impulse... steel ball falling on steel slab

by offbeatjumi
Tags: ball, falling, impulse, slab, steel
offbeatjumi is offline
Nov2-09, 12:42 PM
P: 29
1. The problem statement, all variables and given/known data

mass of steel ball = 40g = .04 kg
h1 = 2.00 m
h2 = 1.60 m
A steel ball is dropped from height of 2.00 m onto horizontal steel slab, rebounds to height of 1.60m. Calculate impulse of ball during impact.

2. Relevant equations

impulse J = F*dt
J = P2 - P1
I tried using

3. The attempt at a solution

I tried using equations of kinematics to solve for the velocities of the ball before and after impact, to then find the final and initial momentum, and solve for impulse by finding change in momentum.
I have the answer from the back of the book, 0.47 N*s. But I did not get that answer with my method. How should I approach this? Thanks.
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kuruman is offline
Nov2-09, 02:01 PM
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Your method is correct. It is probably its implementation that is incorrect. Maybe you missed a minus sign, maybe you punched in the wrong numbers. If you show the details of your calculation, someone might be able to locate where you went wrong.
offbeatjumi is offline
Nov2-09, 09:33 PM
P: 29
I divided it in two steps, step one before impact and step two after impact.

in step one:

I use v(f)^2 = v(i)^2 - 2gy
v(f)^2 = 0 - (-39.2)
v(f) = -sqrt(39.2) = v(1)
P(1) = mv(1) = -.2504 kg*m/s

in step two:
I use the same kinematic equation
v(f)^2 = v(1)^2-2gy
v(f)^2 = 39.2 - 2*(-9.8)*1.60 = 70.56
v(f) = 8.4 m/s = v(2)
P(2) = mv(2) = 0.336

P(2) - P(1) = 0.336 - (-.2504) = 0.5863 N*m

answer should be 0.47 N*m

kuruman is offline
Nov3-09, 08:20 AM
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impulse... steel ball falling on steel slab

The complete kinematic equation is


Here vf is the speed when the object is at position xf and vi is the speed when the object is at xi.

We assume that position is measured from the slab.

For step 1, xi=2.00 m, xf=0, vi=0, a = -9.8 m/s2 and we are looking for vf, the speed of the ball just before it hits the slab on its way down. This gives

[tex]v^{2}_{f}=0^{2}+2(-9.8\;m/s^2)(0 - 2.00\;m)[/tex]

If you do the calculation, you get -.2504 kg*m/s for the momentum as you have already found.

The problem with your solution is the identification of the variables after the bounce. After the bounce the speed of the ball at xi = 0 is the unknown speed vi. What is the speed vf when the position is xf = 1.60 m?
offbeatjumi is offline
Nov3-09, 06:11 PM
P: 29
thank you very much

yeah i did not assign the correct positions/velocities of the ball to the equation, i see now that the final velocity in the second situation is zero

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