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Distance to traveled to reach 0m/s not matter in impulse?

  • #1
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Homework Statement


I figured out this homework problem, after many tries, but I'm confused about the correct final answer. Here's the question:
The rebounding ball, mass of 0.06 kg, traveling horizontally at 7.6 m/s, is caught by a player who brings it to rest. During the process, her hand moves back 0.60 m. What is the impulse received by the player?

Homework Equations


I = Δp
I = FΔt
F = ma

The Attempt at a Solution


So in the end I just used I = Δp. I did I = (mvf) - (mvi) = (0.06)(0) - (0.06)(-7.6) = 0.456 kg⋅m/s.
Before I was using the kinematics equations to solve for the acceleration of the ball when it slowed down from 7.6 m/s to 0 m/s in the 0.6 m the player's hand moved when catching it. Then I used the acceleration to get the time taken for this process. Then I plugged the acceleration value and the mass of the ball into F = ma to find the force on the ball by the player's hand. Finally, I plugged in this force into I = Favg.Δt to find the impulse. But doing so I got 0.493 N⋅s, which was wrong.

Why is the answer solved without taking into account the distance traveled by the player's hand?
 

Answers and Replies

  • #2
BvU
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Your Δp calculation is fine. The error is in the checking procedure (I do get .456 that way).
Can you show the checking calculation in detail ?
 
  • #3
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With the kinematic equations, this is what I did:
vf2 = vi2 - 2a(xf-xi)
So vi2 = 7.62 = 57.76 m/s.
And vf2 = 02 = 0 m/s.
xf - xi = 0.6 m.

Solving for a:
-57.76 = -2a(0.6)
a = 48.133 m/s/s
a = -48.133 m/s/s because it is in the opposite direction of the ball's velocity.

With a, I find the time it takes to travel the 0.6 m.
So I use xf = xi + vit + 0.5at2
So xf - xi = 0.6 m.

0.6 = 7.6t - 0.5(48.133)t2
0 = -24.0665t2 + 7.6t - 0.6

Solving the quadratic, I get 0.1575 and 0.1583 for t.

Using F = ma, I solve for the force on hand by ball:
F = (0.06)(-48.133) = -2.88798N

And the impulse would be FΔt, (-2.88798)(0.1575) = 0.455 Ns
Using the other value for t, I get (-2.88798)(0.1583) = 0.457 Ns

Oh, well I guess I did make a mistake somewhere my first time. The numbers are pretty close to 0.456. But if you used the quadratic formula, how would you know which value to use? They each produce a different result.
 
  • #4
haruspex
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With the kinematic equations, this is what I did:
vf2 = vi2 - 2a(xf-xi)
So vi2 = 7.62 = 57.76 m/s.
And vf2 = 02 = 0 m/s.
xf - xi = 0.6 m.

Solving for a:
-57.76 = -2a(0.6)
a = 48.133 m/s/s
a = -48.133 m/s/s because it is in the opposite direction of the ball's velocity.

With a, I find the time it takes to travel the 0.6 m.
So I use xf = xi + vit + 0.5at2
So xf - xi = 0.6 m.

0.6 = 7.6t - 0.5(48.133)t2
0 = -24.0665t2 + 7.6t - 0.6

Solving the quadratic, I get 0.1575 and 0.1583 for t.

Using F = ma, I solve for the force on hand by ball:
F = (0.06)(-48.133) = -2.88798N

And the impulse would be FΔt, (-2.88798)(0.1575) = 0.455 Ns
Using the other value for t, I get (-2.88798)(0.1583) = 0.457 Ns

Oh, well I guess I did make a mistake somewhere my first time. The numbers are pretty close to 0.456. But if you used the quadratic formula, how would you know which value to use? They each produce a different result.
They really cannot produce a different result. Try doing it all algebraically, not using any actual numbers. You should find you get the same algebraic expression. Any numerical difference you are seeing must be from arithmetic error or rounding error.
Note also that your method using distance necessarily makes an assumption about the deceleration profile - constant deceleration in this case. It doesn't matter because it turns out to be irrelevant in this problem, but you should be wary of making such assumptions generally.
 

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