a) If two protons start out at rest relativeto each other, separated by 0.1 nano
ID: 1663235 • Letter: A
Question
a) If two protons start out at rest relativeto each other, separated by 0.1 nanometers in vacuum, and then theyare released, calculate their relative positions after 10seconds.
b) Consider the two protons above, initiallyseparated by an equilibrium distance of 0.1 nanometers. This time,however, they are connected by linear spring having a springconstant kspr that is keeping them at theirequilibrium distance. How fast should this system rotate about itscenter to generate a displacement of 10% (that is, a displacementof 0.01 nanometers) away from the equilibrium distance between theprotons?
Explanation / Answer
right now i have limited time, so i will try to answer part a. a) using the center of mass (mid-point between the 2 protons as theorigin of the coordinate system), the 2 protons are at (-d,0,0) and(d,0,0) at time t=0. separation = 2d = .1 nm = 10^-10 m since the problem is symmetric in 2 protons, let the locations be(x,0,0) and (-x,0,0) at time t. we have to solve the equation of motion for only 1 proton; theother can be found by symmetry. force on proton at (x,0,0) is F = /(2x)^2 where 2x is the separation, and =q^2/(40) = 9*10^9 *(1.6*10^-19)^2 =2.3*10^-28 Nm^2 Equation of motion is mx'' = F = /(2x)^2 >> x' meansspeed dx/dt, x'' means acceleration, etc. x'' = /x^2 >>> = /(4m)= 2.3*10^-28 /(4*1.67*10^-27) =.0344 2x'x'' = d/dt (x'^2) = 2x'/x^2 = -d/dt(2/x) integrating, (x')^2 = A - 2/x >>> A is aconstant since at t=0, x=d and x'=0, A = 2/d x' = (2(1/d - 1/x)) dx/(1/d-1/x) = (2) dt using variable z = x/d, dz/(1-1/z) = (2/d^3) dt integrating, (z(z-1)) + ln (z + (z-1)) =(2/d^3) t + B >>> B is a constant since at t=0, x=d, z=1, B =0 (z(z-1)) + ln (z + (z-1)) =(2/d^3) t at t =10 sec, (z(z-1)) + ln (z + (z-1)) =(2/d^3)*10 = 10*(2*.0344/(.5*10^-10)^3) =7.42*10^15 Although this is a transcendental equation for z, since RHS is verylarge, z is very large. Then log terms can be neglected and,because z ˜z-1, z = x/d = 7.42*10^15 x = 7.42*10^15 * .5*10^-10 = 3.71*10^5 m which means that the 2 protons are 7.42*10^5 m apart.
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