Compute how much kinetic energy was "lost" in this inelastic collision.

Question

A 9,100-kg truck runs into the rear of a 1,000-kg car that was stationary. The truck and car are locked together after the collision and move with speed 8 m/s. Compute how much kinetic energy was "lost" in this inelastic collision.

Answer 1

I am assuming no friction.

Let

M = 9100 kg be the mass of the truck,

m = 1000 kg be the mass of the car,

V be the initial velocity of the truck (unknown),

v = 8 m/s be the velocity of the two after the collision.

The kinetic energy of the pair before the collision is

E = M V² / 2

The kinetic energy of the pair after the collision is

e = (M+m)v² / 2

The amount "lost" is the difference E - e

We calculate V from conservation of momentum.

The momentum of the pair before the collision is

P = MV

The momentum of the pair after the collision is

p = (M+m)v

By conservation of momentum, P = p, therefore

MV = (M+m)v

Therefore

V = v(M+m)/M

Substituting V into the "lost" energy we get

E - e =

M v² (M+m)² / (2 M²⁾ - (M+m) v² / 2 =

(M+m) v² ((M+m)/M - 1) /2 =

(M+m) v² m / (2 M)

Note: E - e = e(m/M)

Substituting actual numbers:

E - e = 10100 kg x 64 m²/s² x 1000 kg / 18200 kg = 35516 J