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
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) $\frac{3}{2}$
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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