1. Given data

• Particle 1: mass $m$, velocity $2v$ along $x$-axis $\vec u_1 = 2v\,\hat i$
• Particle 2: mass $2m$, velocity $v$ along $y$-axis $\vec u_2 = v\,\hat j$
• Collision is perfectly inelastic, so both stick together after collision.

2. Initial momentum of the system

Momentum of particle 1: $\vec p_1 = m(2v\hat i)=2mv\hat i$

Momentum of particle 2: $\vec p_2 = 2m(v\hat j)=2mv\hat j$

Total initial momentum: $\vec P = 2mv\hat i+2mv\hat j$

After collision, total mass: $M = m+2m=3m$

Let common velocity after collision be $\vec V$. Then $3m\vec V = 2mv\hat i+2mv\hat j$ $\vec V = \frac{2v}{3}\hat i+\frac{2v}{3}\hat j$

So, $V^2=\left(\frac{2v}{3}\right)^2+\left(\frac{2v}{3}\right)^2 = \frac{8v^2}{9}$

3. Initial kinetic energy

For particle 1: $K_1 = \frac12 m(2v)^2 = 2mv^2$

For particle 2: $K_2 = \frac12 (2m)(v)^2 = mv^2$

Thus, $K_i = 2mv^2+mv^2=3mv^2$

4. Final kinetic energy

$K_f = \frac12 (3m)V^2$ $K_f = \frac12 (3m)\cdot \frac{8v^2}{9} = \frac{4}{3}mv^2$

5. Loss in kinetic energy

$\Delta K = K_i-K_f = 3mv^2-\frac{4}{3}mv^2 = \frac{5}{3}mv^2$

Percentage loss: $\frac{\Delta K}{K_i}\times 100 = \frac{\frac{5}{3}mv^2}{3mv^2}\times 100$ $= \frac{5}{9}\times 100 \approx 55.56\%$

This is closest to $56\%$

6. Option check

• A: $56\%$ ✅
• B: $62\%$ ❌
• C: $44\%$ ❌
• D: $50\%$ ❌

Hence the correct option is A.