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?
Question
Answer:
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