Question 59: Centre of Mass and Rotational Motion

Question

As shown below, bob A of a pendulum having massless string of length 'R' is released from 60° to the vertical. It hits another bob B of half the mass that is at rest on a frictionless table in the center. Assuming elastic collision, the magnitude of the velocity of bob A after the collision will be (take g as acceleration due to gravity.)

Accepted Answer

Velocity of a just before hitting :
$$\mathrm{u}=\sqrt{2 \mathrm{~g} \frac{\mathrm{R}}{2}}=\sqrt{\mathrm{gR}}$$
Just after collision, let velocity of A and B are $\mathrm{v}_1$ and $v_2$ respectively
$$	herefore 	ext { by COM: }$$
$$\begin{aligned}
& \mathrm{mu}=\mathrm{mv}_1+\frac{\mathrm{m}}{2} \mathrm{v}_2 \
& 2 \mathrm{v}_1+\mathrm{v}_2=2 \mathrm{u} \quad	ext{...... (i)}\
& \mathrm{e}=1=\frac{\mathrm{v}_2-\mathrm{v}_1}{\mathrm{u}} \
& \Rightarrow \mathrm{v}_2-\mathrm{v}_1=\mathrm{u} \quad	ext{...... (ii)}
\end{aligned}$$
From (i) -(ii)
$$\Rightarrow 3 \mathrm{v}_1=\mathrm{u} \Rightarrow \mathrm{v}_1=\frac{\mathrm{u}}{3}=\frac{1}{3} \sqrt{\mathrm{gR}}$$

Answer

$\frac{4}{3}\sqrt{Rg}$

Answer

$\frac{1}{3}\sqrt{Rg}$

Answer

$\sqrt{Rg}$

Answer

$\frac{2}{3}{\sqrt{Rg}}$

Centre of Mass and Rotational Motion

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