myxomatosii
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SOLVED
Let me start by saying my ONLY question is. Why can I not use..
vf=(2aΔs).5
Why do I have to use
vf=(2mgH).5
to get the vf? Why have things changed now that I am dealing with momentum problems, isn't vf just vf?
With the mgH formula my velocity is much lower. This doesn't make sense and its bothering me that I am missing something so fundamental.
But I'll post the problem anyway.
A 160 g ball is dropped from a height of 2.8 m, bounces on a hard floor, and rebounds to a height of 2.1 m. Figure P9.28 shows the impulse received from the floor. What maximum force does the floor exert on the ball?
http://img441.imageshack.us/img441/8416/p928.gif
vf=(2aΔs).5 (I thought but I guess not)
vf=(2mgH).5
Jx=Δpx=FmaxΔt
I am solving it using the method written in the link below.
Which brought me to the question listed above.
http://answers.yahoo.com/question/index?qid=20081026102629AAdYIDa
"""""
You start by defining what "impulse" is.
In math talk, it's dP = F dt; where dP = p1 - p0 = m(v1 - v0), the change in momentum, F is the average force over the time interval dt = t1 - t0 in which the momentum changed. F = ? is the force you are looking for.
Solve for F = m(v1 - v0)/dt; where m = .15 kg, dt = .005 sec, you need to find v0, the velocity at impact and v1 the velocity at rebound.
To find v0, use the conservation of energy so that v0^2 = 2mgH; where H = 1.7 m. Then v0 = sqrt(2mgH) = sqrt(2*.15*9.81*1.7) = 2.236761051 mps. Here the potential energy mgH is converted to kinetic energy.
To find v1, use the same conservation of energy so that v1 = sqrt(2*.15*9.81*1) = 1.715517415 mps, where the height h = 1.0 m on rebound. Here the kinetic energy of the rebound is converted to potential energy PE = mgh at height h.
Recognize that the directions of the two velocities are directly opposite; so they are additive, in other words v1 - (-v0) = v1 + v0 = 3.952278466
Solve for F = dP/dt = m(v1 + v0)/dt = .15*3.95/.005 = 118.5 kg.m/sec^2.
"""""
Homework Statement
Let me start by saying my ONLY question is. Why can I not use..
vf=(2aΔs).5
Why do I have to use
vf=(2mgH).5
to get the vf? Why have things changed now that I am dealing with momentum problems, isn't vf just vf?
With the mgH formula my velocity is much lower. This doesn't make sense and its bothering me that I am missing something so fundamental.
But I'll post the problem anyway.
A 160 g ball is dropped from a height of 2.8 m, bounces on a hard floor, and rebounds to a height of 2.1 m. Figure P9.28 shows the impulse received from the floor. What maximum force does the floor exert on the ball?
http://img441.imageshack.us/img441/8416/p928.gif
Homework Equations
vf=(2aΔs).5 (I thought but I guess not)
vf=(2mgH).5
Jx=Δpx=FmaxΔt
The Attempt at a Solution
I am solving it using the method written in the link below.
Which brought me to the question listed above.
http://answers.yahoo.com/question/index?qid=20081026102629AAdYIDa
"""""
You start by defining what "impulse" is.
In math talk, it's dP = F dt; where dP = p1 - p0 = m(v1 - v0), the change in momentum, F is the average force over the time interval dt = t1 - t0 in which the momentum changed. F = ? is the force you are looking for.
Solve for F = m(v1 - v0)/dt; where m = .15 kg, dt = .005 sec, you need to find v0, the velocity at impact and v1 the velocity at rebound.
To find v0, use the conservation of energy so that v0^2 = 2mgH; where H = 1.7 m. Then v0 = sqrt(2mgH) = sqrt(2*.15*9.81*1.7) = 2.236761051 mps. Here the potential energy mgH is converted to kinetic energy.
To find v1, use the same conservation of energy so that v1 = sqrt(2*.15*9.81*1) = 1.715517415 mps, where the height h = 1.0 m on rebound. Here the kinetic energy of the rebound is converted to potential energy PE = mgh at height h.
Recognize that the directions of the two velocities are directly opposite; so they are additive, in other words v1 - (-v0) = v1 + v0 = 3.952278466
Solve for F = dP/dt = m(v1 + v0)/dt = .15*3.95/.005 = 118.5 kg.m/sec^2.
"""""
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