Question 19: Centre of Mass and Rotational Motion

Question

In the figure, pendulum bob on left side is pulled a side to a height $$h$$ from its initial position. After it is released it collides with the right pendulum bob at rest, which is of same mass. After the collision, the two bobs stick together and rise to a height

Accepted Answer

When bob $$A$$ strikes to the bob $$B$$, then
$$\begin{aligned}
m u & =(m+m) v^{\prime} \
\Rightarrow \quad v^{\prime} & =\frac{u}{2} \quad 	ext{.... (i)}
\end{aligned}$$
The potential energy of $$A$$ at height $$h$$ gets converted into kinetic energy of this mass, at point $$O$$, i.e.
$$\begin{aligned}
m g h & =\frac{1}{2} m u^2 \
\Rightarrow \quad u & =\sqrt{2 g h} \quad 	ext{[From Eq. (i)]}\
	herefore \quad v^{\prime} & =\frac{\sqrt{2 g h}}{2}=\sqrt{\frac{g h}{2}}
\end{aligned}$$
Let the combined mass moves to a height $$h^{\prime}$$, then total mass $$=2 m$$
Then, $$2 m g h^{\prime}=\frac{1}{2}(2 m) v^{\prime 2}$$
$$\Rightarrow \quad g h^{\prime}=\frac{g h}{4} \Rightarrow h^{\prime}=\frac{h}{4}$$

Answer

$$\frac{3 h}{4}$$

Answer

$$\frac{2 h}{3}$$

Answer

$$\frac{h}{2}$$

Answer

$$\frac{h}{4}$$

Centre of Mass and Rotational Motion

In the figure, pendulum bob on left side is pulled a side to a height $h$ from its initial position. After it is released it collides with the right pendulum bob at rest, which is of same mass. After the collision, the two bobs stick together and rise to a height

Choose the correct answer: