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 0.593 kg. a) If the balls are released from a height where the bottom of the basketball is at 0.941 m above the ground, what is the absolute value of the basketball

Explanation / Answer

(a) and (b) Solve using conservation of energy: PE + KE at the top = PE + KE at the bottom. PE at top = mgh KE at top = .5mv^2 = 0 since they are not moving PE at the bottom = 0 since height is zero KE at the bottom = .5mv^2 So, PEtop = KEbottom mgh = .5mv^2 gh = .5v^2 9.8(.941) = .5(v^2) BOTH the golf ball and the basketball have the same speed just before they hit the ground. (c) Elastic collision means kinetic energy is conserved. Therefore the basketball hits the floor and bounces back up with the same speed. The golf ball's speed changes as a result: Vg,f = [Mg - Mb][v]/[Mg + Mb] Where Mg = mass of golf ball, Mb = mass of basketball, v is their speed upon hitting the floor. Vg,f = [.0459 - .941][-v] / [.0459 + .941] *NOTE: V becomes negative here because it was assumed to be positive as they fell down. Now they are going up. Vg,f is the speed of the golf ball after the collision. The momentum is given by: Vg,f * Mg. (D) How will it go? Use conservation of energy again. mgh = .5m(Vg,f)^2 gh = .5(Vg,f)^2 Solve for h and add .2385 meters to represent the diameter of the basketball.

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