Answer to Question #104290 in Physics for Samantha Woody

Question

A 2,700-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 3 m/s. Compute how much kinetic energy was "lost" in this inelastic collision.

Answer 1

From the conservation of momentum:

$$Mv=(m+M)u\to 2700v=(1000+2700)3$$

$$v=4.11\frac{m}{s}$$

Kinetic energy was "lost" in this inelastic collision:

$$\Delta K=0.5Mv^2-0.5(m+M)u^2$$

$$\Delta K=0.5(2700)(4.11)^2-0.5(1000+2700)3^2=\\=6200\ J=6.2\ kJ$$