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(C) For a perfectly inelastic collision,the coefficient of restitution $e = 0$. Using the law of conservation of linear momentum: $M\mu_1 + M\mu_2 = (

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$\frac{M}{4}(\mu_1 - \mu_2)^2$

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(C) For a perfectly inelastic collision,the coefficient of restitution $e = 0$. Using the law of conservation of linear momentum: $M\mu_1 + M\mu_2 = (M + M)V$ $M(\mu_1 + \mu_2) = 2MV$ $V = \frac{\mu_1 + \mu_2}{2}$ Initial kinetic energy $(K_i)$: $K_i = \frac{1}{2}M\mu_1^2 + \frac{1}{2}M\mu_2^2 = \frac{M}{2}(\mu_1^2 + \mu_2^2)$ Final kinetic energy $(K_f)$: $K_f = \frac{1}{2}(2M)V^2 = M \left( \frac{\mu_1 + \mu_2}{2} \right)^2 = \frac{M}{4}(\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2)$ Loss in energy $(\Delta K = K_i - K_f)$: $\Delta K = \frac{M}{2}(\mu_1^2 + \mu_2^2) - \frac{M}{4}(\mu_1^2 + \mu_2^2 + 2\mu_1\mu_2)$ $\Delta K = \frac{M}{4} [2\mu_1^2 + 2\mu_2^2 - \mu_1^2 - \mu_2^2 - 2\mu_1\mu_2]$ $\Delta K = \frac{M}{4}(\mu_1^2 + \mu_2^2 - 2\mu_1\mu_2)$ $\Delta K = \frac{M}{4}(\mu_1 - \mu_2)^2$

Two identical masses $M$ moving with velocities $\mu_1$ and $\mu_2$ collide perfectly inelastically. The loss in energy is

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