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NTA Abhyas JEE Main2020PhysicsCenter of Mass, Momentum and CollisionPractice

A ball starts falling under the effect of gravitational force from a height of 45 m . When it reaches a height of 25 m it explodes into two pieces of mass ratio 1 : 2 . There is no change in the vertical motion of the pieces after the explosion but they acquire horizontal velocity. If the heavier piece gains a horizontal velocity of 10 m s - 1 , then the distance between the two pieces when both of them strike the gr

Options

  1. A30 m
  2. B10   m
  3. C20   m
  4. D15   m

Correct answer

A. 30 m

Step-by-step solution

Let us assume the mass of the ball is 3 m and the velocity of the lighter piece in horizontal direction after the explosion is v . We know that just after the collision, the horizontal velocity of the heavier piece is 10   m   s - 1 . Using conservation of momentum in the horizontal direction we get m v = 2 m × 10 ⇒ v = 20   m   s - 1 Since there is no change in vertical motion, the time taken by the pieces to reach the ground can be calculated in the following way. Total time to fall

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