Center of Mass, Impulse, Momentum, and Collisions

Question

You are driving a truck of mass 5 650 kg at 13 m/s. You accidentally hit a parked car. The car and the passengers have a total mass of 984 kg. (a) What is the speed of the car and the truck after the impact assuming that they are locked together? (b) If the impact lasts for 0.5 s, what force is experienced by an 80 kg passenger? Use GUESS METHOD

Answer 1

Let

M = 5650 kg be the mass of the truck,

V = 13 m/s be the velocity of the truck before the colision,

m = 984 kg be the mass of the car,

v (to be calculated) be the velocity of both after the collision,

t = 0.5 s be the duration of the collision.

Before the collision the truck has momentum MV, and the car has momentum 0.

After the collision the car and truck together have momentum (m+M)v.

By conservation of momentum

(m+M)v = MV

Hence

v = MV/(m+M)

Substituting actual numbers

v = 5650×13/(5650+984) ≈ 11 m/s

Assuming (guessing) that the passenger experiences constant acceleration,

we can calculate the average acceleration as

a = (v-0)/t = 11/0.5 = 22 m/s²

By Newton's second law the force is

F = ma = 80×22 = 176 N