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Question

A long 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 moved along A with some velocity u. The coefficient of friction between the blocks is μ :
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A
the blocks will reach a final common velocity u3
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B
the work done against friction is two-thirds of the initial kinetic energy of B
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C
before the blocks reach a common velocity, the acceleration of A relative to B is 23μg
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D
before the blocks reach a common velocity the acceleration of A relative to B is 32μg
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Solution

The correct options are
A the work done against friction is two-thirds of the initial kinetic energy of B
C the blocks will reach a final common velocity u3
D before the blocks reach a common velocity the acceleration of A relative to B is 32μg
As there are no external forces acting on the 'A + B' system, its total momentum is conserved. If the masses of A and B are 2m and 2 respectively, and v is the final common velocity, mu=(m+2m)v or v=u/3

Work done against friction = loss in KE =12mu212(3m)v2

=12mu212(3m)u29=12mu2[113]=23×12mu2

The force of friction between the blocks is μmg.

Acceleration of A (to the right) =a1=μmg2m=μg2

Acceleration of B (to the left) =a2=μmgm=μg

Acceleration of A relative to B =a1(a2)=32μg.

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