## Assignment 2, solutions

#### Problem 1:

A system consists of two blocks of known masses M and m (M > m) connected by a light spring.  The system slides toward a wall along a frictionless floor at the speed v as shown.  What is the maximum potential energy U of the system after it bounces off the wall?

Solution:

• Concepts:
Elastic collisions, energy and momentum conservation
• Reasoning:
Assuming that the collision between the mass M and the wall is perfectly elastic and rapid (such that there is hardly any time for the spring to compress or transmit information about the collision to the smaller mass), the immediate post-collision state of the masses-spring system is shown on the right.  The larger mass M undergoes a velocity reversal and is now traveling to the right at +v, while for that instant the smaller mass m is still traveling to the left, with velocity -v, with the spring still uncompressed.
• Details of the calculation:
On the frictionless surface the system momentum will remain constant because no net external force acts on the system.  The center-of-mass velocity =is vCM = v (M - m)/(M + m), where vCM > 0 since M > m.  The system moves to the right, away from the wall.  In the CM frame the two masses vibrate about their center of mass.  In the CM frame, the velocities of the masses right after the collision with the wall are given by
(v - vCM) = v 2m/(M + m)  for M,  (-v - vCM) = -v 2M/(M + m)  for m.
Therefore, in the CM frame, the initial kinetic energy of the system is given by
KCM = ½M(v - vCM)2 + ½M(-v - vCM)2 = 2Mmv2/(M + m).
All this kinetic energy will have been converted into spring potential energy when the spring has its maximum compression and the masses are instantaneously at rest in the CM frame.
The maximum potential energy U of the system after it bounces off the wall is 2Mmv2/(M + m).

#### Problem 2:

Two point sources emit spherical waves with wave number k and frequency ω into 3D space.  One source is located at the origin and one source a small distance r away from the origin.   A detector is located far away at R, with R >> r.
For each source, let Ψi = A exp(i(kri - ωti + φ))/ri, where ri is the distance from the source and A is real.  For the source at the origin choose φ = 0, while for the source at r choose a non-zero φ.
(a)  Write down an expression |Ψtotal|2 (proportional to the intensity) at the detector.  Make the far-field approximation, i.e. lines from each source to the detector are parallel to each other.
(b)  Let k = kR/R.  What condition must hold for k·r, so that the average intensity at the detector is maximized?
(c)  What happens if another point source is added at position 2r with phase 2φ.
(d)  For case (c), what condition must hold for k·r, so that the average intensity at the detector zero?

Solution:

• Concepts:
Wave interference
• Reasoning:
The intensity at R is proportional to |Ψ1(R) + Ψ2(R)|2
• Details of the calculation:
(a)  Ψ1(R) = Aexp(i(kR - ωt))/R,  Ψ2(R) = Aexp(i(kR - k·r - ωt + φ))/|R - r|.
Since R >> r we can neglect the difference between R and |R - r| in the denominator, but not in the phase.
1(R) + Ψ2(R)|2
= (2A2/R2) + (A2/R2) [exp(i(kR - ωt))exp(-i(kR - k·r - ωt + φ) + exp(-(kR - ωt))exp(i(kR - k·r - ωt + φ))]
= (2A2/R2) + (A2/R2) [exp(i(k·r - φ)) + exp(-i(k·r - φ))]
= (2A2/R2) + (2A2/R2) cos(k·r - φ).
(b)  We need cos(k·r - φ) = 1 for maximum intensity.
We need k·r - φ = n2π, of k·r = n2π + φ,  n = 0, 1, 2, ... .
(c)  |Ψ1(R) + Ψ2(R) + Ψ3(R)|2 = (3A2/R2) + (2A2/R2)[ 2cos(k·r - φ) + cos(2(k·r - φ))].
We need cos(k·r - φ) = 1 and cos(2(k·r - φ)) = 1 for maximum intensity.
This still happens when  k·r = n2π + φ,  n = 0, 1, 2, ... .
(d)  We need Ψ1(R) + Ψ2(R) + Ψ3(R) = 0.
exp(i(kR - ωt)) + exp(-i(kR - k·r - ωt + φ) + exp(-i(kR - 2k·r - ωt + 2φ) = 0.
1 + exp(i(k·r - φ)) + exp(i(2k·r - 2φ)) = 0.
Let x = k·r - φ.
1 + e-ix + e-i2x = e-ix(1 + eix + e-ix) = e-ix(1 + 2 cosx) = 0.
(1 + 2 cosx) = 0,  cosx = -½,  x = 2π/3 + n2π or x = 4π/3 + n2π.
We need k·r - φ = 2π/3 + n2π or 4π/3 + n2π for zero intensity.

#### Problem 3:

Moisture condenses at the constant rate λ units of mass per unit time on a falling raindrop.  If the drop falls from rest and has initial mass M, find the distance it has fallen in time t.  Neglect air resistance.

