Question 331: Centre of Mass and Rotational Motion

Question

A particle of mass '$$m$$' 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 $$\mathrm{m} / \mathrm{s}$$ )

Accepted Answer

Momentum of particle moving towards east
$$\overrightarrow{\mathrm{p}_1}=\mathrm{mv \hat {i }}$$
Momentum of particle moving towards North
$$\overrightarrow{\mathrm{p}_2}=\mathrm{mv} \hat{\mathrm{j}}$$
Momentum after collision,
$$\vec{p}=2 m\left(v_x \hat{i}+v_y \hat{j}ight)$$
Applying momentum conservation,
$$\begin{array}{ll} 
& \overrightarrow{p_1}+\overrightarrow{p_2}=\vec{p} \
& m v \hat{i}+m v \hat{j}=2 m\left(v_x \hat{i}+v_y \hat{j}ight) \
	herefore \quad & 2 m v_x=m v \
& v_x=\frac{v}{2}
\end{array}$$
Similarly, $$\mathrm{v}_{\mathrm{y}}=\frac{\mathrm{v}}{2}$$
$$v_R=\sqrt{v_x^2+v_y^2}=\sqrt{\left(\frac{v}{2}ight)^2+\left(\frac{v}{2}ight)^2}=\frac{v}{\sqrt{2}}$$

Answer

$$\mathrm{V}$$

Answer

$$2 \mathrm{~V}$$

Answer

$$\frac{\mathrm{V}}{2}$$

Answer

$$\frac{\mathrm{V}}{\sqrt{2}}$$

Centre of Mass and Rotational Motion

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