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(A) Initial kinetic energy of the system:$K_i = \frac{1}{2} m (2v)^2 + \frac{1}{2} (2m) v^2 = 2mv^2 + mv^2 = 3mv^2$Initial momentum of the system:$\ve

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$56$

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(A) Initial kinetic energy of the system:$K_i = \frac{1}{2} m (2v)^2 + \frac{1}{2} (2m) v^2 = 2mv^2 + mv^2 = 3mv^2$Initial momentum of the system:$\vec{p}_i = m(2v)\hat{i} + (2m)v\hat{j} = 2mv\hat{i} + 2mv\hat{j}$Since the collision is perfectly inelastic,the particles stick together and move with a common velocity $\vec{V}$.By the law of conservation of linear momentum:$\vec{p}_f = \vec{p}_i$$(m + 2m)\vec{V} = 2mv\hat{i} + 2mv\hat{j}$$3m\vec{V} = 2mv(\hat{i} + \hat{j})$$\vec{V} = \frac{2v}{3}(\hat{i} + \hat{j})$Final kinetic energy of the system:$K_f = \frac{1}{2} (m + 2m) |\vec{V}|^2 = \frac{1}{2} (3m) \left( \sqrt{(\frac{2v}{3})^2 + (\frac{2v}{3})^2} ight)^2$$K_f = \frac{3m}{2} \left( \frac{4v^2}{9} + \frac{4v^2}{9} ight) = \frac{3m}{2} \left( \frac{8v^2}{9} ight) = \frac{4}{3} mv^2$Loss in energy:$\Delta K = K_i - K_f = 3mv^2 - \frac{4}{3} mv^2 = \frac{5}{3} mv^2$Percentage loss in energy:$	ext{Percentage Loss} = \frac{\Delta K}{K_i} 	imes 100 = \frac{\frac{5}{3} mv^2}{3mv^2} 	imes 100 = \frac{5}{9} 	imes 100 \approx 55.55\% \approx 56\%$

$A$ particle of mass $m$ 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 $..........\%$

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