Question

Two balls are placed as shown in the figure on a weightless support formed by two smooth inclined planes each of which forms an angle of $\alpha$ with the horizontal. The support can slide without friction along a horizontal plane. The upper ball of mass $m_1$ is released. Determine the condition under which the lower ball of mass $m_2$ starts climbing up the support.

• A. $m_1\cos 2\alpha >m_2$
• B. $m_2\cos 2\alpha >m_1$
• C. $m_1\sin 2\alpha >m_2$
• D. $m_1\cos 2\alpha >m_2$

Answer 1

The correct option is A $m_1\cos 2\alpha >m_2$

Free body diagram of system is given as

Resolving the forces

$N_1\sin \alpha =N_2\sin \alpha \Rightarrow N_1=N_2..\text{(i)}$

Conditions
1. Net acceleration of the ball with mass $m_1$ should be non-existent perpendicular to the surface 2
2. Net acceleration of the ball with mass $m_2$ should be existent to the surface 2

Now separately analysing the free body diagrams
For ball with mass $m_1$
From FBD
$N_1=m_{1}g\cos \alpha$ ....$\text{(ii)}$
Now for ball with mass $m_2$
$N_2\cos 2\alpha -m_{2}g\cos \alpha >0\Rightarrow N_2\cos 2\alpha >m_{2}g\cos \alpha$
Now from eq$\text{(i)}$ anad eq$\text{(ii)}$ we get
$m_{1}g\cos \alpha \cos 2\alpha >m_{2}g\cos \alpha \Rightarrow m_1\cos 2\alpha >m_2$