What is the maximum value of the contact force during the collision?

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Homework Help Overview

The discussion revolves around determining the maximum value of the contact force during the collision of a tennis ball with a wall. The scenario involves a 42-g tennis ball rebounding off a wall with an initial velocity of 35 m/s, and participants are analyzing a graph representing contact force versus time during the collision.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants are attempting to calculate the maximum force by analyzing the area under the force-time graph and relating it to momentum changes. There are discussions about the calibration of the vertical axis and how to derive values from the graph. Questions about the total change in momentum and impulse are raised, along with clarifications on the relationships between variables.

Discussion Status

Several participants have offered guidance on interpreting the graph and calculating areas, while others are exploring different interpretations of the problem. There is an ongoing dialogue about the correct approach to finding the maximum force, with some participants expressing uncertainty about their calculations and assumptions.

Contextual Notes

Participants note the absence of specific numerical values on the vertical axis of the graph, which complicates the calculations. There is also mention of potential errors in area calculations that could affect the outcome.

huybinhs
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Homework Statement



The Figure shows an approximate representation of the contact force versus time during the collision of a 42-g tennis ball with a wall. The initial velocity of the ball is 35 m/s perpendicular to the wall; it rebounds with the same speed, also perpendicular to the wall. What is the maximum value of the contact force during the collision?

1-7.png



Homework Equations



P = mv

The Attempt at a Solution



I know P = 35 (42/1000) = 1.47 N*m

What is next step? Thank you!
 
Last edited:
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You're trying to find the value of the maximum force, which is the value where the graph flattens out.

Solve for J, using the same equation

Po + J = Pf

J is the area of the figure.

Think about it for a few minutes and post again if you still need more help. I can't just spoonfeed you with the answer :-p
 
but in the problem, there are no number is given on vertical axis!
 
huybinhs said:
but in the problem, there are no number is given on vertical axis!

That's what the problem asks you to figure out :cool:
 
so the 1st Area is 0.5 * 0.005 * F

2nd Area is 0.5 * 0.004 * F

3rd Area is 0.5 * 0.001 * F

So I add all them up: 0.005F, then 1.47 - 0.005F = (42/1000)vf

now I have 2 unknown ?
 
Last edited:
huybinhs said:
so the 1st Area is 0.5 * 0.005 * F

2nd Area is 0.5 * 0.004 * F

3rd Area is 0.5 * 0.001 * F

So I add all them up: 0.005F, then 1.47 - 0.005F = (42/1000)vf

now I have 2 unknown ?

You did a good job with the 0.005F. What does 0.005F equal? (Hint: read the previous posts)
 
Last edited:
0.005 F = mv right?
 
Hint #2: It's in my first response :wink:
 
0.005 F = Pf - P(o) = > 0.005 F = m [vf - v(o)], right?
 
  • #10
huybinhs said:

Homework Statement



The Figure shows an approximate representation of the contact force versus time during the collision of a 42-g tennis ball with a wall. The initial velocity of the ball is 35 m/s perpendicular to the wall; it rebounds with the same speed, also perpendicular to the wall. What is the maximum value of the contact force during the collision?

1-7.png



Homework Equations



P = mv

The Attempt at a Solution



I know P = 35 (42/1000) = 1.47 N*m

What is next step? Thank you!
The maximum force is easy to locate on the graph but the problem is we do not know how the vertical axis is calibrated - ie how many Newtons per square. You have to work that out from the information given.

What is the total change in momentum? How is that related to the total impulse? What aspect of the graph represents the total impulse? How does that enable you to determine the vertical scale of the graph?

AM
 
  • #11
but 0.005 F = (42/1000) * 35 => F = 294 N => Wrong!
 
  • #12
huybinhs said:
0.005 F = Pf - P(o) = > 0.005 F = m [vf - v(o)], right?

This looks right. I deleted my other post because I quoted the wrong response.

Remember that 0.005F would be negative in this case because it's a force opposite the motion.
 
  • #13
but what is vf ?
 
  • #14
it rebounds with the same speed

quoted directly from the problem
 
  • #15
Oh! I see! So -0.005F = -2 v m => F = 588 N ?
 
  • #16
huybinhs said:
Oh! I see! So -0.005F = -2 v m => F = 588 N ?

Yep, that's the answer I have.
 
  • #17
588 N = incorrect :(
 
  • #18
I went back to your calculation of your area:2nd Area is 0.5 * 0.004 * FThat 0.5 shouldn't be there.
 
  • #19
thrill beat me to it..
 
  • #20
-5*10^-9 F = -2 m v ?
 
  • #21
huybinhs said:
-5*10^-9 F = -2 m v ?

...no.

The area calculation is correct except for the tiny mistake I pointed out.
 
  • #22
GOT IT! Thanks for your patience again! ;)
 
  • #23
I'm sorry I didn't catch that mistake a few posts back. Today was just a tiring day for me, in fact I'm about to go to bed now :smile:

Again I'm glad it worked out well for you :approve:
 
  • #24
Good night! See u tomorrow LOL
 
  • #25
huybinhs said:
Good night! See u tomorrow LOL

More problems?? :smile:

Just learn how to manipulate the equation based on the given info and what you're looking for. Most of these problems are similar in nature and ask about one specific thing or another.
 
  • #26
yeah! I'm really bad at the graph problem :(
 

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