Question

To make a bounce pass, a player throws a 0.55-kg basketball toward the floor. The ball hits the floor with a speed of 5.5 m/s at an angle of 58 ∘ to the vertical.

Part A

If the ball rebounds with the same speed and angle, what was the magnitude of the impulse delivered to it by the floor?

Express your answer using two significant figures.

I =

______

  kg⋅m/s  

Answer 1

Given that mass of the ball m = 0.55kg speed of the ball v=v_i=v_f = 5.5 m/s in this case the change in momentum dp_x = 0 and dp_y = positive (initial py and down y and final py and up y) , so we can only take vertical components i.e. dV = v_i(cos58)+v_f(cos58) = 2vcos58 so change in momentum dP = m(2vcos58) ==> 2mvcos58 = 2(0.55)(5.5)(0.529) = 3.20 kgm/s From linear momentum and impulse theory, impulse = change in linear momentum therefore impulse = 3.20kgm/s