Physics 250, Sections 611 & 612
Introductory Physics I
Summer, 2016
Dr. A. J. Viescas
134 Woodland
ajv3@psu.edu
I. MATERIALS
Required: Webassign subscription.
Suggested: College Physics, 10th Edition
Solutions to Exam 1A
1. A motorist drives north for 36.0 minutes at 92.0 km/h and then stops for 15.0 minutes. He then
continues north, traveling 130.0 km in 2.50 h.
(a) What is the total time for the
Solutions to Exam 2A
1. A child and sled with a combined mass of 54.0 kg slide down a frictionless slope. The sled starts from rest
and has a speed of 2.20 m/s at the bottom. (Note: You CANNOT use New
Chapter 4 Suggested Problem Solutions
3. m = 6.0 kg. a = 2.0 m/s2.
(a) Fnet = ma = (6.0 kg)( 2.0 m/s2) = 12 N.
(b) Fnet = 12 N. m = 4.0 kg.
a = Fnet / m = (12 N) / (4.0 kg) = 3.0 m/s2.
5. wE = 5.00 lb
Solutions to Exam 1A
1. A jet plane lands with a speed of 89.0 m/s and accelerates at a rate of -4.70 m/s2 as it comes to
rest.
(a) From the instant the plane touches the runway, what is the time need
47. For a body in circular orbit the period of motion is called T and is the time it takes for the body to revolve
around its primary. T is related to the angular velocity in that for an orbiting body
Old Exam 3
1. A large grinding wheel in the shape of a solid cylinder of radius 0.330 m is free to rotate on a frictionless,
vertical axle. A constant tangential force of 220 N applied to its edge cau
Solutions to Exam 1C
1. A motorist drives north for 32.0 minutes at 60.5 km/h and then stops for 15.0 minutes. He then
continues north, traveling 130.0 km in 2.30 h.
(a) What is the total time for the
Solutions to Exam 1B
1. A motorist drives north for 34.0 minutes at 89.0 km/h and then stops for 15.0 minutes. He then
continues north, traveling 130.0 km in 2.20 h.
(a) What is the total time for the
Equations and Constants
KINEMATICS
DYNAMICS
One and Two Dimensional Motion
Newtons First Law
Fnet,x = F1x + F2x + . . . + FNx = 0
x = xf xi
y = yf yi
Fnet,y = F1y + F2y + . . . + FNy = 0
vav,x = x / t
Chapter 8 Suggested Problem Solutions
9. m = 2.00 kg. B = 15o. (see figure)
rg = 33.0 cm = 0.330 m.
rB = 8.00 cm = 0.0800 m.
Assuming the angle between the arm and
the gravitational force is 90o, the
15. m = 7.80 g = 7.80 10-3 kg. v i = 575 m/s. x = 5.50 cm = 0.0550 m. v f = 0.
Assuming that the only non-conservative force on the bullet is the force of friction, f k , and also assuming that
the bu
Old Exam 2
1. A 2.10 103 kg car starts from rest at the top of a 4.20-m-long driveway that is inclined at
24.0 with the horizontal. An average friction force of 4.50 103 N impedes the motion of the
ca
29. x = 40.0 m. t = 8.50 s. vx = 2.80 m/s.
(a) For constant acceleration: vx2 = v0x2 + 2axx and vx = vx0 + axt
From the first of these equations: ax = (vx2 v0x2) / (2x)
From the other equation: ax = (
63. Let us call v BW the velocity of the boat with respect to the water, vWE the velocity of the water with respect
to the Earth, and v BE the velocity of the boat with respect to the Earth. From the
Chapter 13 Suggested Problem Solutions
11. xi = 2.00 cm = 0.0200 m (compressed). xf = 0. m = 100 g = 0.100 kg.
vi = 0. (it starts from rest) At the top of the trajectory: vf = 0. y = 60.0 cm = 0.600 m
Chapter 9 Suggested Problem Solutions
9. m = 200 kg. L0 = 4.00 m. A = 0.200 10-4 m2. Y = 8.00 1010 N/m2.
F = mg = (200 kg)(9.807 N/kg) = 1.96 103 N.
F / A = YL / L0
L = L0F / (AY) = 4.90 10-3 m.
(2.02
Chapter 1 Suggested Problem Solutions
7. L = 9.72 m. W = 5.3 m.
A = LW = (9.72 m)(5.3 m) = 51. (Round final answer to 2 significant figures.)
14.
(a) 756 + 37.2 + 0.83 + 2.5 = 797. (Round to zero deci
The strengths of the couplings between pairs of CSFs whose energies cross are
evaluated through the SC rules. CSFs that differ by more than two spin-orbital occupancies
do not couple; the SC rules giv
Chapter 12
Along "reaction paths", configurations can be connected one-to-one according to their
symmetries and energies. This is another part of the Woodward-Hoffmann rules
I. Concepts of Configurati
of the electrons (the contribution to the dipole operator arising from the nuclear charges - a
Zae2 Ra does not contribute because, when placed between 1 and 2 , this zero-electron
operator yields a v
< |1s1s2p02p0| H |1s1s2p-12p1| > = < 2p02p0 | 2p-12p1 >
< |1s1s2p12p-1| H |1s1s2p-12p1| > = < 2p12p-1 | 2p-12p1 >.
Certain of these integrals can be recast in terms of cartesian integrals for which
eq
In all of the above examples, the SC rules were used to reduce matrix elements of
one- or two- electron operators between determinental functions to one- or two- electron
integrals over the orbitals w
1/2 cfw_< 2s 2s | x x > + < 2s 2s | y y > +i < 2s 2s | y x > -i < 2s 2s | x y > =
< 2s 2s | x x > = K2s,x
(here the two imaginary terms cancel and the two remaining real integrals are equal);
< 2s 2s
and
Kij = <ij | e2/r12 |ji>
are the orbital-level one-electron, coulomb, and exchange integrals, respectively.
Coulomb integrals Jij describe the coulombic interaction of one charge density ( i2
above
9. What is the electric dipole matrix elements between the
1 = | 1 1| state and the 1 = 2-1/2[| 1 -1| +| -1 1|] state?
<2-1/2[| 1 -1| +| -1 1|] |r| 1 1|>
= 2-1/2[< -1|r| 1> + < -1|r| 1>]
=21/2 < -1|r|
<| H|> = <|> = <*|*>
(note, again this is an exchange-type integral).
6. What is the Hamiltonian matrix element coupling | and
2-1/2 [ |*| - |*|]?
The first determinant differs from the 2 determinant
*i(r) *j(r') g(r,r') k(r)l(r') drdr'.
The notation < i j | k l> introduced above gives the two-electron integrals for the
g(r,r') operator in the so-called Dirac notation, in which the i and k indice