Question

Masses $m$ and $M$on pulley move $0.6m$in$4s$. What is the ratio of$\dfrac{m}{M}$?

Answer 1

In order to solve this question, we are going to first calculate the acceleration of the two masses on the pulley using the second equation of motion. After that the formula for the acceleration taking gravity in the consideration is used to calculate the ratio of the two masses.

Formula used:
According to the second equation of motion,
$s = ut + \dfrac{1}{2}a{t^2}$
The acceleration of the masses is given by the formula
$a = \left( {\dfrac{{{m_2} - {m_1}}}{{{m_2} + {m_1}}}} \right)g$

Complete step-by-step solution:
It is given in the question that there is a pulley with the two masses
{m_1} = m \\\ {m_2} = M \\\
These masses move the distance equal to $s = 0.6m$
The time taken for the masses to cover the distance is$t = 4s$.
Now, according to the second equation of motion,
$s = ut + \dfrac{1}{2}a{t^2}$
Now, the initial velocity is assumed to be equal to zero, i.e.
$u = 0$
Putting this in the above equation, we get
$s = \dfrac{1}{2}a{t^2}$
Rearranging the terms for the value of the acceleration of the masses we get,
$a = \dfrac{{2s}}{{{t^2}}}$
Now putting the values of the distance and time as given to find the acceleration,
$a = \dfrac{{2 \times 0.6}}{{{4^2}}}$
Solving, we get
$a = \dfrac{{1.2}}{{16}} = \dfrac{3}{{40}}$
Now, in the pulley problem, taking the acceleration due to gravity in consideration, the acceleration of the masses is given by the formula
$a = \left( {\dfrac{{{m_2} - {m_1}}}{{{m_2} + {m_1}}}} \right)g$
Putting the values in this equation, we get
$a = \left( {\dfrac{{M - m}}{{M + m}}} \right)g$
Diving the denominator and numerator of the right hand side by $M$
$a = \left( {\dfrac{{1 - \dfrac{m}{M}}}{{1 + \dfrac{m}{M}}}} \right)g$
Putting the value of the acceleration obtained above and solving,

\Rightarrow 3 + 3\dfrac{m}{M} = 400 - 400\dfrac{m}{M} \\\ \Rightarrow 3\dfrac{m}{M} + 400\dfrac{m}{M} = 400 - 3 \\\ \Rightarrow 403\dfrac{m}{M} = 397 \\\ \Rightarrow \dfrac{m}{M} = \dfrac{{397}}{{403}} \\\ $$ **Thus, the ratio of the two masses is equal to$$\dfrac{{397}}{{403}}$$.** **Note:** It is important to note that the masses of the pulley are undergoing a uniformly accelerated motion, hence all the laws of motion and the equations of motion are applicable. In the presence of gravity by applying the Newton’s and the conservation of energy laws, we find the relation between$$a$$ and$$g$$.