Question

A 500 kg automobile travelling at 30 m/s strikes a 100kg stationary pedestrian. The impact duration is 0.1 s and it slows the automobile by 1m/s. Calculate:
a. the velocity imparted to the pedestrian
b. the force exerted by the vehicle on the pedestrian.(Ans=5000N)

Answer 1

initial momentum = 500 * 30 + 100 * 0 =15,000 kg m/s
final momentum = 500 * 29 + 100 * u = 14,500 + 100 u
but they are the same, no momentum was removed from the system.
so
15,000 = 14,500 + 100 u
u = 5 m/s
It did not hit the pedestrian very hard. In fact they would be more likely to stick together.
force = rate of change of momentum
= 100 *5 / .1 = 5000 kg m/s^2 = 5,000 Newtons

Answer 2

Given:
M1 = 500 kg, V1 = 30 m/s.
M2 = 100 kg, V2 = 0.
V3 = 30-1 = 29 m/s = Velocity of car after collision.
V4 = Velocity of person after collision.

a. Momentum before = momentum after.
M1*V1 + M2*V2 = M1*V3 + M2*V4.
500*30 + 100*0 = 500*29 + 100*V4,
V4 = 5 m/s.

b. V 4 = V2 + a*t .
5 = 0 + 0.1a,
a = 50 m/s^2.

F = M*a = 100 * 50 = 5,000 N.