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This print-out should have 30 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.
001 10.0 points The graph is a potential energy diagram for a particle in some system as a function of position x.
−x 0 x 0 2 x 0 3 x 0 4 x 0 5 x 0
A
B
U
(x
x
What is the force F on the particle at x = 3 x 0?
1. F = −
B
4 x 0
- F =
A
5 x 0
- F = 0
4. F =
A
4 x 0
- F =
B
5 x 0
- F =
B − A
4 x 0
- F =
B
4 x 0
correct
8. F = −
B − A
4 x 0
- F = −
A
4 x 0
- F = −
B
5 x 0
Explanation:
F = −
d U dr
0 − B
5 x 0 − x 0
B
4 x 0
002 10.0 points Halley’s comet moves about the Sun in an elliptical orbit, with its closest approach to the Sun being 0.604 AU and its greatest dis- tance being 16.7 AU (1 AU=the Earth-Sun distance). If the comet’s speed at closest approach is 54 .7 km/s, what is its speed when it is farthest from the Sun? You may assume that its angular momen- tum about the Sun is conserved.
Correct answer: 1.97837 km/s. Explanation: Using conservation of angular momentum, we have
Lapogee = Lperihelion , or (m ra^2 ) ωa = (m rp^2 ) ωp , thus m ra^2
va ra
= m rp^2
vp rp
, giving ra va = rp vp , or va =
rp ra
vp
(0.604 AU)
(16.7 AU)
(54.7 km/s)
= 1 .97837 km/s.
003 10.0 points A force F = F 0
ˆı + ˆ + kˆ
acts on a rigid body at a point r = r 0 (ˆı − ˆ) away from the axis of rotation.
What is the resulting torque on the body?
- r 0 F 0
ˆk − 2ˆı − ˆ
- r 0 F 0
ˆk − ˆı − ˆ
- r 0 F 0
2 ˆk − ˆı − ˆ
. correct
- r 0 F 0
ˆk − 2ˆı − 2 ˆ
- r 0 F 0
2ˆı − ˆk − ˆ
- r 0 F 0
2 ˆ − ˆı − kˆ
- Zero.
- r 0 F 0
ˆk − ˆı − 2 ˆ
- r 0 F 0
ˆı + ˆ + kˆ
Explanation:
~τ = r × F = F 0 r 0
ˆı × ˆ + ˆı × ˆk − ˆ × ˆı − ˆ × ˆk
= F 0 r 0
ˆk − ˆ + ˆk − ˆı
= F 0 r 0
2 ˆk − ˆ − ˆı
004 10.0 points A bowling ball of mass M is coming in at a speed of vb and hits a stationary ping pong ball of mass m.
vb vb vp
Assuming that M >> m and the speed of the bowling ball is essentially unchanged, what would the final speed of the ping pong ball, vp, be closest to? The collision is elastic. (Hint: In the frame of the bowling ball, the ping pong ball bounces back with the same speed it came in with.)
- None of the above.
vb
- Not enough information.
vb
- vb
vb
- 3 vb
- 2 vb correct
- 4 vb
Explanation: We can shift the frame of reference to the bowling ball’s frame, in which the ping pong ball is coming in at a speed of vb and the bowling ball is stationary.
vb vp
vb vb 2 vp
After the collision the bowling ball is still stationary, while the ping pong ball bounces back with the same incoming speed vb. When we shift back to the rest frame, where the ping pong ball is initially stationary, we can see that the final speed of the ping pong ball is 2vb.
005 10.0 points
t
x
0 tQ tR tS tP
Explanation: The displacement is the integral of the ve- locity with respect to time:
~x =
~v dt.
Because the velocity increases linearly from zero at first, then remains constant, then de- creases linearly to zero, the displacement will increase at first proportional to time squared, then increase linearly, and then increase pro- portional to negative time squared. From these facts, we can obtain the correct answer.
t
x
0 tQ tR tS tP 007 10.0 points In otherwise empty space is a system of 4 particles, each of the same mass. The acceler- ations of the particles are as follows: a 1 = ˆi a
a 2 = ˆj
a 2
a 3 = −(ˆi + ˆj) (2 a) a 4 = (ˆi − ˆj) (3 a) ,
where a > 0 is a constant. What is the acceleration of the center of mass of the system of 4 particles?
