Worked example 6.1: Cannon in a railway carriage

Question: A cannon is bolted to the floor of a railway carriage, which is free to move without friction along a straight track. The combined mass of the cannon and the carriage is $M=1200 {\rm kg}$. The cannon fires a cannonball, of mass $m=1.2 {\rm kg}$, horizontally with velocity $v=115 {\rm m/s}$. The cannonball travels the length of the carriage, a distance $L=85 {\rm m}$, and then becomes embedded in the carriage's end wall. What is the recoil speed of the carriage right after the cannon is fired? What is the velocity of the carriage after the cannonball strikes the far wall? What net distance, and in what direction, does the carriage move as a result of the firing of the cannon?

Answer: Conservation of momentum implies that the net horizontal momentum of the system is the same before and after the cannon is fired. The momentum before the cannon is fired is zero, since nothing is initially moving. Hence, we can also set the momentum after the cannon is fired to zero, giving

$$0 = M u + m v,$$

where $u$ is the recoil velocity of the carriage. It follows that

$$u = - \frac{m}{M} v = -\frac{1.2\times 115}{1200} = -0.115 {\rm m/s}.$$

The minus sign indicates that the recoil velocity of the carriage is in the opposite direction to the direction of motion of the cannonball. Hence, the recoil speed of the carriage is $\vert u\vert= 0.115 {\rm m/s}$.

Suppose that, after the cannonball strikes the far wall of the carriage, both the cannonball and the carriage move with common velocity $w$. Conservation of momentum implies that the net horizontal momentum of the system is the same before and after the collision. Hence, we can write

$$M u + m v= (M+m) w.$$

However, we have already seen that $M u + m v=0$. It follows that $w=0$: in other words, the carriage is brought to a complete halt when the cannonball strikes its far wall.

In the frame of reference of the carriage, the cannonball moves with velocity $v-u$ after the cannon is fired. Hence, the time of flight of the cannonball is

$$t = \frac{L}{v-u} = \frac{85}{115+0.115} = 0.738 s.$$

The distance moved by the carriage in this time interval is

$$d = u t = -0.115\times 0.738 = -0.0849 {\rm m}.$$

Thus, the carriage moves $8.49 {\rm cm}$ in the opposite direction to the direction of motion of the cannonball.