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Problem 1. A Jelly-Filled Universe (10 points total)

a) With no other external forces on the particle, a = 0. We can show this using Newton’s second law:

- ka = ma,

which yields a = 0. Therefore,

x = x0 + v0t.

b) Again, using Newton’s laws we have F - ka = ma, which yields a = F/(k + m). We see that the pseudo-drag force acts by contributing an effective mass to the particle. Integrating, we get

2016PUPC普林斯顿大学物理竞赛真题答案

so

2016PUPC普林斯顿大学物理竞赛真题答案

c) We know that the net force on the particle is constant. Defining the acceleration a0 = F/(k + m) from part b, we find

2016PUPC普林斯顿大学物理竞赛真题答案

The net work done is therefore the change in kinetic energy during the time T :

2016PUPC普林斯顿大学物理竞赛真题答案

d)Kinetic friction is a constant braking force, so in a given time interval PUPC普林斯顿大学物理竞赛真题t the particle experiences a decrease in speed ofPUPC普林斯顿大学物理竞赛真题v α-PUPC普林斯顿大学物理竞赛真题 t. That is, the speed decreases at a constant rate. For viscous drag, the force increases linearly with speed, soPUPC普林斯顿大学物理竞赛真题v α-vPUPC普林斯顿大学物理竞赛真题 t, which implies that the speed decreases exponentially fast v∼exp( -kt). Unlike the other forces, the pseudo-drag force only acts in the presence of other forces. Alone, it has no effect on the system. In the presence of other forces, it acts by increasing the effective mass of the particle to m →m + k.

e)The force of gravity is Fg = −mg, where m is the particle’s gravitational mass. Using Newton’s second law, we find

2016PUPC普林斯顿大学物理竞赛真题答案

so the integrated equations give

2016PUPC普林斯顿大学物理竞赛真题答案

Solving for the time elapsed, we find

2016PUPC普林斯顿大学物理竞赛真题答案

f)  It is indeed possible to replace g with an e↵ective gravitational field ge↵. Studying Newton’s second law, we see that

2016PUPC普林斯顿大学物理竞赛真题答案

which implies that we should set

2016PUPC普林斯顿大学物理竞赛真题答案

Taking the limits, we get

2016PUPC普林斯顿大学物理竞赛真题答案

These results make sense. As m→0, the particle becomes massless, so it is no longer affected by gravity. So, it doesn’t matter whether there is a gravitational field (g ≠ 0) or not (geff= 0). In the second case, we have that the particle is infinitely massive. We no longer expect the the pseudo-force, which contributes a constant, finite term k to the effective mass, to matter. Thus, we expect geff= g, since the pseudo-drag force no longer contributes.

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