A simple pendulum, made of a string of length l and a bob of mass m, is released from a small angle θ _ 0 . It strikes a block of mass M, kept on horizontal surface at its lowest point of oscillations, elastically. It

Correct answer: B. m( θ _ 0 + θ _ 1 θ _ 0 - θ _ 1 ) u = 2gl(1 - cos θ _ 0 ) .......(i) v = velocity of ball after collision v = ((m - M)/(m + M))u Since ball rises up to angle θ _ 1 v = 2gl(1 - cos θ _ 1 ) = ((m - M)/(m + M))u ......(ii) From (i) and (ii) (m - M)/(m + M) = 1 - cos θ _ 1 1 - cos θ _ 0 = ( θ _ 1 2 ) ( θ _ 0 2 ) ⇒ (M)/(m) = θ _ 0 - θ _ 1 θ _ 0 + θ _ 1 ⇒ M = ( θ _ 0 - θ _ 1 θ _ 0 + θ _ 1 )m

JEE Main 201912 JanuaryMorning Shift

Exam: JEE Main
Year: 2019
Date: 12 January
Shift: Morning Shift
Subject: Physics
Chapter: Center of Mass, Momentum and Collision

JEE Main 2019 · 12 January · Morning Shift · Physics · Center of Mass, Momentum and Collision

A simple pendulum, made of a string of length l and a bob of mass...

A simple pendulum, made of a string of length

l

and

a

bob of mass

m,

is released from a small angle

{\theta}_{0} .

It strikes a block of mass

M,

kept on horizontal surface at its lowest point of oscillations, elastically. It bounces back and goes up to an angle

{\theta}_{1} .

Then

M

is given by: A simple pendulum, made of a string of length l and a bob of mass m, is released from a small angle θ _ 0 . It strikes

Options

A $m\left(\dfrac{{\theta}_{0} - {\theta}_{1}}{{\theta}_{0} + {\theta}_{1}}\right)$
B $m\left(\dfrac{{\theta}_{0} + {\theta}_{1}}{{\theta}_{0} - {\theta}_{1}}\right)$
C $\dfrac{m}{2}\left(\dfrac{{\theta}_{0} + {\theta}_{1}}{{\theta}_{0} - {\theta}_{1}}\right)$
D $\dfrac{m}{2}\left(\dfrac{{\theta}_{0} - {\theta}_{1}}{{\theta}_{0} + {\theta}_{1}}\right)$

Correct answer

B. $m\left(\dfrac{{\theta}_{0} + {\theta}_{1}}{{\theta}_{0} - {\theta}_{1}}\right)$

Step-by-step solution

u = \sqrt{2gl\left(1 - cos{\theta}_{0}\right)}

.......(i)

v =

velocity of ball after collision

v = \left(\dfrac{m - M}{m + M}\right)u

Since ball rises up to angle

{\theta}_{1}

v = \sqrt{2gl\left(1 - cos{\theta}_{1}\right)} = \left(\dfrac{m - M}{m + M}\right)u

......(ii)

From (i) and (ii)

\dfrac{m - M}{m + M} = \sqrt{\dfrac{1 - cos{\theta}_{1}}{1 - cos{\theta}_{0}}} = \dfrac{\sin \left(\dfrac{{\theta}_{1}}{2}\right)}{\sin \left(\dfrac{{\theta}_{0}}{2}\right)}

<br /><br />

⇒ \dfrac{M}{m} = \dfrac{{\theta}_{0} - {\theta}_{1}}{{\theta}_{0} + {\theta}_{1}} ⇒ M = \left(\dfrac{{\theta}_{0} - {\theta}_{1}}{{\theta}_{0} + {\theta}_{1}}\right)m

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