CALC? A proton with mass m moves in one dimension. The potential-energy function is U(x) = ?/x2) – (?/x), where ? and ? are positive constants. The proton is released from rest at x0 = ?/?. (a) Show that U(x) can be written as Graph U(x). Calculate U(x0) and thereby locate the point x 0 on the graph. (b) Calculate v(x), the speed of the proton as a function of position. Graph v(x) and give a qualitative description of the motion. (c) For what value of x is the speed of the proton a maximum? What is the value of that maximum speed? (d) What is the force on the proton at the point in part (c)? (e) Let the pro-ton be released instead at x1 = 3?/?. Locate the point x1 on the graph of U(x). Calculate v(x) and give a qualitative description of the motion. (f) For each release point (x = x0 and x = x1), what are the maximum and minimum values of x reached during the motion?

Solution 87CP Step 1: 2 The potential energy fraction is U(x) = /x (/x). Where and are positive constants. The proton is released from rest at x = / . 0 2 2 U(x) = /x 0(x /x0 (x /x)0……..(1) 2 By factoring /x 0 on R.H.S of eq……..(1).where = /x 0 U(x) = /x .x /x /x x 0 0 0 U(x) = /x 0x /x0 /x 02 x0/x 2 2 U(x) = /x (0 /x 0 x /x) 0 The curves of U(x) and V (x) are shown in above graphs. U(x) = 0,while U(x) is positive for x < x and U(x) is negative for x > x . 0 0 Step 2: (b).At x the proton is released from rest,which implies that V (x) = K(x ) = 0,since 0 0 U(x) = 0 We find that the total mechanical energy of proton is E = K + U = 0 2 Since here energy is conserved K = U The proton moves in the positive x -direction, speeding up until it reaches a maximum speed, and then slows down but it never stops. The proton cannot be found at x < x 0ince the quantity under the square root would then be negative. It will therefore only be found where U(x) < 0 when x > x 0 . A graph of V (x) is shown in above graphs. Step 3: (c) The velocity will be at a maximum when the kinetic energy will also be maximum, which from equation (2) implies that the potential energy will then be at a minimum. A close examination of above graphs shows that the minimum for U(x) happens when dU/dX = 0. We therefore calculate 2 2 3 2 dU/dX = /x (0x /x0 x /x ) 0 dU/dX = /x x0(2x /x0 1) When dU/dX = 0 at x = 2x 0 The maximum speed is given by vmax = /2mx 2 0