1)A ball of mass 180 g is released from rest at a height of 2.50 m above the flo
ID: 1629570 • Letter: 1
Question
1)A ball of mass 180 g is released from rest at a height of 2.50 m above the floor and it rebounds straight up to a height of 0.700 m .part c) If the contact time with the floor was 0.0750 s , what was the average force the floor exerted on the ball? i cant get the answer i have tried 25.7, 68 N and 82 N
2)A 20 g bullet moving horizontally at 400 m/s penetrates a 3.1 kg wood block resting on a horizontal surface.(Figure 1) part A)If the bullet slows down to 300 m/s after emerging from the block, what is the speed of the block immediately after the bullet emerges?
After Before 400 m/sI 300 m/sExplanation / Answer
1.) Force = Rate of change of momentum
mass of ball = 0.18 kg;
Velocity just before hitting the floor = 7m/s; by conservation of energy
velocity after hitting the ground = 3.70 m/s
Force =[ 0.18(7+3.70)] / 0.075
Force = 25.68 N
2)
The block has consumed 100 m/s velocity of bullet.
if bullet had been fred with 100 m/s it would have come to rest within
applying the law of conservation of energy
the lose in kinetic energy of bullet will lead t gain in velocity of the block
0.5x0.02(70,000) = 0.5x3.12xV^2
V = 21.18 m/s Answer
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