A particle of mass m moving with velocity 1 m/s collides perfectly elastically with another stationary particle of mass 2m. If the incident particle is deflected by 90°, the heavy mass will make an angle $\theta$ with the initial direction of m equal to

From conservation of momentum along incident direction

$m\times 1=2m{v}_{2}Cos\theta \left(i\right)$

From conservation of momentum along perpendicular direction,

$\begin{array}{l}m\times 0+2m\times 0=m{v}_1-2m{v}_{2}sin\theta \\ \Rightarrow v_1=2v_{2}sin\theta \left(ii\right)\end{array}$

From energy conservation, $\frac{1}{2}m\times {\left(1\right)}^2=\frac{1}{2}m{v}_1^2+\frac{1}{2}2m{v}_2^2$

Solving we get $\theta$ = 30°.