# How is a ballistic pendulum affected by two objects with different masses?

Imagine that bullet A (made of lead) has twice the mass of bullet B

It is shot at the same muzzle velocity as bullet B onto a ballistic pendulum; all other initial variables are the same: size and mass of brass ring, size of bullet, length of the pendulum rod, etc.

What would be different about the final state of the apparatus, i.e. the angle of the pendulum?

Can someone explain how two objects with different masses can have the same initial velocity?

• Can you please explain why two objects with different masses cannot have identical velocities? Or do you think that "initial velocities" are somehow special? – WhatRoughBeast Jul 6 '15 at 22:23

Can someone explain how two objects with different masses can have the same initial velocity?

Since the kinetic energy $E_{kin}=m\cdot v^2/2$ you need more energy = more gunpowder to get the same velocity for a heavier bullet.

What would be different about the final state of the apparatus, i.e. the angle of the pendulum?

The pendulum would swing higher, since the higher kinetic energy provided by the heavier bullet would be transformed into a higher potenial energy when the pendulum gets kicked up.

In the ballistic pendulum, it is the momentum that matters (that is conserved).

For a pendulum with mass $M$, being hit by a bullet with mass $m$ and velocity $v$, the velocity $v_1$ immediately after the impact is (from conservation of momentum):

$$v_1 = \frac{m}{m+M}v$$

And when you have an object with twice the mass but the same velocity, you get the pendulum velocity $v_2$ after impact:

$$v_2 = \frac{2m}{2m+M}v$$

This velocity allows you to calculate the energy after the collision (always less than the initial energy - because it's an inelastic collision).

$$E = \frac12 (M+m) v_1^2$$

The pendulum will stop deflecting when all the kinetic energy has been turned into potential energy:

$$(M+m) g \ell (1-\cos\theta) = \frac12 (M+m) v_1^2$$

A little rearranging will show you that the heavier bullet will cause a greater deflection of the pendulum. Exactly how much greater will depend on the mass of the pendulum itself.

$$1 - \cos\theta = \frac{(M+m)v_1^2}{2(M+m)g\ell} = \frac{\left(\frac{m}{m+M}v\right)^2}{2g\ell}$$

Now assuming that the deflection $\theta$ is small so we can approximate $\cos\theta = 1-\frac12 \theta^2$, then

$$\theta = \frac{1}{\sqrt{2g\ell}}\frac{m}{M+m}v$$

As you can see, for $m << M$ the deflection angle will be proportional to mass (and impact velocity). As the mass of the bullet gets larger, the deflection will be slightly less than proportional to $m$.