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A long block A of mass M is at rest on a smooth horizontal surface. A small block B of mass $\frac{M}{2}$ is placed on A at one end and projected along A with some velocity v. The coefficient of friction between the blocks is $μ$ . The accelerations of blocks A and B before reaching a common velocity will be respectively

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$\frac{μg}{2}$ (towards right), $μg$ (towards left)

$μg$ (towards right), $\frac{μg}{2}$ (towards left)

$μg$ (towards right), $μg$ (towards left)

$\frac{μg}{2}$ (towards right), $\frac{μg}{2}$ (towards left)

answer is C.

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Detailed Solution

The force causing the motion of A is frictional force between A and B,

${μM}_{B} g = M_{A} a_{A}$

So, acceleration of A,

$a_{A} = μ (\frac{M_{B}}{M_{A}}) g$

$= \frac{μg}{2} (towards right)$

Block B experiences friction force toward left,

$M_{B} a_{B} = {μM}_{B} g$

$\Rightarrow a_{B} = μg (towards left)$

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