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# 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 a 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) $m\left( {\dfrac{{{\theta _0} + {\theta _1}}}{{{\theta _0} - {\theta _1}}}} \right)$B) $\dfrac{m}{2}\left( {\dfrac{{{\theta _0} - {\theta _1}}}{{{\theta _0} + {\theta _1}}}} \right)$C) $\dfrac{m}{2}\left( {\dfrac{{{\theta _0} + {\theta _1}}}{{{\theta _0} - {\theta _1}}}} \right)$D) $m\left( {\dfrac{{{\theta _0} - {\theta _1}}}{{{\theta _0} + {\theta _1}}}} \right)$

Last updated date: 29th May 2024
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Hint: In question it is given that it collides with the block elastically. So, recall the properties of elastic collision, velocity of approach and velocity of separation and also remember that in elastic collision the energy is conserved. So, apply energy conservation and law of momentum conservation here.

Firstly apply law of energy conservation before collision,
Potential energy at the point m is equal to kinetic energy at that point.
$mg\left( {l - l\cos {\theta _0}} \right) = \dfrac{1}{2}m{v^2}$
Where, $m$ is the mass of the bob
$l$ is the length of the pendulum
$g$ is the acceleration due to gravity
$v$ is the velocity of the pendulum just before strike
${\theta _0}$ is the angle at which the pendulum is released
So, from this we get the velocity of the bob just before it collides with the mass $M$,
$v = \sqrt {2gl\left( {1 - \cos {\theta _0}} \right)}$
Momentum of the system before collision is given by,
momentum of the pendulum + momentum of the block of mass $M$ which is at rest
$\Rightarrow m\sqrt {2gl\left( {1 - \cos {\theta _0}} \right)} + M\left( 0 \right)$
$\Rightarrow m\sqrt {2gl\left( {1 - \cos {\theta _0}} \right)}$ ……………(i)
Now apply energy conservation after collision,
Potential energy after collision at point m is equal to the kinetic energy after collision
$mg(l - l\cos {\theta _1}) = \dfrac{1}{2}m{v'^2}$
Where, $m$ is the mass of the bob
$l$ is the length of the pendulum
$g$ is the acceleration due to gravity
$v'$ is the velocity of the pendulum after collision
${\theta _1}$ is the angle at which the pendulum reached after collision
So, from this we get the velocity of the bob after it collides with the mass $M$,
$v' = \sqrt {2gl(1 - \cos {\theta _1})}$
Momentum after collision is given by,
$- m\sqrt {2gl(1 - \cos {\theta _1})} + M{v_m}$ ………..(ii)
Here, ${v_m}$ is the velocity of the block of mass $M$ after collision.
And here the minus sign shows the direction of the bob after the collision, it moves towards the left.
Now, by applying the law of conservation of momentum,
Momentum before collision = momentum after collision
$m\sqrt {2gl\left( {1 - \cos {\theta _0}} \right)} = M{v_m} - m\sqrt {2gl(1 - \cos {\theta _1})}$
$\Rightarrow m\sqrt {2gl\left( {1 - \cos {\theta _0}} \right)} + m\sqrt {2gl\left( {1 - \cos {\theta _1}} \right)} = M{v_m}$
Now taking $m\sqrt {2gl}$ common from the equation we get,
$\Rightarrow m\sqrt {2gl} \left[ {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right] = M{v_m}$ ………(iii)
We know that whenever pendulum collides elastically ,
$e = 1$
This means, velocity of approach = velocity of separation
$\sqrt {2gl(1 - \cos {\theta _0})} = {v_m} - \left[ { - \sqrt {2gl(1 - \cos {\theta _1})} } \right]$
$\Rightarrow \sqrt {2gl(1 - \cos {\theta _0})} = {v_m} + \sqrt {2gl(1 - \cos {\theta _1})}$
Now from this we get the value of ${v_m}$,
${v_m} = \sqrt {2gl} [\sqrt { - \cos {\theta _0}} - \sqrt {1 - \cos {\theta _1}} ]$
Put value of ${v_m}$ in equation (iii),
$m\sqrt {2gl} \left[ {\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} } \right] = M\sqrt {2gl} [\sqrt { - \cos {\theta _0}} - \sqrt {1 - \cos {\theta _1}} ]$
On solving we get,
$\Rightarrow \left[ {\dfrac{{\sqrt {1 - \cos {\theta _0}} + \sqrt {1 - \cos {\theta _1}} }}{{\sqrt {1 - \cos {\theta _0}} - \sqrt {1 - \cos {\theta _1}} }}} \right] = \dfrac{M}{m}$
By property of componendo and dividendo,
$\Rightarrow \dfrac{{\sqrt {1 - \cos {\theta _0}} }}{{\sqrt {1 - \cos {\theta _1}} }} = \dfrac{{M + m}}{{M - m}}$
$\Rightarrow \dfrac{{\sqrt 2 \sin \dfrac{{{\theta _0}}}{2}}}{{\sqrt 2 \sin \dfrac{{{\theta _1}}}{2}}} = \dfrac{{M + m}}{{M - m}}$
As ${\theta _0}$ and ${\theta _1}$ are very small. So, on further solving we get,
$\Rightarrow \dfrac{{\dfrac{{{\theta _0}}}{2}}}{{\dfrac{{{\theta _1}}}{2}}} = \dfrac{{M + m}}{{M - m}}$
$\Rightarrow \dfrac{{{\theta _0}}}{{{\theta _1}}} = \dfrac{{M + m}}{{M - m}}$
Again applying componendo and dividendo, we get
$\Rightarrow \dfrac{{{\theta _0} + {\theta _1}}}{{{\theta _0} - {\theta _1}}} = \dfrac{M}{m}$
On further solving we get the mass of the block,
$M = m\left[ {\dfrac{{{\theta _0} + {\theta _1}}}{{{\theta _0} - {\theta _1}}}} \right]$

Therefore, the correct option is (A).

Note: Law of conservation of momentum states that for two or more bodies in an isolated system acting upon each other, their total momentum remains constant unless an external force is applied. Therefore, momentum can neither be created nor destroyed. The law of conservation of energy states that the total energy of an isolated system remains constant and is said to be conserved over time.