# Homework Help: I'm soo confused

1. Dec 11, 2005

### F.B

I am still doing my lab and now im stuck on this collision. Can anyone tell me if i did this right.

*U=initial velocity and V=final velocity

Mass 1 = Mass 2 = 0.541 kg
U(1)= 0.29 m/s [40 N of E]
U(2)= 0.29 m/s [35 N of W]
V(1) =0.26 m/s [36 N of W]
V(2) = 0.26 m/s [30 N of E]

Oh these values were measured they werent given to me.

Ok here are my calculations.

Ekbefore=Ekafter
1/2m(1)U(1)^2 + 1/2m(2)U(2)^2 = 1/2m(1)V(1)^2 + 1/2m(2)V(1)^2
0.29^2 + 0.29^2 = 0.26^2 + 0.26^2
0.1682=0.1352

How would you find the percentage of error and what percentage would be reason to be off by.

Pbefore=Pafter
m(1)U(1) + m(2)U(2) = m(1)V(1) + m(2)V(2)
0.29[40 N of E] + 0.29[35 N of W] = 0.26[36 N of W] + 0.26[30 N of E]

X component of Pbefore.
0.29cos40 - 0.29cos35 = -0.0154

Y-component
0.29sin40 + 0.29sin35 = 0.353

Pbefore = square root (0.353^2 + 0.0154^2) = 0.353

Pafter x = -0.26cos36 + 0.26cos30 = 0.0148
Pafter y = 0.26sin36 + 0.26sin30 = 0.253

Pafter = square root (0.253^2 + 0.0148^2) = 0.253

Is that all right or am i doing something wrong.

2. Dec 11, 2005

### Chi Meson

Kinetic energy should not be assumed to be constant. Only in perfectly elastic collisions will KE be conserved. Perfectly elastic collisions only occur for small gas molecules. Assume tha KE before is NOT equal to KE after. KE will be partially converted to heat and sound during the collison, so KE after will be less.

Your math seems to be correct (I don't have my calcualtor, but I assume you entered your digits correctly). Obviously 0.353 is not equal to 0.253. These shoudl be the same number. My guess is measurements of angles would be the main source of error (most likely). How certain are you of the measured angles? Better than plus or minus 1 degree? Do a sample recalculation while changing the angles all by plus one degree and notice how the answers change.

Are you supposed to propagate errors here?

3. Dec 11, 2005

### F.B

Yes we are supposed to take into account error. But i have one question how do we determine the error percentage between the befores and afters

4. Dec 11, 2005

### Chi Meson

OK.

For KE, there is supposed to be a difference before and after. With your calculated results you can calculate the percentage of KE lost:

(KEbefore-KEafter)/before
convert this decimal into percent if you wish. Unless you estimated the uncertainty of each measurement, all you can do is verify that this is a reasonable amout of KE to lose (based on the sound created during impact and the amount of friction present).

For the conservation of momentum, you have a problem. No matter what, momentum must be conserved, so your before and after disparity shows that measurements MUST have been off. Find the percentage difference the same way [ (before-after)/before which can also be considered (theoretical - actual)/ theoretical ] .

Since the momentum definitely shows that measurements were off by a bit, then you can definitely say that the amount of KE lost must also be off by a similar (but not equal) percentage.