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ANSWER::

Two friends A and B (each weighing 40 kg) are sitting on a frictionless PLATFORM some distance d apart. A rolls a ball of mass 4 kg on the platform towards B which B catches. Then B rolls the ball towards A and A catches it. The ball keeps on moving back and forth between A and B. The ball has a fixed speed of 5 m/s on the platform.

(a) CASE - 1 => Total momentum of the man A and the ball will remain constant

Therefore ,

0 = 4 x 5 - 40 x v

v = 0.5 m/s towards left

(b) CASE - 2 => When B catches the ball , the momentum between the B and the ball will remain constant.

4 x 5 = 44v

v = 20/44 m/s

CASE - 3 => When B throws the ball , then applying LAW of Conservation of Linear Momentum (L.C.L.M.)

44 x (20/44) = -4 x 5 + 40 x v

20 = -20 + 40v

v = 40/40 = 1 m/s towards right

CASE - 4 => When A catches the ball , then applying L.C.L.M.

-4 x 5 + (-0.5) x 40 = -44 v

-20 - 20 = -44 v

v = 10/11 m/s towards left

(c) CASE - 5 => When A throws the ball , then applying L.C.L.M.

44 x (10/11) = 4 x 5 - 40 x v

v = 60/40

v = 3/2 m/s towards left

CASE - 6 => When B receives the ball , then applying L.C.L.M.

40 x 1 + 4 x 5 = 44 x v

v = 60/44

v = 15/11 m/s towards right

CASE - 7 => When B throws the ball , then applying L.C.L.M.

44 x (66/44) = -4 x 5 + 40 x v

v = 80/40

v = 2 m/s towards right

CASE - 8 => When A catches the ball , then applying L.C.L.M.

-4 x 5 -40 x (3/2) = -44 v

v = 80/44

v = 20/11 m/s towards left

Similarly after 5 round trips

Velocity of A will be 50/11 and velocity of B will be 5 m/s

(d) Since after 6th round trip , the velocity of A is 60/11 m/s > 5 m/s .

So , it can't catch the ball .

So , it can only roll the ball six.

(e) LET the ball and the body A at the initial POSITION be at origin.

Centre of mass of the system =( 40 x 0 + 4 x 0 + 40 x d ) / (40 + 40 + 4 )

= 40d / 84

= 10d/11

Hope it helps!