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 friction less table in the centre. Assuming elastic collision, the magnitude of the

STUDY MATERIALS
COURSES
MORE
SUCCESS STORY
ADMISSION
STORE

Back

JEE ADVANCE

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

JEE MAIN

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

NEET

CLASS XI - XII
CRASH COURSE

CLASS XI - XII

CRASH COURSE

BITSAT

CLASS XI - XII

CLASS XI - XII

CBSE BOARD

CLASS IX

CLASS X

CLASS XI

CLASS XII

CLASS VI

CLASS VII

CLASS VIII

UP BOARD

CLASS IX
CLASS X
CLASS XI
CLASS XII

CLASS IX

CLASS X

CLASS XI

CLASS XII

CUET

CRASH COURSE

CRASH COURSE

Courses

CLASSROOM COURSES
ONLINE COURSES

ONLINE COURSES

More

Solution

Q.
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 friction less table in the centre. Assuming elastic collision, the magnitude of the velocity of bob A after the collision will be (Take g as acceleration due to gravity) [2025]

1 $\frac{1}{3} \sqrt{R g}$

2 $\sqrt{R g}$

3 $\frac{4}{3} \sqrt{R g}$

4 $\frac{2}{3} \sqrt{R g}$

Ans.
(1)

Velocity of a just before hitting:

$u = \sqrt{2 g \frac{R}{2}} = \sqrt{g R}$

Just after collision, let velocity of A and B are $v_{1}$ and $v_{2}$ respectively

By COM: $m u = m v_{1} + \frac{m}{2} v_{2}$

$2 v_{1} + v_{2} = 2 u$ ... (i)

$e = 1 = \frac{v_{2} - v_{1}}{u}$

$\Rightarrow v_{2} - v_{1} = u$ ... (ii)

From (i) and (ii)

$\Rightarrow 3 v_{1} = u \Rightarrow v_{1} = \frac{u}{3} = \frac{1}{3} \sqrt{g R}$

Want Us to Email you About Special Offers & Updates?