- AThe total momentum of the $‘A$ plus $B’$ system is $p$ before and after the impact, and $(p-J)$ during the impact.
- BDuring the impact $A$ gives impulse $J$ to $ B$
- CThe coefficient of restitution is $\frac{{2J}}{p} - 1$
- ✓Both $(B)$ and $(C)$
Let $v_{1}, v_{2}=$ speeds of $A$ and $B$ after impact.
$u=v_{1}+v_{2}$ and $v_{1}-v_{2}=-e u$
$u=v_{1}+v_{2}$ and $v_{1}-v_{2}=-e u$
$\therefore v_{1}=\frac{1}{2} u(1-e)$ and $v_{2}=\frac{1}{2} u(1+e)$
$J=m v_{2}=m\left[\frac{1}{2} u(1+e)\right]=\frac{1}{2} p(1+e)$
$\Rightarrow e=\frac{2 J}{p}-1$
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$\left( g =980 \,cm / s ^{2}\right)$
$(i){\vec V_C} - {\vec V_A} = 2\left( {{{\vec V}_B} - {{\vec V}_C}} \right)$
$(ii){\vec V_C} - {\vec V_B} = {\vec V_B} - {\vec V_A}$
$(iii)\left| {{{\vec V}_C} - {{\vec V}_A}} \right| = 2\left| {{{\vec V}_B} - {{\vec V}_C}} \right|$
$(iv)\left| {{{\vec V}_C} - {{\vec V}_A}} \right| = 4\left| {{{\vec V}_B}} \right|$
$A$: standing on the horizontal surface
$B$: standing on the block According to the observer $A$
