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Need it Fast please

  • Thread starter googooloo
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  • #1
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Need it Fast please!!!

Homework Statement



The question is:
A ball with 58 Grams of mass strikes against a wall.
The initial speed is 35 m/s. The path of the strike is always vertical to the wall (90 degrees of angle between them).
figure 1 represents F(force) diagram.
What is the maximum force from wall to the ball?


Homework Equations





The Attempt at a Solution


well i am not sure and neither familiar with the question.
formula 1 is been attached.
 

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Answers and Replies

  • #2
gneill
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Consider what the area under the F vs time curve represents.
 
  • #3
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Consider what the area under the F vs time curve represents.
Well, it represents delta P...but as u know delta P = p(2) - p (1)
i have the mass, and same as v(1) but i dont have v (2)....so i cant find F(max) because of the lack of v(2)....
 
  • #4
gneill
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20,801
2,778


I think you'll have to assume that the ball bounces back with the same speed. This would tend to be born out by the fact that the curve is symmetric.
 
  • #5
125
1


elastic collision?
 
  • #6
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Is this correct?
by the area under the F vs time =>
integrate of F.dt= delta p and intagrate of F.dt = delta U = - W = -delta K.
so
Delta P = -Delta K => mv(f) - mv(i) = [(1/2)*m*v(i)^2] - [(1/2)*m*v(f)^2]
=>m(vf-vi)= (1/2)m[v(i)^2 - v(f)^2]
=>2Vf-2Vi=vi^2-vf^2 => by vi=34
=>2vf-68=1156-v(f)^2 => v(f)^2+2vf-1224=0
vf=34 and vf=-36

so

integrate of F.dt =the under the F vs t = F(max)*(6+2)*(1/2)*(1/10(because the time was per (ms) not (S))=Delta P = m(vf-vi)
=58*(-36-34) = > F=-10150
 
  • #7
125
1


No... momentum and energy are measured in different units =)
F.dt = dp, but F.ds = -dU.
dU = -dK if dE = 0 (total energy is conserved = elastic collision)
 
  • #8
8
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No... momentum and energy are measured in different units =)
F.dt = dp, but F.ds = -dU.
dU = -dK if dE = 0 (total energy is conserved = elastic collision)
Oh, then how can i solve this question?
 
  • #9
gneill
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20,801
2,778


Oh, then how can i solve this question?
Write an expression for the area under the force vs time curve for the time period covering the duration of the impact (total 6ms). This expression will have Fmax as a parameter. Note that you should be able to do this by inspection, since the geometry of the curve is so simple.

The area represents the total momentum change, delta P as you said previously, so equate it to what you determine from the mass of the ball and its velocity before and after the impact. Solve for Fmax.
 

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