After falling from rest from a height of 35 m, a 0.52 kg ball rebounds upward, reaching a height of 25 m. If the contact between ball and ground lasted 2.5 ms, what average force was exerted on the ball?
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f =10047.47952 N; rounded to two sig figs = 1.0 x10 ^4 N
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Using Kinematics:
d=1/2gt^2; use this formula to find time: 35m =1/2(9.8m/s^2)t^2 = 2.672612 seconds to fall
v=gt; now that we have time solve for velocity v= ( -9.8m/s^2) 2.672612 = -26.19160 m/s down
d=1/2gt^2; use this to find the time it took for the ball to bounce up so we can then find the velocity in which it went back up after the bounce.
-25 m =-4.9m/s^2T^2 = 2.2587697 s now use this time to find the upward velocity after it bounced.
v=gt: 9.8m/s^2 * 2.2587697 s = + 22.13594 m/s up
The ball had a change in momentum when it bounced; a change in momentum is called an impulse.
One formula for impulse is m(vf - v0) = f*t; (.52 kg)[22.13594m/s - (-26.19160m/s)] = f(2.5 x 10^-3s)
solve for f =10047.47952 N = 1.0 x10 ^4 N
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Check this using law of conservation of energy:
P.E. = K.E or mgh = 1/2mv^2
.52Kg(9.8m/s^2)(35m) = 1/2(.52kg)v^2; v = -26.19160 m/s falling down
.52Kg(9.8m/s^2)(25m) = 1/2(.52kg)v^2; v = 22.13594 m/s going up
plug these into the impulse formula and it will be the same answer.
Throw me a bone here!
***************************************…
Using Kinematics:
d=1/2gt^2; use this formula to find time: 35m =1/2(9.8m/s^2)t^2 = 2.672612 seconds to fall
v=gt; now that we have time solve for velocity v= ( -9.8m/s^2) 2.672612 = -26.19160 m/s down
d=1/2gt^2; use this to find the time it took for the ball to bounce up so we can then find the velocity in which it went back up after the bounce.
-25 m =-4.9m/s^2T^2 = 2.2587697 s now use this time to find the upward velocity after it bounced.
v=gt: 9.8m/s^2 * 2.2587697 s = + 22.13594 m/s up
The ball had a change in momentum when it bounced; a change in momentum is called an impulse.
One formula for impulse is m(vf - v0) = f*t; (.52 kg)[22.13594m/s - (-26.19160m/s)] = f(2.5 x 10^-3s)
solve for f =10047.47952 N = 1.0 x10 ^4 N
***************************************…
Check this using law of conservation of energy:
P.E. = K.E or mgh = 1/2mv^2
.52Kg(9.8m/s^2)(35m) = 1/2(.52kg)v^2; v = -26.19160 m/s falling down
.52Kg(9.8m/s^2)(25m) = 1/2(.52kg)v^2; v = 22.13594 m/s going up
plug these into the impulse formula and it will be the same answer.
Throw me a bone here!
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Thanks, I know you are not taking my AP Physics, because we are only still in Kinematics. I have to speed them up, but I have an AP and intro Paul Hewitt combo class. It is difficult to teach 2 courses in one, but I only had 18 AP physics students this year. They cancelled the class,
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