A horizontal block $A$ is at rest on a smooth horizontal surface. $A$ small block $B$ , whose mass is half of $A$ , is placed on $A$ at one end and project along other end with some velocity $u$ . The coefficient of friction between blocks is $µ$ . Then

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before the block reach a common velocity, the acceleration of $A$ relative to $B$ is $(3 / 2) µ g$

the work done against friction is two-third of the initial kinetic energy of $B$

before the blocks reach a common velocity, the acceleration of $A$ relative to $B$ is $(2 / 3) µ g$

the blocks will reach a final common velocity $u / 3$

answer is A, B, D.

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Detailed Solution

From conservation of momentum,

$(\frac{m}{2}) u = (m + \frac{m}{2}) v ∴ v = \frac{u}{3}$

Work done against friction $= E_{i} - E_{f}$

$= \frac{1}{2} (\frac{m}{2}) u^{2} - \frac{1}{2} (\frac{3 m}{2}) {(\frac{u}{3})}^{2} = \frac{1}{6} m u^{2} = \frac{2}{3} (\frac{1}{4} m u^{2})$

Force of friction on the two blocks before the blocks reach a common velocity is as shown below,

$a_{1} = μ g a n d a_{2} = \frac{μ}{2} g ∴ a_{r} = \frac{3}{2} μ g$

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