How Cheenta works to ensure student success?

Explore the Back-StoryTry this problem on pulley problem on inclined plane from NSEP 2015 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) $\frac{4}{3}$

Newton's Laws of Motion

Idea of accelerations, velocity and displacement

Suggested Reading

Source of the Problem

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) $\frac{3}{2}$

Hint 1

Hint 2

Hint 3

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

Try this problem on pulley problem on inclined plane from NSEP 2015 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) $\frac{4}{3}$

Newton's Laws of Motion

Idea of accelerations, velocity and displacement

Suggested Reading

Source of the Problem

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) $\frac{3}{2}$

Hint 1

Hint 2

Hint 3

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

Cheenta is a knowledge partner of Aditya Birla Education Academy

Advanced Mathematical Science. Taught by olympians, researchers and true masters of the subject.

JOIN TRIALAcademic Programs

Free Resources

Why Cheenta?

Google