NTA Abhyas JEE Main2020PhysicsCenter of Mass, Momentum and CollisionPractice
A massive vertical wall is approaching a man at a speed u . When it is at a distance of 10   m , the man throws a ball with speed 10   m / s at an angle of 37 ° , which after having a completely elastic collision with the wall, reaches back directly into the hands of the man. The velocity of the wall is
Options
- A13 3 m/s
- B18 2 m/s
- C26 4 m/s
- D31 5 m/s
Correct answer
A. 13 3 m/s
Step-by-step solution
As the vertical motion remains unaffected T = 2 u sin θ g ⇒ T = 2 × 10 × 3/5 10 T = 6 5 s     ...(1) from coefficient of restitutation along the line of collision e = relative velocity of separation after collision relative velocity of approach before collision 1 = v - u 8 + u v - 2u = 8 ...(2) by the time (t 1 ) ball collide with wall, distance travelled by man = 10 - ut 1 t 1 = 10 8 + u ...(3) by the time t 2 ball reach the hand of man t 2 = 10 - u 1 v ...(4) Since T = t