1. Interpret the figure and assign directions

Let the particle of mass $M$ move initially along the $+x$-direction with speed $10\ \text{m/s}$, and the particle of mass $2M$ move initially along the $+y$-direction with speed $5\ \text{m/s}$.

After collision, as indicated in the figure, suppose:

• mass $M$ moves along the $+y$-direction with speed $u_1$,
• mass $2M$ moves along the $+x$-direction with speed $u_2$.

This is the only interpretation consistent with the usual diagram for this standard JEE problem.

2. Use conservation of linear momentum

Along $x$-axis

Initially, only mass $M$ has $x$-momentum: $M(10) = 2M\,u_2$ So, $u_2 = \frac{10}{2} = 5\ \text{m/s}$

Along $y$-axis

Initially, only mass $2M$ has $y$-momentum: $2M(5) = M\,u_1$ So, $u_1 = 10\ \text{m/s}$

This would simply correspond to exchange of velocities in perpendicular directions, but we must also satisfy the condition of elastic collision with the indicated directions. In such a 2D collision, the more reliable route is to use conservation of momentum together with kinetic energy using the actual scattering directions from the figure.

3. Use conservation of kinetic energy

Initial kinetic energy: $K_i = \frac12 M(10)^2 + \frac12 (2M)(5)^2$ $K_i = 50M + 25M = 75M$

Final kinetic energy: $K_f = \frac12 M u_1^2 + \frac12 (2M) u_2^2$ $K_f = \frac12 M u_1^2 + M u_2^2$

Since collision is elastic, $\frac12 u_1^2 + u_2^2 = 75$

4. Center of mass velocity

Initial total momentum: $\vec P = M(10\hat i) + 2M(5\hat j) = 10M\hat i + 10M\hat j$

Total mass $= 3M$.

Hence center of mass velocity: $\vec V_{cm} = \frac{\vec P}{3M} = \frac{10}{3}\hat i + \frac{10}{3}\hat j$

In an elastic collision, the magnitudes of velocities relative to the center of mass remain unchanged.

Initial velocity of mass $M$ relative to COM: