A particle of mass ' \(\mathrm{m}\) ' moving east ward with a speed ' \(v\) ' collides with another particle of same mass moving north-ward with same speed ' \(v\) '. The two particles coalesce after collision. The new particle of mass ' \(2 \mathrm{~m}\) ' will move in north east direction with a speed (in \(\mathrm{m} / \mathrm{s}\) )

Momentum of particle moving towards east
\(\overrightarrow{\mathrm{p}_1}=\mathrm{mv \hat {i }}\)
Momentum of particle moving towards North
\(\overrightarrow{p_2}=m v \hat{j}\)
Momentum after collision,
\(\vec{p}=2 m\left(v_x \hat{i}+v_y \hat{j}\right)\)
Applying momentum conservation,
\(\overrightarrow{p_1}+\overrightarrow{p_2}=\vec{p} \)
\(m v \hat{i}+m v \hat{j}=2 m\left(v_x \hat{i}+v_y \hat{j}\right) \)
\(\therefore 2 m v_x=m v \)
\(v_x=\frac{v}{2}\)
Similarly, \(\mathrm{v}_{\mathrm{y}}=\frac{\mathrm{v}}{2}\)
\(\mathrm{v}_{\mathrm{R}}=\sqrt{\mathrm{v}_{\mathrm{x}}^2+\mathrm{v}_{\mathrm{y}}^2}=\sqrt{\left(\frac{\mathrm{v}}{2}\right)^2+\left(\frac{\mathrm{v}}{2}\right)^2}=\frac{\mathrm{v}}{\sqrt{2}}\)