# Ballistic pendulum problem

1. Apr 2, 2004

### psruler

Hi can anybody verify if im using the correct formula,
Here is the problem first,

A bullet embeds itself in a pendulum bob, and the bullet+bob together swing up to some maximum height, which is easy to measure. IF the bullet has a mass of 15g, the bob's mass is 4kg, and the combination reaches a height of 23cm after impact, what was the speed of the bullet just before impact?

Do I use this formula:

v= (mass of bullet/ mass of pendulum bob + mass of bullet)(2(9.80m/s^2)(h))^1/2

THANKS AGAIN!

2. Apr 2, 2004

Ah! I know we had this question before!

Let's see if the search feature works this time...

3. Apr 2, 2004

Nope. No luck with the search.

I'll give you a hint, though. The collision is not elastic, and energy is not conserved. Use a momentum argument to solve this.

4. Apr 2, 2004

### psruler

hmmm... i don't quite understand cookiemonster. I guess my formula above is wrong?

5. Apr 2, 2004

Ah, I didn't look at it. It's too much work to look at non-formatted math that's half symbolic and half numerical and I'm too tired. Sorry.

What I mean by my previous post is that if you found the kinetic energy of the bullet before the collision and tried to equate that to the potential energy of the block, then you're not going to get the correct answer.

6. Apr 2, 2004

### Chen

If you know the mass of the bob and bullet together and the height they reach, you can find their joint velocity at the bottom of pendulum, right after the impact, using energy conservation. Then use the conservation of momentum to find the speed of the bullet right before the impact.

7. Jun 3, 2004

### e(ho0n3

Glad I found this thread. I'm just doing this problem at the moment. There is an example in the book I have about this problem, but it assumes that the change in distance of the bob as the bullet enters the bob is small enough to be negligible to make the problem easier. I'm supposed to try this problem considering the latter as not negligible.

Given the masses of the bullet (m) and the bob (M), and the height (H) the pendulum travels from the rest position, find the velocity (v) of the bullet before collision. I'm also given the distance (d) the bullet travels within the bob.

There are two 'phases' in this problem: the collision phase and the post-collision phase.

What is happening during the collision phase? When the bullet is colliding with the bob, the velocity of the bullet slows down due to it entering the bob while the bob is gaining velocity and increasing its height. At the same time the bullet is also increasing its height since it's moving with the bob (vertically speaking).

The impulse on m during this collision is given by
$$mu_i - mv_i = \bar{F_i}\Delta{t}$$​
where $$\bar{F_i}$$ is average force on m during the collision (of duration $$\Delta{t}$$) (Note: I'm using index notation for the vectors.). All I need to do then is find $$u_i$$, $$\bar{F_i}$$, and $$\Delta{t}$$, and then solve for v and I'm done. Right? But what about gravity? How do I handle gravity during the duration of the collision? Should it be ignored or do I have to take it into account (but then my impulse equation wouldn't be valid)?

e(ho0n3

Last edited: Jun 3, 2004
8. Jun 5, 2004

### e(ho0n3

How can you say energy is not conserved! Mechanical energy is not conserved, but the total energy remains the same.
But how can I use conservation of momentum when there is an external force involved (namely gravity).

The only way I can solve this problem (as I see it) is to use energy conservation, but a problem arises during the collision phase since I would have to figure out the work done by friction to stop the bullet within the bob and the thermal energy involved. I could simplify things if I considered the impulse force as the force of friction but I'm not sure whether this is legal.

9. Jun 5, 2004

### turin

I believe cookiemonster meant the mechanical energy of the bullet pendulum system. This energy is not conserved (the system may heat up, there may be sonic emissions, etc., the point is that the collision is largely inelastic).

10. Jun 8, 2004

### Chen

Gravity only acts along the vertical axis. There are no external forces along the horizontal axis, so momentum in that axis is conserved. No?