Question

A basketball has a mass of 606 g. Moving to the right and heading downward at an angle of 29° to the vertical, it hits the floor with a speed of 3 m/s and bounces up with nearly the same speed, again moving to the right at an angle of 29° to the vertical. What was the momentum change Δp→ ? (Take the +x axis to be to the right and the +y axis to be up. Express your answer in vector form.)

Answer 1

m=606 g =0.606 kg 0=29° v=3 m/s The initial momentum of the basketball is, P; = mv sin i – mv cos Oj =(0.606 kg)(3 m/s)(sin 29°) { – (0.606 kg)(3 m/s)(cos 29°) = (0.606 kg)(3 m/s)(0.484809) ì -(0.606 kg)(3 m/s)(0.87462) j = (0.88 kgm/s)ì -(1.58 kg . m/s)ì The final momentum of the basketball is, Ď, = mvsin 0 i + mv cos oſ = (0.606 kg)(3 m/s)(sin 29° ) +(0.606 kg)(3 m/s)(cos 29°) = (0.606 kg)(3 m/s)(0.484809) { +(0.606 kg)(3 m/s)(0.87462) | =(0.88 kg-m/s)i +(1.58 kg. m/s) j The change in momentum of the basketball is, Ap=P:-P = ((0.88 kg. m/s) î +(1.58 kg.m/s) j)-((0.88 kg.m/s) î – (1.58 kg. m/s) j) = (0.88 kg.m/s) { +(1.58 kg.m/s) -(0.88 kg.m/s) { +(1.58 kg.m/s) j =(0 kg. m/s) î +(3.16 kg. m/s)ſ Therefore, the change in momentum of the basketball in vector form (0 kg.m/s) { +(3.16 kg-m/s);|. is