Question

A particle of mass $m$ moving eastwards with a speed $v$ collides with another particle of same mass moving northwards with same speed $v$. The two particles coalesces on collision. Then, the new particle of mass $2m$ will move with velocity:

• A. $\frac{v}{2}\text{North-East}$
• B. $\frac{v}{\sqrt{2}}\text{South-West}$
• C. $\frac{v}{2}\text{North-West}$
• D. $\frac{v}{\sqrt{2}}\text{North-East}$

Answer 1

The correct option is D $\frac{v}{\sqrt{2}}\text{North-East}$

Considering $X-$axis along East direction, $Y-$axis along North direction.
$\Rightarrow$In this case particles are colliding obliquely, but linear momentum will be conserved since $F_{\text{ext}}=0$


Initial momentum of system of particles is given as,
${\overrightarrow{P}}_i=mv\hat{i}+mv\hat{j}$
final momentum of system after collision,
${\overrightarrow{P}}_f=2m\times {\overrightarrow{v}}^{\prime }$
From $P.C.L.M$:
${\overrightarrow{P}}_i={\overrightarrow{P}}_f$
$mv\hat{i}+mv\hat{j}=2m\times {\overrightarrow{v}}^{\prime }$
$\overrightarrow{v^{\prime }}=\frac{v}{2}\hat{i}+\frac{v}{2}\hat{j}$
$\Rightarrow$ Magnitude of the final velocity is given as:
$|\overrightarrow{v^{\prime }}|=\sqrt{{\left(\frac{v}{2}\right)}^2+{\left(\frac{v}{2}\right)}^2}$
$\therefore |\overrightarrow{v^{\prime }}|=\frac{v}{\sqrt{2}}$
Now direction of motion after collision is:
$\tan \theta =\frac{v_y^{\prime }}{v_x^{\prime }}=\frac{\frac{v}{2}}{\frac{v}{2}}=1$
$\therefore \theta ={45}^{\circ }$
$\Rightarrow$Direction is $\text{North-East}$