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

In summary: But thanks for your patience and sorry for taking your time!In summary, the conversation was about finding the maximum value of contact force during the collision of a 42-g tennis ball with a wall. The initial velocity of the ball was given as 35 m/s and it rebounded with the same speed. The equation P = mv was used to calculate the momentum and the area under the graph was used to determine the vertical scale. The final answer was determined to be 588 N.
  • #1
huybinhs
230
0

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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  • #2
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:
 
  • #3
but in the problem, there are no number is given on vertical axis!
 
  • #4
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:
 
  • #5
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:
  • #6
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:
  • #7
0.005 F = mv right?
 
  • #8
Hint #2: It's in my first response :wink:
 
  • #9
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 :rofl:

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?? :rofl:

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 :(
 

1. What is contact force?

The contact force is the force that acts between two objects when they are in direct physical contact with each other. It is also known as the normal force or the reaction force.

2. How is the maximum value of contact force determined during a collision?

The maximum value of contact force during a collision can be determined by analyzing the forces acting on the objects involved, including the initial velocities, masses, and the duration of the collision.

3. Does the maximum value of contact force depend on the materials of the colliding objects?

Yes, the maximum value of contact force can vary depending on the materials of the objects involved in the collision. Different materials have different properties that can affect the forces experienced during a collision.

4. Is there a limit to the maximum value of contact force during a collision?

There is no specific limit to the maximum value of contact force during a collision. It depends on the properties and characteristics of the colliding objects, as well as the external forces acting on them.

5. How does the maximum value of contact force affect the outcome of a collision?

The maximum value of contact force can determine the magnitude of the impact and the resulting movement or deformation of the objects involved. It can also affect the amount of energy transferred during the collision.

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