Q: Two smooth blocks A of mass 1 kg and B of mass 2 kg are connected by alight string passing over a smooth pulley as shown. The block B is sliding down with a velocity 2 m/s. A force F is applied on the block A so that the block B will reverse its direction of motion after 3 s.

(i) Find the acceleration of block A.

(a) 1 m/s²

(b) 2/3 m/s²

(c) 3 m/s²

(d) 4/9 m/s²

Ans: (b)

(ii) Find the tension in the string.

(a) 10 N

(b) 11.33 N

(c) 13.67 N

(d) 40 N

Ans: (b)

Sol:(i) When direction of motion will reverse at the instant, the velocity becomes zero.

∵ v = u + at

Or 0 = 2 + a×3

a = -2/3 m/s^{2}

Thus, acceleration of A is 2/3 m/s^{2} in rightward direction.

Sol:(ii)

From FBD of A,

F -T = m_{1} a = 1 × 2/3 ….(i)

From FBD of B, ….(ii)

T – m_{2} g sinθ = m_{2} a

= 2 × 2/3 = 4/3

or, T-10 = 4/3

T= 10 + 4/3 = 34/3

=11.33 N