Q5.5 Making use of the centre of mass II (S)
A cannon is fixed at one end of an (initially stationary) enclosed railway truck, 20 m long. The combined mass of the cannon and the truck is 1000 kg. A cannonball of mass 10 kg is placed next to the cannon at the end of the truck. |
- The cannonball is placed in the cannon and fired at 20 ms-1. What is the speed of the truck just after the cannonball is fired?
- The ball comes to rest within the truck. What is the maximum distance that the truck can have moved? (Note: you can solve this problem using relative velocities. That’s not the aim here - we want you to (as the title says) make use of the centre of mass)
Hint
Reveal

Hint
Two things to think about here:- Is momentum conserved in the truck/cannon/cannonball system?
- What external force acts on the center of mass of the system? If zero, then how far does the center of mass go?
Solution
Reveal

Solution
The main assumption we shall make is that the truck/cannon/cannonball is a isolated system, (this means we neglect any friction between the rails and wheels). Therefore we will assume that momentum is conserved for this system.
All the action takes place along the direction of the cannonball trajectory, so the problem is one dimensional.
Since momentum is conserved, we equate momentum before and after the cannon is fired. Denote the cannon + truck system’s mass and velocity by M and V, the cannonball’s by m and v.
Momentum conservation then gives:
0=MV+mvTherefore the magnitude of the truck’s speed is (immediately after firing) V=-mv/M = 0.2 ms-1. The direction is, of course opposite to the direction of the cannonball.
When the cannonball comes to rest, (after many bany bounces off the walls, etc), the overall momentum of the system must be zero, and the center of mass must be unchanged, since the total external force acting on the system is zero throughout (Key Point 4.3). To make the truck move a maximum distance, the cannonball must end up as far as possible away from its start point.
First, we need a good diagram.
Let’s choose the coordinate system and its origin.
- In the frame of the track, the center of mass of the whole system will be fixed. Therefore we should use the track as our coordinate system.
- We shall choose the initial position of the cannonball as the origin of the coordinate system.
Next, let’s calculate the center of mass of the whole system before the cannonball is fired. We will denote this point B with coordinate xB in the track coordinate system. This is the point that will remain fixed during and after the firing. The center of mass of the cannon + truck system, mass M, is at point A in the truck coordinate system, and initially at point xA1 in the track system.
We can then write down an expression for xB
(M+m)xB=M×xA1+m×0After firing, the truck will move a distance d, and the cannonball will be at the other end of the truck. We can then calculate the center of mass, which will still be at xB. However, point A will have moved with respect to the track, and is now at xA2, and the cannonball is now at l-d with respect to the origin, where l is the length of the truck. Hence:
(M+m)xB=MxA2+m(l-d)Now we can solve for d:
MxA1=MxA2+m(l-d)⇒M(xA1-xA2)=m(l-d).Since xA1-xA2=d, then