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(D) Let $u$ be the speed of $A$ before impact. Thus, $p = mu$.Let $v_1$ and $v_2$ be the velocities of $A$ and $B$ after impact respectively.By conser

Vedclass · Accepted Answer

Both $(B)$ and $(C)$.

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(D) Let $u$ be the speed of $A$ before impact. Thus, $p = mu$.Let $v_1$ and $v_2$ be the velocities of $A$ and $B$ after impact respectively.By conservation of momentum: $mu = mv_1 + mv_2 \Rightarrow u = v_1 + v_2$.The coefficient of restitution $e$ is defined as $e = \frac{v_2 - v_1}{u - 0} \Rightarrow v_2 - v_1 = eu$.Solving these equations: $v_2 = \frac{u(1+e)}{2}$ and $v_1 = \frac{u(1-e)}{2}$.The impulse $J$ given by $B$ to $A$ is the change in momentum of $A$: $J = m(v_1 - u) = m(\frac{u(1-e)}{2} - u) = m(\frac{u - ue - 2u}{2}) = -\frac{mu(1+e)}{2} = -\frac{p(1+e)}{2}$.Taking the magnitude of impulse: $J = \frac{p(1+e)}{2}$.Rearranging for $e$: $2J = p + pe \Rightarrow pe = 2J - p \Rightarrow e = \frac{2J}{p} - 1$.By Newton's third law, if $B$ gives impulse $J$ to $A$, $A$ gives impulse $J$ to $B$ in the opposite direction. Thus, both $(B)$ and $(C)$ are correct.

In a one-dimensional collision between two identical particles $A$ and $B$, $B$ is stationary and $A$ has momentum $p$ before impact. During impact, $B$ gives impulse $J$ to $A$.

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