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A block A of mass $M_{A} = 1 k g$ is kept on a smooth horizontal surface and attached by a light thread to another block B of mass $M_{B} = 2 k g .$ Block B is resting on ground , and thread and pulley are massless and frictionless. A bullet of mass $m = 0.25 k g$ moving horizontally with velocity of $u = 200 m / s$ penetrates through block A and comes out with a velocity of $100 m / s$ $(g = 10 m s^{- 2})$
Column I Column II (Values are in their respective SI units)
i velocity of the 2 kg block just after the bullet
comes out
p 50/3
ii Maximum displacement of 1 kg block in left
direction
q 25
iii Impulse by the string on block B r 25/3
iv Impulse by the particle on block A s 5.2

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$(i) s; (i i) r; (i i i) p; (i v) q$

$(i) s; (i i) q; (i i i) r; (i v) p$

$(i) r; (i i) s; (i i i) p; (i v) q$

$(i) p; (i i) q; (i i i) r; (i v) s$

answer is B.

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Detailed Solution

$M_{A} = 1 k g, M_{B} = 2 k g$

$F o r b u l l e t, \int F d t = m (v_{f} - v_{i}) = 0.25 [(- 100) - (- 200)] = 25 k g m / s ....... (i)$

This is also impulse by particle on A,

For block A, $\int (F - T) d t = M_{A} (v - 0) \dots . . (i i)$

For block B, $\int T d t = M_{B} (v - 0) \dots \dots \dots . . (i i i)$

From Eqs.(ii) and (iii), we get

$\int F d t = (M_{A} + M_{B}) v = 25 = 3 v o r v = \frac{25}{3} m / s$

$\int T d t = M_{B} v = \frac{2 \times 25}{3} = \frac{50}{3}$

Now applying the law of conservation of energy, $M_{B} g h = + \frac{1}{2} (M_{A} + M_{B}) v^{2} = h = \frac{1}{2} \frac{(M_{A} + M_{B}) v^{2}}{M_{B} g} = 5.21 m$

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