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NTA Abhyas JEE Main2020PhysicsCenter of Mass, Momentum and CollisionPractice

Two masses A and B connected with an inextensible string of length l lie on a smooth horizontal plane. A is given a velocity of v m s - 1 along the ground perpendicular to line AB as shown in the figure. Find the tension in a string during their subsequence motion.

Options

  1. A2 m v 2 3 l
  2. B3 m v 2 2 l
  3. Cm v 2 2 l
  4. D4 m v 2 3 l

Correct answer

A. 2 m v 2 3 l

Step-by-step solution

From conservation of linear momentum ​ v cm = m × 0 + 2mv m + 2m ⇒ v cm = 2v 3 position of the centre of mass y cm = 0 × 2m + m × l 2m ⇒ y cm = l 3 blocks will perform circular motion in the center of mass frame velocity of A in COM frame is v A ′ = v - 2 v/3 ⇒ v A ′ = v/3 Tension will provide the required centripetal force T = 2m v A ′ 2 ℓ / 3 ⇒ T = 2m v/3 2 ℓ / 3 ⇒ T = 2mv 2 3 ℓ

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