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 0.619 kg.

Explanation / Answer

Given data The mass of the basketball is, M = 0.619 kg. The mass of the golf ball is, m = 0.0459 kg Basket ball is initially at a height h = 0.634 m above the ground (a) The velocity of the basket ball just before it hits the ground is,                           v = 2gh                              = 2(9.8 m/s2) (0.634 m)                              = 3.52 m/s The momentum of the basket ball is,                         p1 = M v                              = ( 0.619 kg) (3.52 m/s)                              = 2.17 kgm/s ------------------------------------------------------------------------------ (b) The momentum of the golf ball is,                         p2 = m v                              = ( 0.0459 kg) (3.52 m/s)                              = 0.16 kgm/s ------------------------------------------------------------------------------- (c) Since, the collision is elastic, the momentum remains conserved. That is, the initial momentum of the golf ball is equal to its final momentum. Hence, the final momentum = 0.16 kgm/s -------------------------------------------------------------------------------- (d) Let the height reached by the golf ball after collision be h'      Now, the ball rebounces with a velocity of v = 3.52 m/s      So, by using kinematics the equations, we can determine h'                      vf2 - v2 = 2gh'          0 - (3.52 m/s)2 = 2(-9.8 m/s2 ) h'                                h' = 0.63 m                         p2 = m v                              = ( 0.0459 kg) (3.52 m/s)                              = 0.16 kgm/s ------------------------------------------------------------------------------- (c) Since, the collision is elastic, the momentum remains conserved. That is, the initial momentum of the golf ball is equal to its final momentum. Hence, the final momentum = 0.16 kgm/s -------------------------------------------------------------------------------- (d) Let the height reached by the golf ball after collision be h'      Now, the ball rebounces with a velocity of v = 3.52 m/s      So, by using kinematics the equations, we can determine h'                      vf2 - v2 = 2gh'          0 - (3.52 m/s)2 = 2(-9.8 m/s2 ) h'                                h' = 0.63 m                       

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