# Homework Help: Impulse/Force/Bounce Problem

1. Mar 31, 2009

### myxomatosii

SOLVED

1. The problem statement, all variables and given/known data

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 [Broken]

2. Relevant equations

vf=(2aΔs).5 (I thought but I guess not)

vf=(2mgH).5

Jx=Δpx=FmaxΔt

3. 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.

"""""

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.

"""""

Last edited by a moderator: May 4, 2017
2. Apr 1, 2009

### myxomatosii

The answer from the method I used from Yahoo came out wrong.

Completely lost.. as I am as of late.

176.928N no cigar.

3. Apr 1, 2009

### myxomatosii

I solved it, idk what the guy on the yahoo answers was smoking...

Although his idea on the velocities being additive did help me realize something about the momentum.