- acm = ˆia + ˆj
5 a 8
- acm = ˆi
a 4
11 a 8
- acm = ˆi a − ˆj (9 a).
- acm = ˆi (2 a) + ˆj
7 a 8
- acm = ˆi
a 6
− ˆj
7 a 8
- acm = ˆi
a 2
− ˆj
9 a 8
. correct
- acm = ˆi
a 4
− ˆj
9 a 4
- acm = ˆi
a 2
− ˆj
9 a 2
- acm = 0. Explanation: In this case, since the masses are equal,
acm =
a 1 + a 2 + a 3 + a 4 4
=
ˆi a 4
ˆj a 8
(ˆi + ˆj) (2 a) 4
(ˆi − ˆj) (3 a) 4 = ˆi
a 4
a 2
3 a 4
a 8
a 2
3 a 4
= ˆi
a 2
− ˆj
9 a 8
008 10.0 points A large spool is pulled across a level table- top by a constant force applied to a string wrapped tightly around the inner part of the spool.
T
T
If the spool does not slip on the tabletop, and the string does not slip on the spool, in which direction will the spool roll in the two different orientations shown?
- Insufficient information is given.
- To the right; to the right correct
- To the right; to the left
- To the left; to the right
- To the left; to the left Explanation: Pick the spot at which each spool touches the floor. The only torque around this spot is
the torque exerted by the force T, which is in the same direction (into the paper and away from you) on both spools. Therefore, both spools will accelerate in rotation in the same sense, which is the direction of T, implying rightward translational motion.
009 10.0 points A ball of mass m is attached to a spring of negligible mass, and force constant k. The spring is freely pivoted at the opposite end from the ball, and the system is initially at rest and horizontal.
L
L
L
If the ball is allowed to fall, what is its speed when the system is swinging through the vertical and the spring is stretched an
amount ∆L =
L
, where L is the spring’s
initial, unstretched length?
m g L 10
22 g −
k L 10 m
L
correct
k L m
− 11 g
L
g −
k L m
L
k L m
− 10 g
L
- Not enough information is given.
k L 10 m
− 22 g
L
10 g −
k L m
L
k L^2 m g
Explanation: Using conservation of energy with E = K + Us + Ug and taking Ug = 0 at the instant the system is vertical we get
Ei = Ef Ki + Usi + Ugi = Kf + Usf + Ugf
0 + 0 + m g (L + ∆L) =
m v^2 +
k (∆L)^2
m g
11 L
m v^2 +
k
L^2
22 m g L 10
k L^2 100
= m v^2 ( m L 10
22 g −
k L 10 m
= mv^2 √(
22 g −
k L 10 m
L
= v
010 10.0 points A uniform bar of length L and weight W is attached to a wall with a hinge that exerts a horizontal force Hx and a vertical force Hy on the bar. The bar (which makes an angle θ with respect to wall) is held by a cord that makes a 90◦^ angle with respect to bar.
b
b
b
b b
time (s)
velocity (m/s)
What is the position at 9 seconds?
Correct answer: 47.5 m.
Explanation:
b
b
b b
b
b
b (^1 2 3 4 5 6 7 8 9) b
time (s)
velocity (m/s)
The acceleration during the first 2 seconds is
a =
∆v ∆t
9 m/s − 0 m/s 2 s − 0 s = 4.5 m/s^2.
The position at 2 seconds is 10 m plus the area of the triangle (shaded above)
x = 10 m +
(2 s − 0 s)(9 m/s − 0 m/s) = 19 m ;
however, it can also be calculated:
x = xi + vi (tf − ti) +
a (tf − ti)^2 = 10 m + (0 m/s) (2 s − 0 s)
(4.5 m/s^2 )(2 s − 0 s)^2 = 19 m.
