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(B) Initial momentum of the system in $x$-direction: $p_x = m(2v) = 2mv$.Initial momentum of the system in $y$-direction: $p_y = (2m)v = 2mv$.Total in

Vedclass · Accepted Answer

$56$

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(B) Initial momentum of the system in $x$-direction: $p_x = m(2v) = 2mv$.Initial momentum of the system in $y$-direction: $p_y = (2m)v = 2mv$.Total initial kinetic energy: $K_i = \frac{1}{2}m(2v)^2 + \frac{1}{2}(2m)v^2 = 2mv^2 + mv^2 = 3mv^2$.Since the collision is perfectly inelastic,the particles stick together. Let the final velocity be $V_f$ at an angle $	heta$.By conservation of linear momentum: $p_f = \sqrt{p_x^2 + p_y^2} = \sqrt{(2mv)^2 + (2mv)^2} = 2\sqrt{2}mv$.Also,$p_f = (m + 2m)V_f = 3mV_f$.Equating the two: $3mV_f = 2\sqrt{2}mv \Rightarrow V_f = \frac{2\sqrt{2}}{3}v$.Final kinetic energy: $K_f = \frac{1}{2}(3m)V_f^2 = \frac{3m}{2} \left( \frac{8v^2}{9} ight) = \frac{4}{3}mv^2$.Loss in energy: $\Delta K = K_i - K_f = 3mv^2 - \frac{4}{3}mv^2 = \frac{5}{3}mv^2$.Percentage loss: $\frac{\Delta K}{K_i} 	imes 100 = \frac{(5/3)mv^2}{3mv^2} 	imes 100 = \frac{5}{9} 	imes 100 \approx 55.55\% \approx 56\%$.

$A$ 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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