A long block A of mass M is at rest on a smooth horizontal surface. A small block B of mass M/2 is placed on A at one end and projected along A with same velocity v. The coefficient of friction between the block is μ. Then the acceleration of blocks A and B before reaching a common velocity will be respectively

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

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

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

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

Solution

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

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

So, acceleration of A

${μM}_{B} g = M_{A} a_{A} \Rightarrow a_{A} = μ (\frac{M_{B}}{M_{A}}) g = \frac{μg}{2} (towards right)$

Block B experiences frictions force towards the left.

MBaB = μMBg ⇒a_B = μg towards left

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