Question

A ball of mass $m$ moving with a velocity of $3{\text{ms}}^{-1}$ along $x$-axis strikes another ball of mass $2m$ kept at rest. The first ball comes to rest after collision and the other one breaks into two equal pieces, one of the pieces starts moving along $+y$-axis with speed $4{\text{ms}}^{-1}$, what will be the velocity of the other piece ?

• A. $5{\text{ms}}^{-1}$
• B. $1{\text{ms}}^{-1}$
• C. $4{\text{ms}}^{-1}$
• D. $2{\text{ms}}^{-1}$

Answer 1

The correct option is A $5{\text{ms}}^{-1}$


In collision, only internal forces are involved. Hence, linear momentum will be conserved. Here, linear momentum will be conserved individually along $x$ and $y$ axis.



Using the conservation of linear momentum along the $x$-axis we get,

$(P_i{)}_x=(P_f{)}_x$

$(m\times 3)+(2m\times 0)=(m\times vcos\theta )$

$3m=mv\cos \theta$

$\Rightarrow v\cos \theta =3.....\left(1\right)$

Using conservation of linear momentum along the $y$-axis we get,

$(P_i{)}_y=(P_f{)}_y$

$0=4m-mv\sin \theta$

$\therefore v\sin \theta =4.......\left(2\right)$

From equation $\left(1\right)$ and $\left(2\right)$ we get,

$v=\sqrt{3^2+4^2}=5{\text{ms}}^{-1}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, $\left(\text{A}\right)$ is the correct answer.

Why this question ? Key concept: In oblique collision, linear momentum is conserved in mutually perpendicular direction.