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A billiard ball moving with a velocity 'v' collides on a rigid plate moving towards the ball with a velocity 'u'. If the elastic impact lasts for 't' seconds, then the mean elastic force acting on the ball is -


1. \(\frac{2mu}{t}\)
2. \(\frac{2mv}{t}\)
3. \(\frac{2m(v+u)}{t}\)
4. \(\frac{2m(v-u)}{t}\)

1 Answer

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Best answer
Correct Answer - Option 3 : \(\frac{2m(v+u)}{t}\)

The correct answer is option 3) i.e. \(\frac{2m(v+u)}{t}\)

CONCEPT:

  • Elastic collision: Elastic collision is a phenomenon where the collision of objects takes place such that the total linear momentum and kinetic energy of the system are conserved.

​EXPLANATION:

  • Since this is an elastic collision, linear momentum is said to be conserved

As the ball moves forward and the plate moves towards it, the relative velocity of ball = v - (-u) = v + u

After the collision, the relative velocity of the ball = - (v+u) = -v - u

From Newton's second law of motion\(F = \frac{m\Delta v}{t}\)

\(\Rightarrow F = \frac{m[(v+u )-(-v-u)]}{t}\)

\(\Rightarrow F = \frac{2m(v+u )}{t}\)

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