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 projected along A with some velocity u. The coefficient of friction between the block is μ.

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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 the final common velocity,

Using conservation of momentum,

mu=(m+2m)v,

v=u3

Work done by friction is change in loss of kinetic energy, so

12mu2−12(3m)u29=12mu2[1−13]=23×12mu2.

The work done agianst friction is two - thirds of the initial kinetic energy of B

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.

Let v be the final common velocity,

Using conservation of momentum,

mu=(m+2m)v,

v=u3

Work done by friction is change in loss of kinetic energy, so

12mu2−12(3m)u29=12mu2[1−13]=23×12mu2.

The work done agianst friction is two - thirds of the initial kinetic energy of B

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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