PHY509: Mid-Term Solution P1. (a) Substituting F = ku3 into the EOM, we get d2 u/d2 + u = -(mk/l2 )u and therefore d2 u/d2 + 2 u = 0 with = 1 + mk/l2 . Then we have a general solution, u = A cos[( - 0 )]. (b) The energy conservation has 1 2 l2 2 k 2 mr +
PHY509: SOLUTION HW #1. P1. (a) For both (A) and (B), the center of mass motion is given by (2m)vCM = F during 0 < t < t. Therefore, vCM = F t/(2m). (b) B has the larger kinetic energy because in addition to the CM kinetic energy it has a contribution fro
PHY509: SOLUTION HW #2. P1. (a) With u = 1/r, (1/r2 )(dr/d) = -du/d and l2 2m du d
2
+ V (1/u) +
l2 2 u = E. 2m
Differentiating w.r.t. , (l2 /m)u u + (dV /dr)(-1/u2 )u + (l2 /m)uu = 0 and cancelling the commone factor u , we get d2 u m dV m +u= 2 2 = - 2
PHY509: SOLUTION HW #4. P1. (a) Since the boy travels an angle /2 during a time interval t, we have t = /2 i.e., t = /(2). The distance the ball travels is 2R and therefore the speed of the ball in the inertial x y frame is 2R/t = 2 2R/. The velocity in t
PHY509: SOLUTION HW #5. P1. (a) x = Q + s cos , z = -s sin and X = Q. (b) x = s cos , z = -s sin and X = 0. (c) x = Q, z = 0 and X = Q. (d) Q does not change the gravitational potential of the system, therefore QQ = -V /Q = 0. A finite s changes the poten
PHY509: SOLUTION HW #7. P1. (a) Since the bead are equi-spaced in equilibrium the interval between the beads is 2R/N . (b) With the virtual displacement Rl from the equilibrium causes the potenital energy from interaction with neighbors as V (l ) = 1 1 k(
PHY509: SOLUTION HW #8.
z y x
(a) t < 0
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(b) t > 0 v L
L R 0
L
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P1. (a) The angular velocity = ^ since the rotation of the whole system along +x creates x the same motion. From the right-hand rule, Li = mi ri vi for each
PHY509: SOLUTION HW #10. P1. (a) Before the change, the kinetic energy K and the potential energy V are related by
1 Kbefore = - 2 Vbefore for a circular orbit, as can be also verified by the Virial theorem. Since the
change in the mass happens so suddenl