just wanted to illustrate that the bullet can enter into the target. And it can rebound, can penetrate through it, or can be embedded inside the target.
In the problem, the bullet is shot into the block. It is meant to enter and stay inside the block. Otherwise it would be said that the bullet leaves the block with a certain speed or bounces back from the block...

17. May 14, 2015

### moenste

So "What becomes of the lost kinetic energy?" -> it stays in the wooden block? Very sorry for bothering.

I do understand that depending on the object with which the bullet contacts it has different outcomes (rebounding, penetration or going through the object) but KE = the energy which a body possesses solely because it is moving. So the energy which the bullet loses due to the wooden block is given to the wooden block? I would give that as an answer in the exam and from what I got in this thread I did not understand whether this answer is right or wrong and if it's wrong then for what reason. Again very sorry, but maybe due to my English I did not get the tips from posts.

Last edited: May 14, 2015
18. May 14, 2015

### ehild

You have to calculate the kinetic energy that is lost. The kinetic energy here means that of the bullet and the block, together. If you assume inelastic collision, the block and the bullet inside move with the same velocity. You can calculate it and from that, you get the lost kinetic energy. What do you think into what kind of energy does it transform? Do you hear sound at impact? Is there deformation - you said there is. And does the bullet become hot?

19. May 14, 2015

### moenste

KE of bullet - KE of block = 90 J - .35856 J = 89.64 J.
89.64 J is the KE which the bullet loses.

There's sound, deformation and not sure whether the bullet becomes hot.

As I understand the lost KE is transformed in sound, physical change and temperature of the bullet and that should be the answer?

20. May 14, 2015