Two blocks, A and B are on a frictionless table. At t = t0 - ?t, the blocks move

Question

Two blocks, A and B are on a frictionless table. At t = t0 - ?t, the blocks move towrds each other with constant speed as shown. Block B moves with a constant speed twice that of block A. The blocks will collide at t = t0; at time t = t0 + ?t (shortly after the collision) both blocks move to the left with constant speed. The final velocity of block A is that of block B.

From time t = t0 - ?t until t = to + ?t is the magnitude of the acceleration of block A greater than, less than, or equal to the magnitude of the acceleration of block B? Explain.

Explanation / Answer

let the velocity of B is Vb =-v(-ve since travelling towards -ve x direction)

=> the velocity of A is therefore = Va = +2v

now the final velocity of the block A is -u(let, -ve sine travelling left)

and the final velocity of B = -u.

now the acceleration of A is given by aA =Va/T. = -u-2v/2t.

now the accleration of B is given by aB = Vb/T = -u+v/2t.

now the magnitude of acceleration of A is (u+2v/2t) and that of B is (v-u/2t).

it can thus be clearly seen that magnitude of acceleration of A is greater than that of B.

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