Phys 263/Amath 261
Assignment 8
SOLUTIONS
1. (a) The speed of the second ship is dened on the rst ship by u= x t (1)
where x is the measured length of the 2nd ship. This is length contracted by x = 1 L0 (2)
where L0 is the proper length of 100m. Thus u= 1
MIDTERM
Phys 263 / Amath 261 June 16, 2008: 4:30 to 6:30pm
1. For two particles with mass m1 and m2 and positions r1 = (2t2 , 3t, 4) r2 = (1 + t2 , 0, 4t2 ) (a) Find the center of mass coordinates. (b) Find the velocity and acceleration for the center of
Phys 263/Amath 261
Assignment 6
SOLUTIONS
1. (a) To set up this problem, use the gure, where the mass m is a distance r from the center of the ring. Then the potential is given by dM Ga = d (1) b b where b is the distance between dM (on the ring) and m (t
Phys 263/Amath 261
Assignment 6
SOLUTIONS
1. For the orbit r = ke (a) We use the path equation to determine the force law. First note, d1 d e e = = d r d k k and 2 e 2 d2 1 = = d2 r k r then from the path equation F (r) = or F (r) = and inverse cube. (b)
Phys 263/Amath 261
Assignment 5
SOLUTIONS
1. First, consider the center of mass momentum, which reacts to two forces: the tension on the rope (by the hand), and gravity: P = T M g. Now, the mass of the suspended rope is the mass density times the length x
Phys 263/Amath 261
Assignment 4
SOLUTIONS
1. (a) Look at the equilibrium position, and equate mg = kx k= mg 1000 9.8 = = 98000 x 0.1 N/m (1)
(b) The undamped oscillation period is T= (c) At critical damping, = 0 , b = 2m = 1.98 104 (d) Since the mass has
Phys 263/Amath 261
Assignment 3
SOLUTIONS
1. The equation of motion of the oscillator is x + 4x = 0
2 where o = k/m = 4. The general solution, from class, is
(1)
x = A sin(0 t + ) which we can also write x = A sin o t cos + A cos 0 t sin substituting new
Phys 263/Amath 261
Assignment 2
SOLUTIONS
1. The law of conservation of mass does not hold in quantum mechanics and special relativity. 2. The Force dierential equation is dened by F = ma = m in one dimension. Integrating:
v (t)
dv = Cekv dt
(1)
ekv dv
vo
Phys. 263
Assignment 1
SOLUTIONS
1. (a) A + B + C = (5, 1, 3) (b) B C = (1, 4, 5) so that A (B C) = 1 + 8 + 15 = 24. (c) A B = (8, 8, 8) so that C (A B) = 8 + 8 + 8 = 24. (d) A B C = A (B C) = (2, 2, 2) (e) to (h) are =0, since these are constant vectors
1
Review: Vector Analysis
REFERENCES: Arya Chapter 5.
