# Momentum and kinetic energy in collisions with a bullet

1. Mar 28, 2014

### jdawg

1. The problem statement, all variables and given/known data

A bullet of mass 4.2 g strikes a ballistic pendulum of mass 2.0 kg. The center of mass of the pendulum rises a vertical distance of 18 cm. Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.

2. Relevant equations

3. The attempt at a solution

I started by converting everything into the correct units:
Bullet's mass(m)= 0.0042 kg Ballistic pendulum(M)= 2 kg Vertical distance(h)= 0.18 m

Then used this equation to solve for the velocity of the bullet after the collison:
vbullet=$\sqrt{2gh}$
vbullet=$\sqrt{2(9.8)(0.18)}$
vbullet=1.878 m/s

Then this equation to get the bullet's initial speed:
vi=$\frac{(M+m)}{m}$(V)
vi=$\frac{(2+0.0042)}{0.0042}$(1.878)
vi=896.16 m/s

I'm not sure what I'm doing wrong, my online homework only wants 2 significant digits.
Also, if someone could explain to me where these equations came from I would really appreciate it. I'm having some trouble understanding how to manipulate the energy equations to wind up with the ones I used in the problem.

2. Mar 28, 2014

### dauto

The solution seems correct. The first equation you used is energy conservation while the second equation is momentum conservation what value did you did you plug in as your answer to the homework? You said the homework wants 2 sig figs. Did you do the rounding correctly?

3. Mar 28, 2014

### dauto

Also, 42 g = 0.042 kg.

4. Mar 28, 2014

### jdawg

Wouldn't you divide 4.2 by 1000 and get 0.0042 kg?
And for my answer into the homework I tried putting in 896.

5. Mar 28, 2014

### dauto

Sorry. I though the problem said 42g. It actually says 4.2g. Try using 900 for your answer

6. Mar 28, 2014

### jdawg

Thank you so much, it worked!

7. Mar 28, 2014

### swordthrower

Instead of telling us what equation you used, frame it in terms of the physics concepts that you are using to solve the problem. For example, explain how you will apply conservation of momentum and conservation of energy.

8. Mar 28, 2014

### jdawg

I don't understand it well enough to explain what I did :( I don't really know what is going on in this problem.

9. Mar 28, 2014

### dauto

That's a problem. The point of doing all that work is so that you will actually know what is going on in the problem

10. Mar 28, 2014

### jdawg

I know, I'm really lost in this chapter. Could you explain it to me?

11. Mar 28, 2014

### dauto

This problem is a two part problem.

1st part: THE COLLISION

During the collision momentum is conserved (no external forces) but energy isn't (A large fraction of the bullets energy is converted to heat. So you get momentum before collision equal momentum after collision. That provides the equation you used to find the bullet's initial speed. But before doing that you had to find the speed after the collision which brings as to part 2.

2nd part: THE SWING

During the swing momentum is not conserved because the tension in the string holding the block provides an external force that affects the momentum. But that force does no work (force perpendicular to trajectory) so energy is conserved during the swing. Kinetic energy at the bottom of the swing (immediately after the collision) is converted to potential energy at the top of the swing (where h is measured). That provides the equation you used to find the speed after the collision.

12. Mar 28, 2014

### jdawg

Ohhh, thanks! That cleared a lot of things up :)