Question

The diagram above shows a ballistic pendulum. A 10 g bullet is fired into the suspended 2 kg block of wood and
remains embedded inside it (a perfectly inelastic collision). After the impact of the bullet, the block swings up to a
maximum height h. If the initial speed of the bullet was 35 m/s:

A.) What was the momentum of the bullet before the collision?

B.) What was the kinetic energy of the bullet before the collision?

C.) What was the velocity of the bullet-block system just after the collision?

D.) What was the total kinetic energy of the bullet-block system after the collision?

E.) What is the maximum possible potential energy of the bullet-block system when it reaches its maximum height?

F.) What is the maximum possible height of the bullet-block system?

Answer 1

The kinetic energy of the bullet-block system is equal to the potential energy of the bullet-block system when it reaches its maximum height.

Conservation  of Linear momentum

According to the principle of conservation of linear momentum, momentum before collsion is equal to momentum after collision. Let us now answer the questions individually.

1) The momentum of the bullet before collsion = (0.001 Kg * 35 m/s) = 0.35 Kgms-1

2) The kinetic energy of the bullet before collision = 0.5 * 0.001 * (35 m/s)^2 = 0.6125 J

3) Velocity after collsion is obtained from;

(0.001 Kg * 35 m/s) + (2 kg * 0 m/s) = (0.001 Kg + 2 kg) v

v = 0.35/2.001 =

v=0.1749 m/s

4) Total kinetic energy after collison = 0.5 * (0.001 Kg + 2 kg) * (0.1749 m/s)^2

= 0.031 J

5) The maximum possible potential energy of the bullet-block system when it reaches its maximum height = Total kinetic energy after collison = 0.031 J

6) The  maximum possible height of the bullet-block system is obtained from;

PE = mgh

h = 0.031 J/(0.001 Kg + 2 kg) * 9.8 ms-2

h = 0.0015 m or 0.15 cm

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