A bouncing ball of mass 200 ~g falls from the height of 5 ~m on a horizontal ground. After every impact with the ground, the velocity of the ball decreases by 1 2 times. The total momentum, the ball imparts on to the ground after 3 impacts is ( . Let .g=10 ~m / s ^2 )

Step-by-step solution

Given, mass of ball, m=200 ~g =0.2 ~kg and height of the ball, h=5 ~m Velocity of ball at the time of first impact, v₁^ = v₁ 2 = 10 2 =5 ~m / s Momentum imparted by the ball in first impact, aligned p₁ & =m v₁-m (-v₁^ ) & =m v₁+m v₁^ aligned aligned & =0.2 10+0.2 5 p₁ & =3 ~kg ~m / s aligned Similarly, velocity of the ball at the time of second impact and its recoil velocity will be 5 ~m / s and 5 2 ~m / s . Hence, momentum imparted by ball in second impact, aligned p₂ & =m 5-m (- 5 2 ) & =0.2 5+0.2 5 2 =1.5= 3 2 ~