The plunger of a pinball machine has mass 0.440 kg and is attached to a spring o

Question

The plunger of a pinball machine has mass 0.440 kg and is attached to a spring of force constant 132.0 N/m. The spring is compressed a distance 8.0 cm from its equilibrium position x = 0 and released. A ball of mass 0.308 kg is in contact with the plunger when it is released.

1) What is x when the ball loses contact with the plunger? (Assume that the surface is horizontal and frictionless so that the ball slides, but does not roll.)

2) What is the speed vs of the ball when it leaves the plunger?

3)At what distance xf from equilibrium will the plunger come to rest momentarily?

Explanation / Answer

ball loses contact when spring attains it's natural length

x = 0.08 m

at x = 0.08 m total spring energy is converted to kinetic energy

0.5kx2 = 0.5(0.44 + 0.308)v2

v = 1.06 m/s

now the plunger's kinetic energy is converted to spring energy

now elongation in spring is x1

0.5kx12 = 0.5*0.44*1.062

x1 = 0.06 m

xf = x + x1 = 0.14m

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