6 CONSERVATION OF MOMENTUM 6.7 Collisions in 2 - dimensions
M
mL
vtwo equations [i.e., Eqs. (6.64) and (6.65)] and two unknowns (i.e., vf and θf).
Clearly, we should be able to find a unique solution for such a system.
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 kg. The cannon fires a cannonball, of mass m =
1.2 kg, horizontally with velocity v = 115 m/s. The cannonball travels the length
of the carriage, a distance L = 85 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 = −m
v = −1.2 × (^115) = −0.115 m/s.
M 1200
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 |u| = 0.115 m/s.