Impulse steel ball falling on steel slab

AI Thread Summary
The discussion focuses on calculating the impulse of a steel ball dropped from a height of 2.00 m onto a steel slab, which rebounds to a height of 1.60 m. The impulse is determined by the change in momentum before and after impact, with the initial momentum calculated as -0.2504 kg*m/s and the final momentum needing correction. Participants highlight the importance of correctly identifying variables in kinematic equations, particularly the final velocity after the bounce. The original poster realizes they misassigned positions and velocities, leading to an incorrect impulse calculation. The correct impulse value from the textbook is 0.47 N*s, indicating a need for careful variable assignment in the calculations.
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Homework Statement



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.


Homework Equations



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


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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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.
 
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
??
 
The complete kinematic equation is

v^{2}_{f}=v^{2}_{i}+2a(x_{f}-x_{i})

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

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

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?
 
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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