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Unformatted text preview: ticle when it is (a) 4 nuclear radii from the nucleus; (b) 100 nuclear radii from the nucleus; (c) very far fram the nucleus (x -+ oo)? Solution
A.t the turning points, the kinetic energy is (instantaneously) zero, so. the energy prin~iple (with gravitational patential energy U zero at y ::=0 and Wnc = 0 for the normalfarce of the track). gives E = 20 J = mgYmax = mg (4 mwe - b2)ITh . e GX .xmax' quadratic farmula pasitivesolution far x~ax) yields x~ax = (b + b2 + 4aE/mg)/2a == 4.46m2, so. Xmax = x2.i1 m. Solution
In Prablem 16 above, we faund the potential energy difference far a repulsive inverse square force, Problem
73. A child sleds dawn a frictianless hill wllOse vertical drapis 7.2 m. At the battam is a level but raugh stretch where the coefficient of kinetic frictian is 0.51. Haw far daes she slide acrass the level stretch? U(X2) - U(XI) =- =A(~-~),
X2 Xl 1
Xl %2"2dx = A-11/%2 A
The child starts near the hilltap with KA ~ 0 and staps an rough level graund, KB= O,after:lfalling thraugha pat...
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