Two identical blocks A and B, connected by a massless string, are placed on a frictionless horizontal plane. A bullet having same mass, moving with speed u, strikes block B from behind as shown. Id the bullet gets embedded into block B, then match the following.
List 1 List 2
I) The velocity of A, B, C after collision P) $\frac{2 m u}{3}$
II) Impulse on A due to tension in the string Q) $\frac{m u}{3}$
III) Impulse on C due to normal force of collision R) $\frac{u}{3}$
IV) Impulse on B due to normal force of collision S) $- \frac{2 m u}{3}$
T) $- \frac{m u}{3}$

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$(I) \to (T); (II) \to (R); (III) \to (S); (IV) \to (P)$

$(I) \to (S); (II) \to (P); (III) \to (Q); (IV) \to (R)$

$(I) \to (R); (II) \to (Q); (III) \to (S); (IV) \to (P)$

$(I) \to (R); (II) \to (Q); (III) \to (T); (IV) \to (P)$

answer is C.

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Detailed Solution

Bullet get embedded in the block

$V_{B C} = \frac{m a}{2 m} = \frac{u}{2}$

For B, C collision $J_{N} = m u + \frac{m u}{2}, P_{t} = \frac{m u}{2}$

$\begin{array}{l} {(J_{N})}_{c} = \frac{m u}{2}; (J_{N}) \to B \to P_{i} = 0 \\ P_{f} = \frac{m u}{2}; {(J_{N})}_{B} = \frac{m u}{2} \end{array}$

After collision A, B, C move with common velocity J = mv

$\begin{array}{l} J = 2 m (\frac{u}{2}) - 2 m V; m V = m u - 2 m V \Rightarrow = \frac{u}{2} \\ ∴ a \to (V_{f}) = \frac{4}{3}; b \to J_{A} = \frac{m u}{2}; \\ c \to \frac{- 2 m u}{3}; d \to \frac{2 m u}{3} \end{array}$

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List 1	List 2
I) The velocity of A, B, C after collision	P) $\frac{2 m u}{3}$
II) Impulse on A due to tension in the string	Q) $\frac{m u}{3}$
III) Impulse on C due to normal force of collision	R) $\frac{u}{3}$
IV) Impulse on B due to normal force of collision	S) $- \frac{2 m u}{3}$
	T) $- \frac{m u}{3}$