The case of two vehicles crashing at an intersection is an old favourite, much loved by textbook writers and examiners. So, here goes! Note that we have to assume that the two vehicles constitute an isolated system, in other words, we ignore the frictional forces between the vehicles and the road. In real life, this would not be the case.
| A truck of mass 10 000 kg travelling due east at 15 m.s-1 collides with a car of mass 1 500 kg travelling north at 25 m.s-1. The wrecked vehicles tangle together. In what direction do the wrecks move, and with what speed? | |
Since the vehicles stick together after the crash, we are dealing with a perfectly inelastic collision.
Total initial momentum in the easterly direction: peast = ptruck,east + pcar, east = 150 000 + 0 = 150 000 kg.m.s-1.
Total initial momentum in the northerly direction: pnorth = ptruck,north + pcar, north = 0 + 37 500 = 37 500 kg.m.s-1.
Since the momentums in each direction are conserved, we can write
Final momentum of combined wrecks in easterly direction pwrecks, east = 150 000 kg.m.s-1, and,
Final momentum of combined wrecks in northerly direction pwrecks, north = 37 500 kg.m.s-1.
If θ is the angle of motion from the horizontal after the crash, tanθ = pwrecks, north/pwrecks, east = 37 500/150 000 = 0.2500, hence θ = 14°, that is, 14° north of east.
The magnitude of the resultant momentum vector is |pwrecks|= (pwrecks, north² + pwrecks, east²)½ = (37 500² + 150 000²)½ = 150 000 kg.m.s-1. We need to divide this by the mass of the combined wrecks, 11 500 kg to give the speed: speed = 150 000/11 500 = 13 m.s-1.
