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NTA Abhyas JEE Main2020PhysicsCenter of Mass, Momentum and CollisionPractice

Two point masses connected by an ideal string are placed on a smooth horizontal surface as shown in the diagram. A sharp impulse of 10   k g  m   s - 1 is given to the 5   k g mass at an angle of 60 ° to the line joining the masses. The velocity of the 10   k g mass just after the impulse will be

Options

  1. A2 3   m   s - 1
  2. B1 3 m s - 1
  3. C2   m   s - 1
  4. Dzero

Correct answer

B. 1 3 m s - 1

Step-by-step solution

Apply impulse equation, along string Impulse = Δ p 10 × 1 2 = 10 + 5 v v = 1 3 m s - 1

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