Elastic/inelastic collision - speed?

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The discussion revolves around calculating the speed of a cart before and after a perfectly inelastic collision using impulse data. The cart's mass is 734.0 g, and the area under the force-vs.-time curve indicates an impulse of 0.4209 N·s. Participants express confusion about how to relate impulse to velocity without knowing the mass of the other object involved in the collision. It is clarified that impulse equals the change in momentum, and the equation J = pf - pi can be used to find the final velocity after the collision. The conversation emphasizes the distinction between impulse and momentum, leading to the conclusion that the calculated velocity before the collision is approximately 0.5734 m/s.
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Homework Statement



The students set up a cart with a spring attached to the force probe.They weigh the cart assembly and find that its mass is 734.0 g. They launch the cart towards the target and collect the following force-vs.-time data.

The students discover that Data Studio has an “area under the curve” feature. They use this and find the total area under the curve to be 0.4209 N·s.

1) Assuming a perfectly inelastic collision, what is your best estimate of the speed of the cart before the collision?


2) Assuming a perfectly inelastic collision, what is your best estimate of the speed of the cart after the collision?

*what would you do for an elastic collision?



Homework Equations



elastic
1/2m1v1^2 + 1/2m2v2^2= 1/2m1v1'^2 +1/2m2v2'^2


The Attempt at a Solution



very confused even on how to start this question...
how would we calculate without the mass of the other object?
cannot find the equations that go from a given impulse (force and time) to velocity. help? :cry:

:confused: :confused: :confused: :confused: :confused:
 
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Besides F \Delta t
what else is impulse equal to?
Hint: Impulse equals the change in...
 
Last edited:
well the data is given in a graph of force vs. time so the area under the curve represents the impulse (0.4209 N·s).
I just don't know how to get from that impulse value to velocity?
 
impulse is also equal to the change in momentum:
delta p = pf -pi
and p = mv

but how would you know pf or pi given just delta p?
 
so for the velocity after an perfectly nonelastic collision,
J = pf -pi
pf = mv
.4209 N·s.= .7340 kg (v)
v = .5734 m/s

correct?
 
rubyred401 said:
so for the velocity after an perfectly nonelastic collision,
J = pf -pi
pf = mv
.4209 N·s.= .7340 kg (v)
v = .5734 m/s

correct?

I don't believe so.

Your implying that the impulse(.4209 Ns) = mv. False, it is the momentum (p) that equals mv, not the impulse.
But if your assuming (for a perfectly nonelastic collision) p(final) = 0, thus J=pi=mv, than I believe your right. I guess you probably meant that.
 

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