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posted by  eatyourveggies on 3/19/2009 11:37:04 AM  |  status: Live  |  Earned Karma: 99

Finding Speed

Course Textbook Chapter Problem Needs by
Calculus Based Physics Physics for Scientists and Engineers A Strategic Approach (2nd) by Knight 10 74CP N/A
Question Details:
The air-track carts in the figure  (Intro 1 figure)  are sliding to the right at 1.0 {\rm{ m/s}}. The spring between them has a spring constant of 150 N/m and is compressed 4.5 cm. The carts slide past a flame that burns through the string holding them together.

What is the speed of 100-\rm g cart?
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(Winona)
posted by Big PA on 11/4/2009 10:51:17 PM  |  status: Live
Clarification Details:
So I have a question. This was solved with slightly different equations then I have learned in class so far. With the Energy unit that this question is from, I believe my professor would have me set this up as:
"Before the string snapped"                                                        "After the string snapped"
Total E=Potential spring energyi + Kinetic Energyi             Total E=Potential spring energyf + Kinetic Energyf
or TE =Usi + Ki                                                                                            TE=Usf +Kf
 Then the equations would be set equal to one another as
Usi + Ki=Usf +Kf
Would it be correct to say that Usf is actually 0 because the string snaps and then there is no force being exerted by the force then? And the Δ s2 would be zero? If that assumption is correct, then you would substitute the equations in so you'd have:
(1/2 K  Δ s2) + (1/2 m1Vi2)= 0+(1/2 m1Vf2)
doing all the inserting of numbers so that I'm solving for Vf  of the 1st mass I get:
Vf= √ (.5 k Δ s2 + .5m1Vi2) / (.5 m1)
Known numbers:
K =120 N/m
Spring compression = 0.045 m (that's Δ s)
Vi= 1 m/s
Mass 1= 0.1 kg
Mass 2= 0.3 kg
I solve and get:
Vf = 1.85 m/s
and this is not the right answer
Help?
Paula M.
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posted by Anonymous on 11/10/2009 3:53:44 PM  |  status: Live
Clarification Details:
This answer isn't correct.

AAnswers:

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Oracle
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posted by Antony08 on 3/19/2009 12:01:41 PM  |  status: Live
Asker's Rating: Helpful   
Response Details:
The air-track carts are sliding to the right at 1.0 m/s
The spring constant of the spring between them is k = 150 N/m
The spring is compressed by x = 4.5 cm = 4.5 * 10-2 m
The force between the masses of the air-track cart is
F = k * x
or m * a = k * x
or a = (k * x/m)
Let the speed of the 100-g cart be v,therefore,we get
v2 - u2 = 2ax
Substituting the value of a in the above equation,we get
v2 - u2 = 2 * (k * x/m) * x
or v2 - u2 = (2 * k * x2/m)
or v2 = u2 + (2 * k * x2/m)
or v = [u2 + (2 * k * x2/m)]1/2
Substituting the values in the above equation,we get
v = [(1.0)2 + (2 * 150 * (4.5 * 10-2)2/(100 * 10-3)]1/2
or v = 2.66 m/s
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