Question

In the figure shown all the surfaces are smooth. All the blocks $A,B$ and $C$ are movable $X$-axis is horizontal and $y$-axis vertical as shown. Just after the system is released from the position as shown:

• A. Acceleration of '$A$' relative to ground is in negative $y$-direction
• B. Acceleration of '$A$' relative to $B$ is in positive $x$-direction
• C. The horizontal acceleration of '$B$' relative to ground is in negative $x$-direction.
• D. The acceleration of '$B$' relative to ground directed along the inclined surface of '$C$' is greater than g sin $\theta$.

Answer 1

Answer: (A), (B), (C), (D)

The correct options are
A Acceleration of '$A$' relative to ground is in negative $y$-direction
B Acceleration of '$A$' relative to $B$ is in positive $x$-direction
C The horizontal acceleration of '$B$' relative to ground is in negative $x$-direction.
D The acceleration of '$B$' relative to ground directed along the inclined surface of '$C$' is greater than g sin $\theta$.

Answers are A, B, C and D
There is no horizontal force on block A, therefore it does not move in x-direction. There is net downward force (mg -N) is acting on it, making its acceleration along negative y-direction. Hence option A is correct.
Block B moves downward as well as in negative x-direction. Downward acceleration of A and B will be equal due to constrain, thus w.r.t. B, A moves in positive x-direction. Hence option B is correct.
Due to the component of normal exerted by C on B, it moves in negative x-direction. Therefore the direction of acceleration is in negative x-axis. Hence option C is correct.
Acceleration of block A and B is due to the force $Mgsin\theta$, where M is the combined mass of A and B. Hence option D is correct.