# Ballistic Pendulum

1. Feb 16, 2006

### nahya

A rifle bullet of mass m = 2.5 g traveling at vb = 243 m/s collides with and embeds itself in a pendulum of mass M = 237.5 g, initially at rest and suspended vertically with massless strings of length L = 2 m.
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i first converted the masses into kilograms.
i found out that net momentum = 0.6075 kg m/s, for the bullet-pendulum mass of 0.24kb. that means that the pendulum moves at first with v = 2.53125.

now... where do i go from here?
i don't even know the angle at which the bullet-pendulum combination reaches the max height. if i knew, it would be 2*tan(theta).

so i guess the challenge is to find the angle at which the thing goes at its max height...? bleh.. i'm lost

edit:

by the energy equation, i guess...
for M = total mass, 1/2*Mv(f)^2 + Mgy(f) = 1/2*Mv(i)^2 + Mgy(i)
y(i) is zero, so the last term cancels out.
v(f) is zero, so the first term cancels out.
Mgy(f) = 1/2*Mv(i)^2
y(f) = Mv(i)^2 / (2Mg)
i get 0.6538, which is not the answer...

Last edited: Feb 16, 2006
2. Feb 16, 2006

### Andrew Mason

What is the question? Do you want to find the angle? If so, you have worked out the right expression for y(f) and from that you can find the angle(see correction below). Do a drawing.

AM

Last edited: Feb 16, 2006
3. Feb 16, 2006

### nahya

i'm trying to find the maximum height that it reaches.
i keep getting the same answer for y(f), and it is, apparently, incorrect.

4. Feb 16, 2006

### Andrew Mason

ok. Your expression for y(f) is not right. Energy is not conserved. The momentum of the bullet before the impact is equal to the momentum of the block + bullet afterward. The velocity immediately after the impact has to be worked out from the bullet momentum ($mv_b$).

So:

$$v_i = \frac{m}{m+M}v_b$$

and:

$$\frac{1}{2}(M+m)v_i^2 = (M+m)gy_f$$

AM

Last edited: Feb 16, 2006