Blocks of masses m, m, 2m, 4m and 8m are arra | vedclass

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(D) Since all collisions are perfectly inelastic,after the final collision,all blocks will move together as a single system with a common velocity $v^

Vedclass · Accepted Answer

$94$

Vedclass · Answer

(D) Since all collisions are perfectly inelastic,after the final collision,all blocks will move together as a single system with a common velocity $v^{\prime}$.Applying the principle of conservation of linear momentum for the entire system:Initial momentum = Final momentum$mv = (m + m + 2m + 4m + 8m)v^{\prime}$$mv = 16mv^{\prime}$$v^{\prime} = \frac{v}{16}$Initial kinetic energy of the system:$E_{i} = \frac{1}{2}mv^{2}$Final kinetic energy of the system:$E_{f} = \frac{1}{2}(16m)(v^{\prime})^{2} = \frac{1}{2}(16m)\left(\frac{v}{16}ight)^{2} = \frac{1}{2}m\frac{v^{2}}{16}$Energy loss $\Delta E = E_{i} - E_{f} = \frac{1}{2}mv^{2} - \frac{1}{2}m\frac{v^{2}}{16} = \frac{1}{2}mv^{2}\left(1 - \frac{1}{16}ight) = \frac{1}{2}mv^{2}\left(\frac{15}{16}ight)$Percentage energy loss $p = \frac{\Delta E}{E_{i}} 	imes 100$$p = \frac{\frac{1}{2}mv^{2}(\frac{15}{16})}{\frac{1}{2}mv^{2}} 	imes 100 = \frac{15}{16} 	imes 100 = 93.75\%$Rounding to the nearest integer,the value of $p$ is close to $94$.

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