In the figure shown, upper block (1 kg) is given a velocity 6 m/s and very long plank (2 kg), velocity 3 m/s. The following quantities are to be matched when both attain the same velocity. List - I List - II (A) Work done by friction on 1 kg block in Joule (P) Positive (B) Work done by friction on 2 kg plank in Joule (Q) Negative (C) Change in momentum in N-s of 2 kg plank (R) 3 (D) Change in K.E. of system consisting of block and plank in joule (S) 7 (T) 2

Common velocity = 3 1 ( 6 ) + 2 ( 3 ) = 4 m / s . Answer: (A) - Q; (B) - P, S; (C) - P, T; (D) - Q, R

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Physics - Center of Mass Question with Solution | TestHub

PhysicsCenter of MassMomentum conservationMedium3 minQB

PhysicsMediummatching list

In the figure shown, upper block (1 kg) is given a velocity 6 m/s and very long plank (2 kg), velocity 3 m/s. The following quantities are to be matched when both attain the same velocity.

List - I	List - II
(A) Work done by friction on 1 kg block in Joule	(P) Positive
(B) Work done by friction on 2 kg plank in Joule	(Q) Negative
(C) Change in momentum in N-s of 2 kg plank	(R) 3
(D) Change in K.E. of system consisting of block and plank in joule	(S) 7
	(T) 2

Options:

Answer:

Solution:

Common velocity $= \frac{1 ( 6 ) + 2 ( 3 )}{3} = 4 m / s$ .

Answer: (A) - Q; (B) - P, S; (C) - P, T; (D) - Q, R

Stream:JEESubject:PhysicsTopic:Center of MassSubtopic:Momentum conservation

⏱ 3mℹ️ Source: QB

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