Question

Two balls, having linear momenta $p _{1}=p \hat{ i }$ and $p _{2}=-p \hat{ i }$, undergo a collision in free space. There is no external force acting on the balls. Let $p _{1}^{\prime}$ and $p _{2}^{\prime}$ be their final momenta. The following option figure (s) is (are) not allowed for any non-zero value of $p, a_{1}, a_{2}, b_{1}, b_{2}, c_{1}$ and $c_{2}$.

• A. $p _{1}^{\prime}=a_{1} \hat{ i }+b_{1} \hat{ j }+c_{1} \hat{ k }, p _{2}^{\prime}=a_{2} \hat{ i }+b_{2} \hat{ j }$
• B. $p _{1}^{\prime}=c_{1} \hat{ k }, p _{2}^{\prime}=c_{2} \hat{ k }$
• C. $p _{1}^{\prime}=a_{1} \hat{ i }+b_{1} \hat{ j }+c_{1} \hat{ k }, p _{2}^{\prime}=a_{2} \hat{ i }+b_{2} \hat{ j }-c_{1} \hat{ k }$
• D. $p _{1}^{\prime}=a_{1} \hat{ i }+b_{1} \hat{ j }, p _{2}^{\prime}=a_{2} \hat{ i }+b_{1} \hat{ j }$

Answer 1

Answer: (A), (D)

Initial momentum of the system $p _{1}+ p _{2}=0$
$\therefore$ Final momentum $p _{1}^{\prime}+ p _{2}^{\prime}$ should also be zero.
Option (b) is allowed because if we put $c_{1}=-c_{2} \neq 0$,
$p _{1}^{\prime}+ p _{2}^{\prime}$ will be zero. Similary, we can check other options.