Two masses A and B having mass m and 2m respectively are kept on a smooth horizontal table as shown in the figure. B is given a velocity v perpendicular to the string parallel to the table. The tension in string is T $= \frac{2 m v ^{2}}{n ℓ}$ . Value of n is

Question

Reconstructed figure from exact-wording Kunduz source: A (m) above B (2m), vertical string ℓ, B velocity v to the right

TestHub · Accepted Answer

Immediately after B receives velocity v perpendicular to the taut string, A is at rest. The string length is ℓ.

The centre of mass lies ℓ/3 from B.

Its velocity is [m(0)+2m(v)]/(3m) = 2v/3.

B moves at speed v−2v/3 = v/3 about the centre of mass, at radius ℓ/3.

The string tension is T = (2m)(v/3)²/(ℓ/3) = 2mv²/(3ℓ).

Comparing with T = 2mv²/(nℓ), n = 3.

Physics - Center of Mass Question with Solution | TestHub

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JEE Advanced 2027 (Class 12+)