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 $5 \, m \, s^{- 1}$ at an angle of $37^\circ $ 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 $sin37^\circ =3/5$ and $cos37^\circ =4/5$ ]

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.

$\left|\Delta K\right|=\frac{1}{2}\times 1.2\times \left(5\right)^{2}-\frac{1}{2}\times 4.8\times \left(1\right)^{2}$
$\left|\Delta K\right|=12.6J$