Suppose a bullet of mass m = 6.82 g is fired into a ballistic pendulum whose bob has a mass of M = 0.650 kg.?

Question

(a) If the bob rises to a height of 0.137 m, what was the initial speed of the bullet?


(b) What was the speed of the bullet-bob combination immediately after the collision takes place?

Answer 1

This is a classic conservation question:

http://hyperphysics.phy-astr.gsu.edu/hbase/balpen.html

Let:
Vi be the initial velocity of the bullet
Vc be the initial velocity of the combination
Mb be the mass of the bullet
Mc be the combined mass

Then the conservation of momentum during the collision gives the equation:

Mb x Vi = Mc x Vc

Once the combined mass starts swinging, its original kinetic energy gets converted to potential energy.

The kinetic energy of a mass M moving at velocity V is
Ek = (1/2) x M x V^2

The potential energy gained by raising a mass M through a height H is given by
Ep = g x M x H where is the gravitational constant (around 9.8 m/s^s on the surface of the Earth)

In this problem, we have M, H, and g so we can compute the potential energy Ep.

Setting Ek = Ep, we can determine Vc^2 and then Vc.

From Vc, Mc, and Mb, we can determine Vi.