Question

A ball of mass 100 kg moving with velocity 72 km / hr strikes a wall normally rebound back with same speed 72 km / hr . Then find out impulse on the ball by the wall.​

Answer 1

Answer:

Initial normal velocity of ball toward horizontal surface, u=4sin30  

o

=2ms  

−1

 

Final normal velocity of ball toward horizontal surface, v=−4sin30  

o

=−2ms  

−1

 

Impulse imparted to ball, I=F.Δt=mΔV=m(v−u)=0.1[2−(−2)]=0.4kgms  

−1

 

Change in momentum, P=mΔV=0.4kgms  

−1

 

Explanation:

Answer 2

Answer:

$4 \times {10}^{3}$

Explanation:

Impulse=Change In Momentum.

Here the ball is colliding normally;I.e;at an angle 90°

So while bouncing back,it must be at 90°.

It means initial velocity is opposite of final.

Vi=-Vf