Here is a popular lecture demonstration that you can perform at home. Place a go

Question

Here is a popular lecture demonstration that you can perform at home. Place a golf ball on top of a basketball, and drop the pair from rest so they fall to the ground. (For reasons that should become clear once you solve this problem, do not attempt to do this experiment inside, but outdoors instead!) With a little practice, you can achieve the situation pictured here: The golf ball stays on top of the basketball until the basketball hits the floor. The mass of the golf ball is 0.0459 kg, and the mass of the basketball is 0.613 kg.

a) If the balls are released from a height where the bottom of the basketball is at 1.101 m above the ground, what is the absolute value of the basketball’s momentum just before it hits the ground?

b) What is the absolute value of the momentum of the golf ball at this instant?


c) Treat the collision of the basketball with the floor and the collision of the golf ball with the basketball as totally elastic collisions in one dimension. What is the absolute magnitude of the momentum of the golf ball after these collisions?


d) Now comes the interesting question: How high, measured from the ground, will the golf ball bounce up after its collision with the basketball? (Hint: do not forget to add the diameter of the basketball, 23.93 cm).

Explanation / Answer

a) The momentum, 'pB', of the basket ball just before it hits ground = 0.603*sq rt(2*9.8*0.861) or pB = 0.603*4.108 = 2.477 kg m./s direction downwards b) The momentum, 'pG', of the golf ball just before it hits ground = 0.0459*sq rt(2*9.8*0.861) or pG =0.0459*4.108 =0.18856 kg m/s direction downwards c and d) Collision of basket ball with floor makes the momentum of the basket ball = 2.477kg m/s upwards At this time the golf ball is still moving downwards with momentum 0.18856 kg m/s Now we consider the collision of golf ball with basket ball Total initial momentum = 2.477 - 0.18856 = 2.288 kg m/s upwards Let Pb and Pg be the respective momenta after collision of basketball and golf balls respectively We have Pb +Pg = 2.288 ----------------------------------------… 1 and conserving kinetic energy, we have Pb^2/(2*0.806) + Pg^2/(2*0.0459) = 2.477^2/(2*0.806) + 0.18856^2/(2*0.04590) or Pb^2/(2*0.806) + Pg^2/(2*0.0459) = 3.806 0 .3873 = 4.1935 ---------------------------- 2 From 1, Pb = 2.288 - Pg; substituting in 2 weget (2.288 - Pg)^2/(1.612) + Pg^2/(0.0918) = 4.1935; Multiplying by (1.612*0.0918), we get 0.0918*[2.288^2 - 2*2.288*Pg + Pg^2] + Pg^2(1.612) = 0.62056 or 0.48057 - 0.4201*Pg + 1.7038*Pg^2 = 0.62056 or 1.70*Pg^2 - 0.42Pg - 0.14 = 0 or Pg = [+0.42+/-sq rt(0.42^2+4*1.70*0.14)]/(3.4) = [+0.42+/-1.062]/(3.4) = 0.436 or - 0.189 The second is trivial showing no collision has occured So we accept 0.436 kg m/s as the momnetum of golf ball upwards after collision Velocity of golf ball = 0.436/0.0459 = 9.5 m/s The height which golf ball reaches above the diameter of basket ball = (9.5^2)/(2*9.8) = 4.60 m or Total height reached by golf ball = 4.60+0.24 = 4.84 m

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