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

A
The blocks will reach the 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 option is D Before the blocks reach a common velocity the acceleration of A relative to B is 32μg.
Let v be common velocity of the blocks
Using conservation of linear momentum, since friction is internal for the 2 block system,
m2u=(m+m2)v
v=u3
Initial kinetic energy of the system
ki=12(m2)u2
ki=14mu2
Using Work energy theorem,
mu24wf=12(3m2)u29
14mu2112mu2=wf
wf=3mu2mu212=mu26=23×14mu2
wf=23ki
Force of friction between blocks
fk=μ(m2)g
Acceleration of A to right.
aA=μmg2(m)=μg2
Acceleration of B to left,
aB=μmg2(m2)=μg
Acceleration of A relative to B,
aAB=aA(aB)
aAB=μg2+μg
aAB=3μg2

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