Impulse steel ball falling on steel slab

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Homework Help Overview

The problem involves a steel ball dropped from a height of 2.00 m onto a horizontal steel slab, which rebounds to a height of 1.60 m. The objective is to calculate the impulse experienced by the ball during the impact.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss using kinematic equations to determine the velocities of the ball before and after impact, and subsequently calculating the change in momentum to find impulse. There are questions regarding the implementation of these equations and the assignment of variables.

Discussion Status

Some participants have provided guidance on checking calculations and ensuring correct variable assignments in the kinematic equations. There is acknowledgment of potential errors in the original poster's calculations, but no consensus on a definitive solution has been reached.

Contextual Notes

Participants note the importance of correctly identifying initial and final positions and velocities in the kinematic equations, as well as the need to clarify assumptions about the motion of the ball during the impact.

offbeatjumi
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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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