A ballistic pendulum is a device used to measure the speed of a bullet. A bullet of mass m is fired at a block of wood (mass M) hanging from a string. The bullet embeds itself in the block, and causes the combined block plus bullet system to swing up a height h. What is v_o, the speed of the bullet before it hits the block?

Is it correct to set the bullet's initial kinetic energy equal to the final gravitational potential energy of the block plus bullet?

1. Yes
2. No

No. A completely inelastic collision takes place, resulting in a loss of mechanical energy. The initial kinetic energy is larger than the final gravitational potential energy - we have to analyze the collision to connect the initial and final states of the system.

Work backwards, starting with the swing of the pendulum just after the collision until it reaches its maximum height. Conservation of mechanical energy applies here, so:

½ (m+M)v_f² = (m+M)gh.

So, the speed of the system immediately after the collision is:

v_f = (2gh)^½

Apply conservation of momentum to determine the relationship between v_f and the bullet's speed before the collision. The wood block is stationary before the collision.

Momentum before the collision = momentum after the collision

mv_o = (m+M)v_f

Therefore

vo

=

M + m m

(2gh)½

Take the case where m = 30 g, M = 870 g, h = 0.74 m. This gives:

v_o = 114 m/s.

Let's figure out the ratio of the mechanical energy after the collision to the mechanical energy before the collision. This is equal to the ratio of the kinetic energy of the block + bullet immediately after the collision to the kinetic energy of the bullet before the collision.

Ef Ei

=

½ (m+M)vf2 ½ mvo2

=

(m+M)vf * vf mvo * vo

From conservation of momentum we know that (m+M)v_f = mv_o, so:

Ef Ei

=

vf vo

=

m m + M

In our numerical example we had m = 30 g and M = 870 g, so E_i/E_f = 0.033.
3.3% of the kinetic energy remains - in other words, 96.7% was transformed into other forms of energy (mostly thermal energy).