University Physics I - Classical Mechanics, 2019

108 CHAPTER 5. INTERACTIONS AND ENERGY
ical functions. This looks complicated, but it just gives you the shapes you want for the velocity
curves. The derivative of the above is
v 1 (t) = 1 (1 + 2 erf(10 − 2t))
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v 2 (t) = 1 (1 − erf(10 − 2t)) (5.21)
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and you may want to try plotting these for yourself; the result should be Figure 3.1.
Now, assume (as I did for figure 4.5) thatm 1 =1kg,andm 2 = 2 kg, and use these values and the
results (5.21) (assumedtobeinm/s)tocalculateK sys as a function of t. ThenU = E sys − K sys ,
with E sys =1/2J:
U = 1 2 − 1 2 m 1v1(t) 2 − 1 2 m 2v2(t) 2 = 1 (
1 − erf 2 (10 − 2t) ) (5.22)
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and now do a parametric plot of U versus x 2 − x 1 ,usingt as a parameter. You will end up with a
figure like the one below:
U (J)
x 2
- x 1
(mm)
Figure 5.5: The potential energy function reconstructed from the information available for the collision
shown in Figs. 3.1, 3.4, 4.5. No information can be gathered from those figures (nor from the explicit
expressions (5.20) and(5.21) above)onthevaluesofU for x 2 − x 1 < 0.28 mm, the distance of closest
approach of the two carts.