JEE MAIN - Physics (2025 - 29th January Morning Shift - No. 16)
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.)
_29th_January_Morning_Shift_en_16_1.png)
Explanation
_29th_January_Morning_Shift_en_16_2.png)
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
$$\therefore \text { by COM: }$$
$$\begin{aligned} & \mathrm{mu}=\mathrm{mv}_1+\frac{\mathrm{m}}{2} \mathrm{v}_2 \\ & 2 \mathrm{v}_1+\mathrm{v}_2=2 \mathrm{u} \quad\text{...... (i)}\\ & \mathrm{e}=1=\frac{\mathrm{v}_2-\mathrm{v}_1}{\mathrm{u}} \\ & \Rightarrow \mathrm{v}_2-\mathrm{v}_1=\mathrm{u} \quad\text{...... (ii)} \end{aligned}$$
From (i) -(ii)
$$\Rightarrow 3 \mathrm{v}_1=\mathrm{u} \Rightarrow \mathrm{v}_1=\frac{\mathrm{u}}{3}=\frac{1}{3} \sqrt{\mathrm{gR}}$$
Comments (0)
