Answer
64.8k+ views
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.
Complete step by step answer:
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.
Complete step by step answer:
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.
Recently Updated Pages
Write a composition in approximately 450 500 words class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Arrange the sentences P Q R between S1 and S5 such class 10 english JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What is the common property of the oxides CONO and class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
What happens when dilute hydrochloric acid is added class 10 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
If four points A63B 35C4 2 and Dx3x are given in such class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
The area of square inscribed in a circle of diameter class 10 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Other Pages
Excluding stoppages the speed of a bus is 54 kmph and class 11 maths JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
In the ground state an element has 13 electrons in class 11 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Electric field due to uniformly charged sphere class 12 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
A boat takes 2 hours to go 8 km and come back to a class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
According to classical free electron theory A There class 11 physics JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)
Differentiate between homogeneous and heterogeneous class 12 chemistry JEE_Main
![arrow-right](/cdn/images/seo-templates/arrow-right.png)