You release a steel ball with a mass of 2 kg. from rest at point A one meter above the center of a steel plate, which has a mass of 8 kg. The plate is suspended by four bungee cords, each having a spring constant k = 100 N/m. The coefficient of restitution between the ball and the plate, e, is 0.9. How far down will the plate travel after impact, and how high above the plate’s initial suspended position will the ball rebound?

Explanation:

Mass of steel ball is 2kg

m=2kg

The steel ball is release from rest at a height h

H=1m

To a hit a steel plate of mass 8kg

M=8kg

Spring constant k=100N/m

Coefficient of restitution e =0.9

Coefficient of restitution is also give as

e =√(P.E after collision / P.E before collision)

Then, the P.E before collision is given as

P.E(before) = mgH=2×9.81×1 =19.62J

P.E(before) =19.62J

The P.E after collision can be model as the potential of the spring and it is given as

P.E(after) = ½kx²

P.E(after) = ½ ×100x²

P.E(after) = 50x²

Then, e = √P.E(after)/P.E(before)

0.9=√(50x²/19.62)

Square both sides

0.9² = 50x²/19.62

0.81 = 2.5484x²

x² = 0.81/2.5484

x² = 0.3178

x=√0.3178

x = 0.564m

The plate travelled 0.564m after impact

2. Coefficient of restitution is also give as

e =√(P.E after collision / P.E before collision)

P.E(after) =mgh

P.E(after) =2×9.81×h

P.E(after) =19.62h

where h is the rebound height

Now, the potential energy before collision is the sum of the spring potential energy and the potential energy of ball

P.E(before) = mgH +½kx²

P.E(before) = 2×9.81×1 + ½×100×0.564²

P.E(before) = 19.62 + 15.89

P.E(before) =35.51J

Now applying the formula

e = √P.E(after)/P.E(before)

0.9 =√19.62h/35.51

Square both sides

0.9² = 0.55h

0.81 =0.55h

h=0.81/0.55

h=1.47m

This is due to the fact that the spring will send the ball higher

f=ma newton's second law 7x8=56n

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