A simple pendulum, made of a… - JEE Main 2019 Physics PYQ

JEE Main 2019 — Physics Mechanics

2019mcqhard

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:

Official previous-year question

Held on 12 Jan 2019 · Verified 6 Jul 2026.

Options

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

Did you get this right?

Solution

JEE Main 2019 Physics, Center of Mass Momentum and Collision — solution figure

$u=\sqrt{2 gl(1-cos {\theta }_{0})}$ .......(i)

$v=$ velocity of ball after collision

$v=(\frac{m-M}{m+M})u$

Since ball rises up to angle ${\theta }_{1}$

$v=\sqrt{2gl(1-cos {\theta }_{1})}=(\frac{m-M}{m+M})u$ ......(ii)

From (i) and (ii)

$\frac{m-M}{m+M}=\sqrt{\frac{1-cos {\theta }_{1}}{1-cos {\theta }_{0}}}=\frac{sin(\frac{{\theta }_{1}}{2})}{sin(\frac{{\theta }_{0}}{2})}$

$\Rightarrow \frac{M}{m}=\frac{{\theta }_{0}-{\theta }_{1}}{{\theta }_{0}+{\theta }_{1}}\Rightarrow M=(\frac{{\theta }_{0}-{\theta }_{1}}{{\theta }_{0}+{\theta }_{1}})m$

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Options

$m(\frac{{\theta }_{0}-{\theta }_{1}}{{\theta }_{0}+{\theta }_{1}})$

$m(\frac{{\theta }_{0}+{\theta }_{1}}{{\theta }_{0}-{\theta }_{1}})$

$\frac{m}{2}(\frac{{\theta }_{0}+{\theta }_{1}}{{\theta }_{0}-{\theta }_{1}})$

$\frac{m}{2}(\frac{{\theta }_{0}-{\theta }_{1}}{{\theta }_{0}+{\theta }_{1}})$

Solution

JEE Main 2019 Physics, Center of Mass Momentum and Collision — solution figure

$u=\sqrt{2 gl(1-cos {\theta }_{0})}$ .......(i)

$v=$ velocity of ball after collision

$v=(\frac{m-M}{m+M})u$

Since ball rises up to angle ${\theta }_{1}$

$v=\sqrt{2gl(1-cos {\theta }_{1})}=(\frac{m-M}{m+M})u$ ......(ii)

From (i) and (ii)

$\frac{m-M}{m+M}=\sqrt{\frac{1-cos {\theta }_{1}}{1-cos {\theta }_{0}}}=\frac{sin(\frac{{\theta }_{1}}{2})}{sin(\frac{{\theta }_{0}}{2})}$

$\Rightarrow \frac{M}{m}=\frac{{\theta }_{0}-{\theta }_{1}}{{\theta }_{0}+{\theta }_{1}}\Rightarrow M=(\frac{{\theta }_{0}-{\theta }_{1}}{{\theta }_{0}+{\theta }_{1}})m$

Options

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Solution

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Options

Did you get this right?

Solution

Your note

JEE Main Mechanics in other years

More JEE Main Mechanics questions

Other Physics topics for JEE Main

Electromagnetism

Modern Physics

Optics

Thermodynamics

Waves & Oscillations

Keep practicing Mechanics