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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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