# Elastic/inelastic collision - speed?

## 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???      ## The Attempt at a Solution

Besides $$F \Delta t$$
what else is impulse equal to?
Hint: Impulse equals the change in...

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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 dont 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?

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.