Solve Force and Impulse Homework: Speed of Tennis Ball After Collision

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Homework Help Overview

The discussion revolves around calculating the speed of a tennis ball after it collides with a wall, using concepts of momentum and impulse. The problem involves analyzing a force versus time graph to determine the impulse experienced by the ball during the collision.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss calculating the impulse by finding the area under the force-time graph and question the implications of the sign of the impulse. There are attempts to relate initial momentum and final momentum using the impulse-momentum theorem.

Discussion Status

Participants are actively engaging with the problem, sharing calculations of areas under the graph and questioning the correctness of their results. Some participants express confusion about the direction of the force and its effect on impulse, while others are clarifying the relationship between impulse and momentum.

Contextual Notes

There is a mention of units and the conversion of grams to kilograms, as well as the time scale of the graph being in milliseconds. Participants are also considering the direction of forces in relation to the motion of the ball.

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Homework Statement



The Figure shows an approximate representation of the contact force versus time during the collision of a 36-g tennis ball with a wall. The initial velocity of the ball is 151.6 m/s perpendicular to the wall. What is the speed of the tennis ball after the collision?

1-6.png


Homework Equations



P = mv

The Attempt at a Solution



I did find 3 Area in the pic the add them up. I got 10200 total, then what's next?
Any help would be appreciated! ;)
 
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P0 + J = Pf
 
P(0) = (36/1000) * 151.6 = 5.4576 N*m

Pf = (36/1000) * vf with Pf = total of 3 Areas, correct?
 
huybinhs said:
P(0) = (36/1000) * 151.6 = 5.4576 N*m

Pf = (36/1000) * vf with Pf = total of 3 Areas, correct?

No, in the equation I told you, J represents the area of the figure. Do you see why?
 
so P(0) = (36/1000) * 151.6 = 5.4576 N*m

then 5.4576 + 10200 = (36/1000) vf => vf = 2.83 * 10^5 m/s, right?
 
huybinhs said:
so P(0) = (36/1000) * 151.6 = 5.4576 N*m

then 5.4576 + 10200 = (36/1000) vf => vf = 2.83 * 10^5 m/s, right?

Time in the graph is given in milliseconds (ms).
 
If so 10200 will be = 10.2

=> 5.4576 + 10.2 = (36/1000) vf => vf = 435 m/s Wrong! Why?
 
huybinhs said:
If so 10200 will be = 10.2

=> 5.4576 + 10.2 = (36/1000) vf => vf = 435 m/s Wrong! Why?

Does the force act on the same direction as the motion?
 
so u mean J in this case is just = 9 ?
 
  • #10
huybinhs said:
so u mean J in this case is just = 9 ?

I mean it's negative, and where did you get 9?
 
  • #11
so I got the 1st area = 0.5 * 0.002 * 1200 = 1.2

2nd = 0.007 * 1200 = 8.4

3rd = 0.001 * 0.5 * 1200 = 0.6

so now why should i do? please be more specific!
 
  • #12
huybinhs said:
so I got the 1st area = 0.5 * 0.002 * 1200 = 1.2

2nd = 0.007 * 1200 = 8.4

3rd = 0.001 * 0.5 * 1200 = 0.6

so now why should i do? please be more specific!

add the 3 = 10.2

I'm saying that since the force of the wall is acting opposite of the motion, then maybe the impulse should be negative (-10.2)
 
  • #13
Got it! Thanks for your patience ;)
 
  • #14
huybinhs said:
Got it! Thanks for your patience ;)

You're most welcome :approve:
 

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