A projectile of mass 1.2 kg undergoes a perfectly inelastic collision with a trolley of mass 3.6 kg as shown in the figure. At the time of the collision, the second of the projectile is `5ms^(-1)` at an angle of `37^(@)` with the horizontal. The trolley is free to move, only along a horizontal rail which coincides with the direction of the projectile's horizontal motion. Assuming that the trolley doesn't topple, calculate the amount of heat energy (in J) released in the collision.
`["Take "sin 37^(@)=3//5 and cos 37^(@)=4//5]`

Answer

`12.6`

Explanation

Using the conservation of linear momentum in the horizontal direction we get
`1.2xx4=4.8xxv`
`impliesv=1ms^(-1)`
The loss in kinetic energy is equal to the amount of heat energy released.
`|DeltaK|=1/2xx1.2xx(5)^(2)-1/2xx4.8`
`|DeltaK|=12.6J`