Question 123 of 130

A particle of mass $\text{m}$ moving in the $x$ direction with speed $2v$ is hit by another particle of mass $2m$ moving in the $y$ direction with speed $v.$ If the collision is perfectly inelastic, the percentage loss in the energy during the collision is close to:

A$62$B$44$C$50$D$56$

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Answer: (D) $56$

The initial momentum of system is $\vec{{P}_{i}}=m\left(2V\right)\hat{i}+\left(2m\right)V\hat{j}$

According to question as



On perfectly inelastic collision the particles stick to each other.

$\vec{{P}_{f}}=3m\vec{{V}_{f}}$

By conservation of linear momentum

$\vec{{P}_{f}}=\vec{{P}_{i}}\Rightarrow 3m\vec{{V}_{f}}=m2V\hat{i}+2mV\hat{j}$

$\Rightarrow \vec{{V}_{f}}=\dfrac{2V}{3}\left(\hat{i}+\hat{j}\right)\Rightarrow |{V}_{f}|=\dfrac{2\sqrt{2}}{3}V$

$∴$ loss in KE. of system $={K}_{initial}-{K}_{final}$

$=\dfrac{1}{2}m{\left(2V\right)}^{2}+\dfrac{1}{2}\left(2m\right){V}^{2}-\dfrac{1}{2}\left(3m\right){\left(\dfrac{2\sqrt{2}V⁡}{3}\right)}^{2}$

$=2m{V}^{2}+m{V}^{2}-\dfrac{4}{3}m{V}^{2}=3m{V}^{2}-\dfrac{4m{V}^{2}}{3}=\dfrac{5}{3}m{V}^{2}$

% Loss in KE $=100\times \dfrac{∆K}{{K}_{i}}=\dfrac{\dfrac{5}{3}m{v}^{2}}{3m{V}^{2}}=\dfrac{5}{9}\times 100=56$