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NTA Abhyas JEE Main2020PhysicsCenter of Mass, Momentum and CollisionPractice

A block A of mass m is in equilibrium after being suspended from the ceiling with the help of a spring of force constant k . The block B of mass m , strikes the block A with a speed v and sticks to it. The value of v for which the spring just attains its natural length is

Options

  1. A6 m g 2 k
  2. B3 m g 2 k
  3. C2 m g 2 k
  4. Dm g 2 2 k

Correct answer

A. 6 m g 2 k

Step-by-step solution

Let the velocity of the combined system just after the collision is v ' , then using conservation of linear momentum we get m v = 2 m v ′ , ⇒ v ′ = v 2 Initial elongation of the spring is x   = m g k Now, using the conservation of mechanical energy we get 1 2 2 m v ' 2 + 1 2 k x 2 = 2 m g x m v 2 4 + m 2 g 2 2 k = 2 m 2 g 2 k v = 6 m g 2 k

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