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The easiest approach to this problem is to solve it in COM frame. The great thing about COM frame that due to the internal forces canceling out it is an inertial frame in this case (there are no external forces).

 

Let the given initial velocity of A and B be 6 m/s = v and C is at rest. 

 

Now, we can find the velocity of COM, which comes out to be 2v/3. This velocity will remain unchanged throughout because there are no external forces acting on the system.

Let us look at the instant just after the collision. The COM is moving by velocity of 2v/3, the board A is moving with velocity v with respect to the ground frame, and thus with velocity v/3 in COM frame. This is the initial velocity of board A. 

 

Finally, the board A comes to rest with respect to the COM frame, thus its final velocity is zero.

The accelaration of block A: The accelaration of block A in ground frame is f/m where f is the force of kinetic friction ,f = (u)mg, where (u) is the coefficient of kinetic friction (u) = 0.3, and thus accelaration turns out to be a = (u)g. The accelaration of block A in COM frame is also a = (u)g, because COM frame is an inertial frame and the accelaration is same in all inertial frames.

The last piece of the puzzle is the distace travelled by the block in COM frame, by looking at the initial and final configurations, it can be found easily to be L/3, where L is the length of a board.



If we put all the values in Newton’s equation of motion v^2 = u^2 + 2 a s, 

 

we get  0 = (v^2)/3 – 2  *(u) *g* (L/3)

 

this gives L = (v^2)/(6(u)g)

 

on putting the values we get L = 2 m.

 

For lucid and elaborate solutions to problems like these please visit,

 

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