Question

After falling from the rest through a height , a body of mass begins to raise a body of mass $M$ $\left( {M > m} \right)$ connected to it through a pulley. Determine the time it will take for the body of mass $M$ to return to its original position.
A. $\dfrac{{2m}}{{M + m}}\sqrt {\dfrac{{2h}}{g}} $
B. $\dfrac{{2m}}{{M - m}}\sqrt {\dfrac{{2h}}{g}} $
C. $\dfrac{{2m}}{{M - m}}\sqrt {\dfrac{h}{g}} $
D. $\dfrac{m}{{M - m}}\sqrt {\dfrac{{2h}}{g}} $

Answer 1

Hint: In this question, we will use the law of conservation of momentum to find the relation for the final velocity after collision. Also, we will use Newton’s second law to find the acceleration. After that, we will use the second equation of motion to find the expression for time.

Complete step by step answer:
Given: The mass of the bodies are $m$ and $M$. A body of mass $m$ raises a body of mass $M$ such that $m$ is less than the $M$.
When a body of mass $m$ falls from the height $h$its potential energy will convert into kinetic energy which can be expressed as:
$\Rightarrow mgh = \dfrac{1}{2}m{u^2}$
Where $u$is the initial velocity of body of mass $m$.
On resolving the above expression, we will get,
$
\Rightarrow {u^2} = 2gh\\
\Rightarrow u = \sqrt {2gh}
$……(i)
Let us assume after collision the mass of the body is $M + m$ and their combined velocity is $v$.
We know that from the law of conservation of momentum, the initial momentum of the body of mass $m$will be equal to the final momentum of the body of mass $M + m$. Since momentum will always be conserved. We also know that momentum is the product of mass and velocity of a body. This can be expressed as:
$\Rightarrow mu = \left( {M + m} \right)v$
 In the above expression, we will substitute the $u$ from the equation (i)
$
\Rightarrow m\left( {\sqrt {2gh} } \right) = \left( {M + m} \right)v\\
\Rightarrow v = \dfrac{{m\sqrt {2gh} }}{{M + m}}
$……(ii)
For the body of mass $M$ suppose tension $T$is acting upward. Also, the weight $Mg$ is acting downward. Then from the Newton’s second law,
$\Rightarrow Mg - T = Ma$……(iii)
For the body of mass $m$ suppose tension $T$ is acting upward. Also, the weight $mg$ is acting downward. Then from the Newton’s second law,
$\Rightarrow T - mg = ma$……(iv)
Now, we will add equation (iii) and (iv), we will get the expression for acceleration.
\[
\Rightarrow Mg - mg = \left( {M + m} \right)a\\
\Rightarrow \left( {M - m} \right)g = \left( {M + m} \right)a\\
\Rightarrow a = \dfrac{{\left( {M - m} \right)g}}{{\left( {M + m} \right)}}
\]……(v)
We know that from the laws of motion,
$\Rightarrow S = ut - \dfrac{1}{2}a{t^2}$
Where $S$ is the displacement,
In this case displacement is zero. Also velocity is $v$ after the collision. Hence, we can rewrite the above expression as:
$
\Rightarrow 0 = vt - \dfrac{1}{2}a{t^2}\\
\Rightarrow vt = \dfrac{1}{2}a{t^2}\\
\Rightarrow v = \dfrac{1}{2}at
$
On rearranging the above expression, we will get
$\Rightarrow t = \dfrac{{2v}}{a}$
We will substitute $v$ from the equation (ii) and $a$from the equation (v).
$
\Rightarrow t = \dfrac{{\dfrac{{2m\sqrt {2gh} }}{{M + m}}}}{{\dfrac{{\left( {M - m} \right)g}}{{M + m}}}}\\
\Rightarrow t = \dfrac{{2m\sqrt {2gh} }}{{\left( {M - m} \right)g}}\\
\Rightarrow t = \dfrac{{2m}}{{\left( {M - m} \right)}}\sqrt {\dfrac{{2h}}{g}}
$
Therefore, option B is the correct answer.

Note: To solve this question, firstly, we will equate the potential energy of the body to the kinetic energy. Since when the body falls from a certain height, its potential energy converts into kinetic energy.