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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) $\frac{4}{3}$

$m_1$ and $m_2$ are interchanged from real problem

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

Physics Olympiad Program at Cheenta

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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) $\frac{4}{3}$

$m_1$ and $m_2$ are interchanged from real problem

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

Physics Olympiad Program at Cheenta

Subscribe to Cheenta at Youtube


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