Solution:

• Concepts:
Newton's 2nd law:  F = dp/dt, dp = mdv + vdm
• Reasoning:
The system consists of the moisture, with zero momentum and the water droplet with momentum p.  In a time interval dt a small amount dm of the moisture receives an impulse vdm.  In the same time interval the velocity of the drop changes by dv.  The total momentum change of the system is dp = mdv +vdm.
• Details of the calculation:
Choose a coordinate system such that the z-axis points downward and F = mgk.  The problem then is a one-dimensional problem.
p = mv,  dp/dt = mdv/dt + vdm/dt = mg.  dv/dt = g - (v/m)dm/dt.
dm/dt = λ is given.
We have to solve the equation dv/dt = g - (λ/m)v = g - vλ/(m0+λt).
Let t' = m0 + λt.  Then dv/dt' = (1/λ) dv/dt,  dv/dt' + v/t' = g/λ.
Look at the equation dx/dt + nx/t = c.  If c = 0, the solution can be found by simple integration.  [dx/x = -ndt/t, x = At-n.]  If c is not equal to zero, the equation has a solution x/t = constant, x = Bt, with B = c/(n + 1).
The most general solution is x = At-n + ct/(n + 1).
For our problem we therefore have v = A/t' + gt'/(2λ).  A is determined by the initial conditions.
v(t) = A/(m0 + λt) + (m0 + λt)g/(2λ), v(0) = 0, A = -gm02/(2λ).
v(t) = (m0 + λt)g/(2λ) - gm02/(2λ(m0 + λt)).
Distance fallen in time t:
d(t) = (1/λ)∫m0m0+λt dt'[A/t' + gt'/(2λ)] = (A/λ)ln(t')|m0m0+λt + gt'2/(4λ2)|m0m0+λt
= -[gm02/(2λ2)] ln((m0 + λt)/m0) + [g/(4λ)](2m0t + λt2).

#### Problem 4:

A flexible chain of length 3L and mass m hangs over a frictionless pulley so that the length of the vertical parts of the chain on either side of the pulley is L.  After a slight disturbance, the chain begins to slide to the right. What is the force exerted by the chain on the pulley at the instant the length of the vertical part of the chain on the right is 3L/2?
Hint:  What is the acceleration of the CM of the chain?

Solution

• Concepts:
Energy conservation, Newton's laws
• Reasoning:
The CM of the chain has vertical and horizontal acceleration components.
• Details of the calculation:
Consider the situation where a length L of the chain is still in contact with the pulley.

Let λ = m/(3L) be the mass per unit length of the chain.
If the right end of the chain has moved down a distance z, the potential energy of the chain has changed by
ΔU = -λz gz = -λgz2.
The kinetic energy then is T = ½ λ3Lv2 = λgz2.
v2 = (2/3)gz2/L.  v = z(2g/(3L))½.
Find the linear acceleration of the chain:
a = dv/dt = (dv/dz)(dz/dt) = (dv/dz)v.
a = (2/3)(g/L)z.
When z = L/2, a = g/3.

Let us now consider the acceleration of the CM of the chain.
The left side is accelerating upward, the right side is accelerating downward, and the section in contact with the pulley has tangential and centripetal acceleration.
By symmetry, the vertical component of the tangential acceleration of this section is zero, while the vertical component of the centripetal acceleration is
-(1/λL)∫0πλRdθ sinθ v2/R = -2v2/L.
The vertical component of the acceleration of the CM therefore is is
ay = [λ(L - z)a - λ(L + z)a - λL(2v2/L)]/(3λL) = -2az/(3L) - 2v2/(3L).
When z = L/2, ay = -g/9 - g/9 = -2g/9.

Only the section in contact with the pulley has tangential acceleration.  By symmetry, the vertical component of the tangential acceleration is zero, while the horizontal component is
(1/λL)∫0πλRdθ a sinθ = 2aR/L = 2a/π.
The horizontal component of the acceleration of the CM is
ax = λL(2a/π)/(3λL) = 2a/(3π).
When z = L/2, ax = 2g/(9π).

The net force on the chain is sum of the force exerted by the pulley and the force exerted by gravity
Fpulley - mg j = 2mg/(9π) i - 2mg/9 j.
Fpulley = 2mg/(9π) i + 7mg/9 j.
By Newton's third law, the force exerted by the chain on the pulley is
F
chain = - Fpulley = -2mg/(9π) i - 7mg/9 j.
|Fchain| = ((2mg/(9π))2 + (7mg/9)2)½ = 0.78 mg.

#### Problem 5:

Use the uncertainty principle, ΔxΔp ≥ ħ/2, to estimate the ground state energy of a particle in a one-dimensional well of the form
(a)  U(x) = U0,  -a/2 < x < a/2,  U(x) = infinite for all other values of x,
(b)  U(x) = k|x|.

Solution:

• Concepts:
The uncertainty principle
• Reasoning:
The ground state energy is the lowest possible energy.
For case (a) Δx is fixed.  E depends only on p.  We use  ΔxΔp = ħ/2 to express Δp in terms of the fixed Δx to estimate the ground state energy.
For case (b) Δx changes with E.  E depends on both x and p.  We use  ΔxΔp = ħ/2 to express Δp in terms of the Δx, and the minimize or estimate for E with respect to Δx.
• Details of the calculation:
(a)  Estimate:  Δx = a, Δp = ħ/(2a),  Emin = Δp2/(2m) + U0 = ħ2/(8ma2) + U0.
(b)  Estimate:  Δp = ħ/(2Δx),  E = ħ2/(8m|Δx|2) +  k|Δx|.
dE/d(|Δx|) = -ħ2/(4m|Δx|3) + k = 0,  |Δx|3 = ħ2/(4mk).
Inserting |Δx|3 = ħ2/(4mk) into the equation for E yields Emin = 0.94 (kħ)2/3/m1/3.