Question

A block of mass m moving horizontally at a speed v collides with a stationary block of mass 2m lying on a smooth horizontal surface. The lighter mass comes to rest after collision. Loss in kinetic energy as a result of collision is

Answer 1

Loss of kinetic energy = 1/3(mu²)

Explanation:

There will be no loss of kinetic if the collision is elastic. But from the given information, we assume that the collision is inelastic. In this case, some energy will be lost as heat, but momentum will be conserved.

Before the collision, a mass m has velocity u. It hits another object of mass 2m which is at rest. So after the collision, the combined bodies will have mass of 3m and velocity v.

Since momentum is conserved, we get:

mu = 3mv

So v = u/3 and momentum is conserved.

The change in KE is:

1/2 mu²  - 1/2*3m*v²  

  = mu²/2 - (3/2)*m *(u/3)²

  = mu²/2 - mu²/6

  = (3mu² - mu²) / 6

  = 2mu²/6

  = mu²/3

Loss of kinetic energy = 1/3(mu²)