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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 $\mathrm{m} / \mathrm{s}$ )
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Verified Answer
The correct answer is:
$\frac{\mathrm{V}}{\sqrt{2}}$
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,
$$
\begin{array}{ll}
& \overrightarrow{p_1}+\overrightarrow{p_2}=\vec{p} \\
& m v \hat{i}+m v \hat{j}=2 m\left(v_x \hat{i}+v_y \hat{j}\right) \\
\therefore \quad & 2 m v_x=m v \\
& v_x=\frac{v}{2}
\end{array}
$$
Similarly, $\mathrm{v}_{\mathrm{y}}=\frac{\mathrm{v}}{2}$
$$
\mathrm{v}_{\mathrm{R}}=\sqrt{\mathrm{v}_{\mathrm{x}}^2+\mathrm{v}_{\mathrm{y}}^2}=\sqrt{\left(\frac{\mathrm{v}}{2}\right)^2+\left(\frac{\mathrm{v}}{2}\right)^2}=\frac{\mathrm{v}}{\sqrt{2}}
$$
$$
\overrightarrow{\mathrm{p}_1}=\mathrm{mv \hat {i }}
$$
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,
$$
\begin{array}{ll}
& \overrightarrow{p_1}+\overrightarrow{p_2}=\vec{p} \\
& m v \hat{i}+m v \hat{j}=2 m\left(v_x \hat{i}+v_y \hat{j}\right) \\
\therefore \quad & 2 m v_x=m v \\
& v_x=\frac{v}{2}
\end{array}
$$
Similarly, $\mathrm{v}_{\mathrm{y}}=\frac{\mathrm{v}}{2}$
$$
\mathrm{v}_{\mathrm{R}}=\sqrt{\mathrm{v}_{\mathrm{x}}^2+\mathrm{v}_{\mathrm{y}}^2}=\sqrt{\left(\frac{\mathrm{v}}{2}\right)^2+\left(\frac{\mathrm{v}}{2}\right)^2}=\frac{\mathrm{v}}{\sqrt{2}}
$$
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