1. Given

• Mass of first particle: $m$
• Initial velocity of first particle: $u\hat i$
• Second particle of mass $3m$ is initially at rest.
• Collision is perfectly elastic.
• After collision, the first particle moves with velocity $v\hat j$.

We need to find $v$.

2. Use conservation of linear momentum

Let the final velocity of mass $3m$ be $\vec V = V_x \hat i + V_y \hat j$

Initial momentum: $\vec p_i = mu\hat i$

Final momentum: $\vec p_f = mv\hat j + 3m(V_x\hat i + V_y\hat j)$

Equating components:

In $x$-direction:

$mu = 3mV_x$ $V_x = \frac{u}{3}$

In $y$-direction:

$0 = mv + 3mV_y$ $V_y = -\frac{v}{3}$

So, $\vec V = \frac{u}{3}\hat i - \frac{v}{3}\hat j$

3. Use conservation of kinetic energy

Since collision is perfectly elastic, $\frac12 mu^2 = \frac12 mv^2 + \frac12 (3m)V^2$

Now, $V^2 = V_x^2 + V_y^2 = \left(\frac{u}{3}\right)^2 + \left(-\frac{v}{3}\right)^2 = \frac{u^2+v^2}{9}$

Substitute into energy equation: $\frac12 mu^2 = \frac12 mv^2 + \frac12 (3m)\cdot \frac{u^2+v^2}{9}$

$\frac12 mu^2 = \frac12 mv^2 + \frac{m}{6}(u^2+v^2)$

Multiply by $6/m$: $3u^2 = 3v^2 + (u^2+v^2)$

$3u^2 = u^2 + 4v^2$

$2u^2 = 4v^2$

$v^2 = \frac{u^2}{2}$

$v = \frac{u}{\sqrt2}$

4. Check options

• A: $\sqrt{\frac23}u$ ❌
• B: $\frac{u}{\sqrt3}$ ❌
• C: $\frac{u}{\sqrt2}$ ✅
• D: $\frac{u}{\sqrt6}u$ ❌ (also dimensionally odd as written in option, but clearly not correct)

5. Final answer

$\boxed{v = \frac{u}{\sqrt2}}$

So the correct option is C.