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# NSEP 2015 Problem 9 | Pulley problem

Try this problem on pulley problem on inclined plane from NSEP 2015 Problem 9.

## NSEP 2015-16 ~ Problem 9

Maases $m_1$ and $m_2$ are connected to a string passing over a pulley as shown. Mass $m_2$ starts from rest and falls through a distance $d$ in time t. Now, by interchanging the masses the time required for $m_1$ to fall through the same distance is $2t$. Therefore, the ratio of masses $m_2 : m_1$

a) $\frac{2}{3}$ b) $\frac{3}{2}$ c) $\frac{5}{2}$ d) $\frac{4}{3}$

### Key Concepts

Newton's Laws of Motion

Idea of accelerations, velocity and displacement

## Suggested Book | Source | Answer

Concept of Physics H.C. Verma

University Physics by H. D. Young and R.A. Freedman

Fundamental of Physics D. Halliday, J. Walker and R. Resnick

National Standard Examination in Physics(NSEP) 2015-2016

Option-(b) $\frac{3}{2}$

## Try with Hints

We know at the beginning the blocks have zero velocities. Using the relation $s= ut+\frac{1}{2}at^2$, we can find the relation between the accelerations for two cases (i.e., when they are interchanged).

Knowing the accelerations we can now use the second law of newton to find the ratio of masses.

From the first hint,

$$\frac{1}{2}a_1t^2 = \frac{1}{2}a_2 (2t)^2 \to a_1 =4 a_2$$

Now, we find the value of $a_1$ and $a_2$ using $a = \frac{F}{M}$

$$\frac{m_2 g - m_1g \sin(30)}{m1+m_2} = 4 \frac{m_1 g - m_2g \sin(30)}{m1+m_2}$$

Rearranging this expression and using $\sin(30) = \frac{1}{2}$,

This gives, $\frac{m_2}{m_1} = \frac{3}{2}$

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