If I drop something that weighs 100 g from a height of 2.0m andwhen it hits the
ID: 1727229 • Letter: I
Question
If I drop something that weighs 100 g from a height of 2.0m andwhen it hits the floor it comes back up (1.5m).I have found the momentum before it hits the floor to be equalto 626N How do I find the momentum after it hits the floor?
I also need to solve for the force of the floor exerted on theobject? If I know that force = change in momentum/change in time then can I say that that Force = momentum final - momentuminitial/change in time, if that is correct than what force do I setit equal to, the force exerted by the floor?
The force of floor I think is solved byF=(100kg)(6.26m/s)/.639 seconds = 979.7N (is that correct?)
I have found the momentum before it hits the floor to be equalto 626N How do I find the momentum after it hits the floor?
I also need to solve for the force of the floor exerted on theobject? If I know that force = change in momentum/change in time then can I say that that Force = momentum final - momentuminitial/change in time, if that is correct than what force do I setit equal to, the force exerted by the floor?
The force of floor I think is solved byF=(100kg)(6.26m/s)/.639 seconds = 979.7N (is that correct?)
Explanation / Answer
mass m = 100 g = 0.1 kg initial height h = 2 m Speed of the object when it just hits the floor u =[ 2gh ] = 6.26 m / s Momentum before just hits the floor P = mu = 0.626 kg m / s (b). Rebound height h ' = 1.5 m So, speed of the object just after hit the floor v = -[ 2gh ' ] since it is oppositeto the initial direction = - 5.422 m / s Momentum of the object just after hit the floor P ' =mv = - 0.5422 kg m / s (c). we know force = change in momentum/ time inrervalto contact the floor = ( P - P ' ) / t F= [1.1682 kg m / s ] / t if we know t then we find force F valueRelated Questions
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