Question

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 speed of the projectile is $5\,ms^{- 1}$ at an angle of $37^\circ $ with the horizontal. The trolley is free to move, only along a horizontal rail that coincides with the direction of the projectile's horizontal motion. Assuming that the trolley doesn't topple, calculate the amount of heat energy (in $mJ$ ) released in the collision. [Take $sin37^\circ =\frac{3}{5}$ and $cos37^\circ =\frac{4}{5}$ ]

Answer 1

Using the conservation of linear momentum in the horizontal direction we get,
$1.2\times 4=4.8\times v$
$\Rightarrow v=1ms^{- 1}$
The loss in kinetic energy is equal to the amount of heat energy released.
$\left|\Delta K\right|=\frac{1}{2}\times 1.2\times \left(5\right)^{2}-\frac{1}{2}\times 4.8\times \left(1\right)^{2}$
$\left|\Delta K\right|=12.6J=12600\,mJ$