Two identical balls A and B of equal mass m a | vedclass

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(D) Let the mass of each ball be $m$. Let $v_A = 16 \,ms^{-1}$ be the initial velocity of ball $A$ and $v_B = 0$ be the initial velocity of ball $B$.A

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$\frac{1}{4}$

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(D) Let the mass of each ball be $m$. Let $v_A = 16 \,ms^{-1}$ be the initial velocity of ball $A$ and $v_B = 0$ be the initial velocity of ball $B$.After collision,let the velocities of $A$ and $B$ be $v_1$ and $v_2$ respectively.By conservation of linear momentum: $m v_A + 0 = m v_1 + m v_2 \implies v_1 + v_2 = 16$.By the definition of coefficient of restitution $e$: $e = \frac{v_2 - v_1}{v_A - 0} \implies v_2 - v_1 = 16e$.Adding the two equations: $2v_2 = 16(1+e) \implies v_2 = 8(1+e)$.For ball $B$ to just reach the top of the inclined plane of height $h = 5 \,m$,its kinetic energy at the bottom must equal its potential energy at the top: $\frac{1}{2} m v_2^2 = mgh$.$v_2^2 = 2gh = 2 	imes 10 	imes 5 = 100 \implies v_2 = 10 \,ms^{-1}$.Substituting $v_2$ in the expression: $8(1+e) = 10 \implies 1+e = \frac{10}{8} = 1.25 \implies e = 0.25 = \frac{1}{4}$.

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