A particle of mass m with an initial velocity $$u\hat{i}$$ collides perfectly elastically with a mass 3m at rest. It moves with a velocity $$v\hat{j}$$ after collision, then, v is given by:

Let the final velocity vector of the mass $$3m$$ be $$\vec{v}_2$$.

Conservation of linear momentum:

$$m(u\hat{i}) + 3m(0) = m(v\hat{j}) + 3m\vec{v}_2$$

$$mu\hat{i} - mv\hat{j} = 3m\vec{v}_2 \implies \vec{v}_2 = \frac{u}{3}\hat{i} - \frac{v}{3}\hat{j}$$

Magnitude squared of $$\vec{v}_2$$: $$v_2^2 = \left(\frac{u}{3}\right)^2 + \left(-\frac{v}{3}\right)^2 = \frac{u^2 + v^2}{9}$$

Conservation of kinetic energy: $$\frac{1}{2}mu^2 + 0 = \frac{1}{2}mv^2 + \frac{1}{2}(3m)v_2^2$$

$$u^2 = v^2 + 3v_2^2$$

$$u^2 = v^2 + 3\left(\frac{u^2 + v^2}{9}\right) \implies u^2 = v^2 + \frac{u^2 + v^2}{3}$$

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

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