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(A) Let the masses of both particles be $m$. The initial kinetic energy is $K_i = \frac{1}{2}mv_0^2$.The final kinetic energy is $K_f = K_i + 0.5K_i =

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$\sqrt{2} v_0$

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(A) Let the masses of both particles be $m$. The initial kinetic energy is $K_i = \frac{1}{2}mv_0^2$.The final kinetic energy is $K_f = K_i + 0.5K_i = 1.5K_i = \frac{3}{2} \left( \frac{1}{2}mv_0^2 \right) = \frac{3}{4}mv_0^2$.Let the final velocities be $v_1$ and $v_2$. By conservation of kinetic energy:$\frac{1}{2}mv_1^2 + \frac{1}{2}mv_2^2 = \frac{3}{4}mv_0^2 \implies v_1^2 + v_2^2 = \frac{3}{2}v_0^2 \quad (i)$By conservation of linear momentum:$mv_0 = mv_1 + mv_2 \implies v_1 + v_2 = v_0 \quad (ii)$Squaring equation $(ii)$:$(v_1 + v_2)^2 = v_0^2 \implies v_1^2 + v_2^2 + 2v_1v_2 = v_0^2$Substituting $(i)$ into this:$\frac{3}{2}v_0^2 + 2v_1v_2 = v_0^2 \implies 2v_1v_2 = v_0^2 - \frac{3}{2}v_0^2 = -\frac{1}{2}v_0^2$The square of the relative velocity is:$(v_1 - v_2)^2 = v_1^2 + v_2^2 - 2v_1v_2 = \frac{3}{2}v_0^2 - \left( -\frac{1}{2}v_0^2 \right) = \frac{3}{2}v_0^2 + \frac{1}{2}v_0^2 = 2v_0^2$Therefore,the magnitude of the relative velocity is $|v_1 - v_2| = \sqrt{2}v_0$.

In a collinear collision,a particle with an initial speed $v_0$ strikes a stationary particle of the same mass. If the final total kinetic energy is $50\%$ greater than the original kinetic energy,the magnitude of the relative velocity between the two particles,after collision,is:

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