A ball of mass 0.120 kg is dropped from rest from a height of 1.25 m. It rebounds from the floor to reach a height of 0.600 m. What impulse was given to the ball by the floor?
2006-10-23 06:58:27 · 2 answers · asked by activegirl 1 in Science & Mathematics ➔ Physics
Impulse is just change in momentum. Start by calculating the ball's speed when it hit the ground. Use v = sqrt(2gy) where g is 9.8 m/s^2 and y is the height it was dropped from. To find the upwards velocity after hitting the floor, use v = sqrt(2gy) again, but with the rebound height of 0.6 m instead. The difference in speeds will be the original downwards velocity minus the new upwards velocity, which will be addition of the magnitudes of the two velocities. Multiply that difference by the mass, which is 0.120 kg, to get the chance in momentum, which is the impulse.
2006-10-23 07:09:32 · answer #1 · answered by DavidK93 7 · 0⤊ 0⤋
stumble on the momentum on the time of the drop, so clean up for the cost appropriate via fact it hits the floor v^2=2(9.8)(a million.05) v=4.53m/s Momentum=mv=.sixty 4*4.fifty 3=2.9kgm/s Then stumble on the cost it is going to have rebounded at to holiday .63m v^2=2(9.8)(.sixty 3) v=3.51m/s Momentum=mv=.sixty 4*3.51m/s=2.25kgm/s Impulse=p0-p1=2.9-2.25=.sixty 5
2016-11-25 00:25:41 · answer #2 · answered by ? 4 · 0⤊ 0⤋