Question

A body of mass $m$ moving with a velocity $v$ in the x-direction collides with a body of mass $M$ moving with a velocity $V$ in the y-direction. They stick together during the collision. Choose the correct statement/statements.

• A. The magnitude of the momentum of the composite body is $\sqrt{\left(m{v}^2\right)+(MV{)}^2}$
• B. The composite body moves in a direction making a angle $\theta =ta{n}^{-1}\left(\frac{MV}{mv}\right)$ with the x - axis
• C. The loss of kinetic energy as a result of collision is $\frac{1}{2}\frac{Mm}{(M+m)}(V^2+v^2)$
• D. All the above choices are correct.

Answer 1

Answer: (B), (C)

The correct options are
B

The composite body moves in a direction making a angle $\theta =ta{n}^{-1}\left(\frac{MV}{mv}\right)$ with the x - axis

C

The loss of kinetic energy as a result of collision is $\frac{1}{2}\frac{Mm}{(M+m)}(V^2+v^2)$

Here, $p=mv$ and $P=MV$. The resultant of $p$ and $P$ is $P_r=\sqrt{p^2+P^2}=\sqrt{(mv{)}^2+(MV{)}^2}$.
Hence, option (a) is not correct.
The angle which the resultant momentum $P_r$ subtends with the x-axis is given by $tan\theta =\frac{P}{p}=\frac{MV}{mv}$, which is choice (b).
Loss of KE $=(\frac{1}{2}m{v}^2+\frac{1}{2}M{V}^2)-\frac{1}{2}\left[\frac{m^2{v}^2+M^2{V}^2}{(M+m)}\right]$
$=\frac{1}{2}\frac{Mm}{(M+m)}(V^2+v^2)$, which is choice (c).