# What is the maximum value of the contact force during the collision? (1 Viewer)

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#### huybinhs

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

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? 2. Relevant equations

P = mv

3. The attempt at a solution

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

What is next step? Thank you!

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#### thrill3rnit3

Gold Member
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 :tongue:

#### huybinhs

but in the problem, there are no number is given on vertical axis!!!

#### thrill3rnit3

Gold Member
but in the problem, there are no number is given on vertical axis!!!
That's what the problem asks you to figure out #### huybinhs

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 ???

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#### thrill3rnit3

Gold Member
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)

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#### huybinhs

0.005 F = mv right?

#### thrill3rnit3

Gold Member
Hint #2: It's in my first response #### huybinhs

0.005 F = Pf - P(o) = > 0.005 F = m [vf - v(o)], right?

#### Andrew Mason

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

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? 2. Relevant equations

P = mv

3. 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

#### huybinhs

but 0.005 F = (42/1000) * 35 => F = 294 N => Wrong!!!

#### thrill3rnit3

Gold Member
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.

#### huybinhs

but what is vf ???

Gold Member

#### huybinhs

Oh! I see!!! So -0.005F = -2 v m => F = 588 N ???

#### thrill3rnit3

Gold Member
Oh! I see!!! So -0.005F = -2 v m => F = 588 N ???
Yep, that's the answer I have.

#### huybinhs

588 N = incorrect :(

#### thrill3rnit3

Gold Member
I went back to your calculation of your area:

2nd Area is 0.5 * 0.004 * F

That 0.5 shouldn't be there.

#### The legend

thrill beat me to it..

#### huybinhs

-5*10^-9 F = -2 m v ????

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