Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
A ball reaches a racket at 60 m s-1 along +x direction and leaves the rocket in the opposite direction with the same speed. Assuming that the mass of the ball as 50 g and the contact time is 0.02 second, the force exerted by the racket on the ball is
PhysicsCenter of Mass Momentum and CollisionNEET
Options:
  • A 300 N along + X direction
     
  • B 300 N along - X direction
     
  • C 300 kN along + X direction
     
  • D 300 kN along - X direction
     
Solution:
1540 Upvotes Verified Answer
The correct answer is: 300 N along - X direction
 
The change in momentum of the ball, Δp=mv--mv=2mv=2×0·05×60=6 kg m s-1

Force exerted on the bell by the rocket, F = Δ p Δ t = 6 0 · 0 2 = 3 0 N

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.