1.1
Vectors
A scalar has only a magnitude. A vector has both a magnitude and a direction. In this course, vectors will be represented in two ways (bold, or overarrow): R = R, (1)
or in the special case
2.5
Nuclear Binding Energy
Clear evidence for the mass-energy equivalence relation is given in the study of mass defects in atomic nuclei. The mass defect is the decit in the mass of a nucleus, compared with the sum of the masses of its constituent nucleo
1.6.2
Proper time
Consider an object moving with a velocity u relative to our rest frame S . Assume two events occur with the object, separated by some time dt . In the S frame, dt = ds c (116)
since dx = 0 in its own rest frame. As discussed before, this
1.6
Four-dimensional Space
The Lorentz transformation treats the xi with i = 1, 2, 3 as equivalent variables. Lets introduce time as simply a forth coordinate: x0 = ct (85)
This four-dimensional space is called Minkowski space. The Lorentz transformation
2. Consider a process where an event at P causes an event at Q. Lets choose the separation coordinates to be x = x2 x1 > 0 and t = t2 t1 > 0, and V = x/t is the speed of the signal. Then t = t2 t1 = (t2 t1 ) = t 1 uV c2 u(x2 x1 ) c2 (61) (62) (63) If V >
1.3
Lorentz Transformation
Since Galilean transformations are inconsistent with Einsteins postulate of the speed of light, we must modify them. Consider our two inertial frames S and S , and let the axes be parallel, with S moving w.r.t. S with a speed u
example: Time dilation. Consider two observers S and S . Observer S is stationary and observes S is moving away with a constant velocity u. S sends a beam of light towards a mirror L away, and receives the beam back after a time interval 2t. To S the beam
1
Special Relativity
Recall the Galilean transformation of Newtonian physics: this relates the the coordinates x, y, z, t in one inertial reference frame S to the coordinates x , y , z , t in a dierent inertial reference frame S , moving with velocity wit
1.2
Gausss law and Poissions Equation
M r r2
Consider a point mass M , with gravitational eld at a distance r of g = G (29)
We can dene a quantity of ux through a sphere of radius r with this point mass M in the center as = 4 r2 g = 4 GM (30)
we will show
1
Gravitation
mM r r2
Newtons law of universal gravitation F = G (1)
where G = 6.6726 0.0008 1011 Nm2 kg2 . This equation applies strictly only to point particles. If one or both of the particles is replaced by a body with a certain extension, we must mak
1.2
Planetary motion
REFERENCE: Arya, Section 7.7 to 7.9
1.3
Rutherford Scattering
The other very important problem involving inverse-square forces is the scattering of charged particles in a Coulomb or electrostatic eld. The potential for this case is V
REFERENCE: Arya, Sections 7.5 to 7.7 EXAMPLES: Arya 7.1 and 7.2 Review the derivation of the following important equations: 1. The angle as a function of radial distance (r) =
2
(L/r2 )dr E V (r)
L2 2r 2
(18)
2. The path equation d2 u + u = 2 2F 2 d Lu
1
Central Force Motion
REFERENCE: Arya, Sections 7.1 to 7.4 Review the derivation of the following important equations: 1. The eective single particle equation: r = F (r) r 2. The force (system of) equations: F (r) = ( r2 ) r 0 = (r + 2r) 3. The integrals
example: A ball of mass m and kinetic energy E is in an elastic collision with a second ball of mass 4m initially at rest. The two balls depart in directions making an angle of 120 degrees with each other. What are the nal energies of the two balls? solut
2
Elastic collisions
We now apply the conservation laws to the interaction of two particles, in particular collisions. In general, two types of collisions are possible: Elastic and Inelastic. By denition 1. Elastic collisions: Pi = Pf and Ki = Kf ; 2. Ine
1.3
Angular Momentum
L = r p = r mr = r mv
The angular momentum of a single particle is dened as (46)
We can extend this denition to a systems of N particles
N N
L=
k=1
(rk pk ) =
k=1
(rk mk r)
(47)
We can take the total angular momentum about any point
example: A chain of uniform mass density , length b, and mass M (where = M/b) hangs from both ends. At time t = 0, the ends are adjacent, but one is released. Find the tension in the chain at the xed point, after the other has fallen a distance x. Assume
1
Systems of particles
REFERENCE: Arya, Chapter 8. The ideas of Newtonian mechanics and the conservation theorems can be straightforwardly extended to systems of N particles.
1.1
Center of Mass
Consider a system of N particles 1, 2, 3.N , with masses m1 ,
1.7
2D oscillations
Consider the motion of a particle with two degrees of freedom. Take the restoring force to be proportional to the distance of the particle from a force center located at the origin, and to be directed toward the origin: F = k r (113) w
example: The equation of motion for a certain driven damped oscillator is x + 3x + 2x = 10 cos t and initially the particle is at rest at the origin. Find the subsequent motion. solution: Compare this equation of motion to equation 73, we see A = 10, = 1,