A ball of mass 0.150 kg is dropped from rest to a height of 0.125 m. It rebounds from the floor to reach a height of 0.960m. What impulse was given by the floor?
please help/explain. Thanks!
2 answers
I know that momemtum (p) is equal to mass times velocity, but I am not sure how to get the velocity.
use the same formula you used to get the velocity falling from 1.25 meters to get the velocity it would get falling from 0.96m. The last is identical to the velocity needed to make it go to that height, by laws of conservation of energy.
In other words, if you dropped an object from a height H and it hit with a speed of V, that is the same set of numbers as if you propelled it upwards with a velocity of V to a height of H.
v = �ã(2gh) = �ã(2(9.8)(1.25)) = 4.95 m/s when it hits the ground
from 0.96 that is �ã(2(9.8)(0.96)) = 4.34 m/s
since the mass doesn't change
I = P1 - P2 = m(V1 - V2) = 0.0915
In other words, if you dropped an object from a height H and it hit with a speed of V, that is the same set of numbers as if you propelled it upwards with a velocity of V to a height of H.
v = �ã(2gh) = �ã(2(9.8)(1.25)) = 4.95 m/s when it hits the ground
from 0.96 that is �ã(2(9.8)(0.96)) = 4.34 m/s
since the mass doesn't change
I = P1 - P2 = m(V1 - V2) = 0.0915
1 answer