Question

A projectile of mass $1.2kg$ undergoes a perfectly inelastic collision with a trolley of mass $3.6kg$ as shown in the figure. At the time of the collision, the speed of the projectile is $5m{s}^{-1}$ 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

Answer: 12600

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