1. A 58 g steel ball is released from rest and falls vertically onto a steel pla
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1. A 58 g steel ball is released from rest and falls vertically onto a steel plate. The ball strikes the plate and is in contact with it for 0.5 ms. The ball rebounds elastically, and returns to its original height. The time interval for a round trip is 7 s. In this situation, the average force exerted on the ball during contact with the plate is closest to:
6620 N
2. A 25 g steel ball is released from rest and falls vertically onto a steel plate. The ball strikes the plate and is in contact with it for 0.5 ms. The ball rebounds elastically, and returns to its original height. The time interval for a round trip is 4 s. In this situation, assume the plate does not deform during contact. The maximum elastic energy stored by the ball is closest to:
A 25 g steel ball is released from rest and falls vertically onto a steel plate. The ball strikes the plate and is in contact with it for 0.5 ms. The ball rebounds elastically, and returns to its original height. The time interval for a round trip is 4 s. In this situation, assume the plate does not deform during contact. The maximum elastic energy stored by the ball is closest to:
9300 N 3980 N 7960 N 5320 N6620 N
2. A 25 g steel ball is released from rest and falls vertically onto a steel plate. The ball strikes the plate and is in contact with it for 0.5 ms. The ball rebounds elastically, and returns to its original height. The time interval for a round trip is 4 s. In this situation, assume the plate does not deform during contact. The maximum elastic energy stored by the ball is closest to:
A 25 g steel ball is released from rest and falls vertically onto a steel plate. The ball strikes the plate and is in contact with it for 0.5 ms. The ball rebounds elastically, and returns to its original height. The time interval for a round trip is 4 s. In this situation, assume the plate does not deform during contact. The maximum elastic energy stored by the ball is closest to:
19 J 9.6 J 4.8 J 14 J 7.2 JExplanation / Answer
1) the time taken for the ball to hit the ground is 7/2 = 3.5 S
speed of the ball when it was hitting the ground is v = u+(g*t)
u is the initial speed = 0 m/s
g is accelaration due to gravity = 9.8 m/s^2
t is the time taken = 3.5 S
then v = 0+(9.8*3.5) = 34.3 m/s
with this same speed it would rebounces back
hence from the Formulas of impulase J = F*dt = m*(v+v) = 2*m*v
Rrequired average force F = 2*m*v/dt = (2*0.058*34.3)/(0.5*10^-3) = 7957.6 N = 7960 N
2) From law of conservation of energy
Total Energy at the releasing point = Total energy at the hitting point of the plate
when the ball was relasing it has maximum elastic energy
when the ball is strikeing the plate it has maximum kinetic enrgy
so From law of conservatuion of energy
maximum elastic energy stroed = maximum kinetic energy = 0.5*m*v^2
v is the speed of the ball when it was striking the plate = u+(g*t)
u is the initila speed = 0 m/s
g is accelaration due to gravity = 9.8 m/s^2
t is the tima taken to reach the plate = 4/2 = 2 S
then v = 0+(9.8*2) = 19.6 m/s
then maximum elastic energy = 0.5*m*v^2 = 0.5*0.025*19.6^2 = 4.8 J
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