# Ballistic Pendulum kinetic energy

## Homework Statement

A 11.0 g rifle bullet is fired with a speed of 360 m/s into a ballistic pendulum with mass 9.00 kg, suspended from a cord 70.0 cm long.

A.) Compute the initial kinetic energy of the bullet

B.) Compute the kinetic energy of the bullet and pendulum immediately after the bullet becomes embedded in the pendulum.

C.) Compute the vertical height through which the pendulum rises.

## The Attempt at a Solution

A.) I just applied 1/2mv^2 to get the initial KE of the bullet to be 713 J (it's correct)

B.) Since the kinetic and potential energy of the pendulum right when the bullet fires is zero, I claim that:

$$713 = \frac{1m_{b}v^{2}_{b}}{2} + \frac{1m_{p}v^{2}_{p}}{2}$$
$$713 = \frac{(.011)v^{2}_{b}}{2} + \frac{(9)v^{2}_{p}}{2}$$
Now I am assuming that the velocity of the bullet and pendulum are the same since they are now attached so:
$$713 = \frac{1}{2}(.011+9)v^{2}$$
$$1426 = (.011+9)v^{2}$$
$$1426 = (9.011)v^{2}$$
$$\frac{1426}{9.011} = v^{2}$$
$$\sqrt{\frac{1426}{9.011}} = v$$
$$v = 12.6 m/s$$

??

C.)
$$\frac{1}{2}(9.011)(12.6)^{2} = (9.011)(9.8)h$$
$$h = 8.1m$$

There is no way this is right, the shaft of the pendulum is only 70cm long!

For part B, KE is not conserved. So you need to use another law of conservation.

And with v, you know how to get KE.

Would I use the conservation of momentum?

$$(0.011)(360) = (9.011)v$$
$$v = 0.439 m/s$$

(Which naturally changed my solution to C to 0.983 cm or 0.00983 m)

Would I use the conservation of momentum?

$$(0.011)(360) = (9.011)v$$
$$v = 0.439 m/s$$

(Which naturally changed my solution to C)

Yes that is what I got and note B is not asking for v.

Yeah, now I just pop that velocity (the bullet with pendulum) into the equation of kinetic energy right?

$$KE = \frac{1}{2}mv^{2}$$

Yeah, now I just pop that velocity (the bullet with pendulum) into the equation of kinetic energy right?

$$KE = \frac{1}{2}mv^{2}$$

Yup. And for C you did as you did with the new KE as you said.

The KE isn't conserved because the bullet used some energy to push itself into the box and in reality some went to friction (heat).

Yup. And for C you did as you did with the new KE as you said.

The KE isn't conserved because the bullet used some energy to push itself into the box and in reality some went to friction (heat).

You predicted my next question!

Suppose, rather than entering the box, the bullet struck it, stuck to it via some magnetic force, or something, and that we could neglect friction, heat, air resistance and all of that.

In that case, would energy be conserved?

You predicted my next question!

Suppose, rather than entering the box, the bullet struck it, stuck to it via some magnetic force, or something, and that we could neglect friction, heat, air resistance and all of that.

In that case, would energy be conserved?

Well I think the bullet would actually have to stop and fall and the box moves forward completely on its own for KE to be conserved..
Because I think also KE isn't conserved because its inelastic and the bullet travels with the box... I'm a little shaky on this so you should confirm it with someone who knows for 100% sure.

But I'm rather sure the bullet would have to stop and not travel with the box.

gneill
Mentor
You predicted my next question!

Suppose, rather than entering the box, the bullet struck it, stuck to it via some magnetic force, or something, and that we could neglect friction, heat, air resistance and all of that.

In that case, would energy be conserved?