The blocks A and B have equal masses. The surface of A is smooth but that of B has a friction coefficient of 0.1 with the floor. Block A is moving at a speed of 10 m/s towards B which is kept at rest. Find the distance travelled by B if (a) the collision is perfectly elastic and (b) the collision is perfectly inelastic. g=10 m/s2 .

a) In a lab frame of reference,

    Velocity of center of mass is given by,

       v'=(mvOA+ mvOB)/2m

Now, in center of mass frame of reference,

Since, there will be perfectly elastic collision. After collision, they exchange their velocity.

In lab frame of reference,

Deceleration due to the friction = µg

From kinematics we have,

VF=VB- µgt

Since the final velocity of B is zero, we get t=10s.

Distance travelled d is given by,

d = VBt - (µgt2)/2

Inserting given values, the distance travelled by block is 50m.

b) After perfectly inelastic collision, both blocks A and B travel as a combined mass.


   We get, initial velocity of the combined mass 5m/s.

    The friction of the block B will have affect whole body.

     Then we have,

     VF=VAB- µgt

     d = VABt - (µgt2)/2

     Solving the equations, the distance travelled by the body is 12.5m.

Work-Energy Theorem can also be used to find the distance.