The acceleration during the second time interval is
a =
∆v ∆t
6 m/s − 9 m/s 6 s − 2 s = − 0 .75 m/s^2.
The position at 6 seconds is 19 m plus the area of the trapezoid
x = 19 m +
(6 s − 2 s)(6 m/s + 9 m/s) = 49 m ;
however it can also be calculated:
x = xi + vi (tf − ti) +
a (tf − ti)^2 = 19 m + (9 m/s) (6 s − 2 s)
(− 0 .75 m/s^2 )(6 s − 2 s)^2 = 49 m.
The acceleration during the third time in- terval is
a =
∆v ∆t
−1 m/s − 0 m/s 9 s − 6 s = − 0 .333333 m/s^2.
Finally the position at 9 seconds is 49 m plus the area of the triangle
x = 49 m +
(9 s − 6 s)(−1 m/s − 0 m/s) = 47.5 m ;
however it can also be calculated:
x = xi + vi (tf − ti) +
a (tf − ti)^2
= 49 m + (0 m/s) (9 s − 6 s)
(− 0 .333333 m/s^2 )(9 s − 6 s)^2
= 47 .5 m.
013 10.0 points A newly discovered planet has the same den- sity as the Earth but twice the radius. How does the gravitational force g′^ on the surface of the new planet compare to g, the gravita- tional force on the surface of the earth?
- g′^ = g/
- g′^ = g/ 4
- g′^ = 2g correct
- g′^ = g/ 8
- g′^ = 2
2 g
- g′^ =
2 g
- g′^ = g/ 2
- g′^ = 4g
- Cannot determine from the information.
- g′^ = 8g
Explanation: From Newton’s second law and the law of universal gravitation, the gravitational force near the surface is
Fg = m g = G
M m R^2 g =
G M
R^2
Now, M = V ρ =
πR^3 ρ and M = V ′ρ = 4 3
πR′^3 ρ. Thus,
g′ g
M ′
M
R
R′
R′
R
R
R′
R′
R
Since R′/R = 2, g′^ = 2g.
014 10.0 points The rotational inertia of a rod of mass M and length L about one end is I =
M L^2.
L
What is the rotational inertia of such a rod about a point in empty space a distance L from the nearest end of the rod, along the line of the rod?
- M (2 L)^2 = 4 M L^2
M L^2
- Still the same, since empty space has no mass and thus no rotational inertia.
M L^2
- Zero.
M L^2 correct
M L^2
M L^2
M L^2
M L^2
(3 ρ + ρw) V g 4
ρw V g 4 Explanation: The buoyant force is the density of the fluid displaced times the volume displaced, so
FB =
ρw V g +
ρ V g =
(ρ + 3 ρw) V g 4
017 10.0 points A steel safe with mass M falls onto a concrete floor. Just before hitting the floor, its speed is vi. It hits the floor without rebounding, and ends up being d shorter than before. Find the magnitude of the (average) force exerted by the floor on the safe while it is being deformed. You can assume this force is a constant during the short time it takes for the safe to deform. Ignore the effects of gravity.
1. F =
2 M v i^2 d^2
- F =
M v i^2 d^2
- F =
M vi d
- F =
M vi d^2
- F =
M v i^2 2 d^2
- F =
M v i^2 2 d
correct
7. F =
M v i^2 d
- F =
M vi 2 d^2
- F =
2 M v i^2 d
- F =
M vi 2 d
Explanation: Since all the movement happens along the vertical axis, we can treat this as a one- dimensional problem, with upward direction
as positive. From momentum principle, we have- ~pf − ~pi = F~ ∆t
M |vf − vi| = F ∆t
F =
M vi ∆t
We can find the time interval using average velocity:
vav =
d ∆t
∆t =
d (vi + vf )/ 2
2 d vi
The force then becomes
F =
M v i^2 2 d
018 10.0 points Consider a microscopic spring-mass system whose spring stiffness is 49 N/m, and the mass is 4. 7 × 10 −^26 kg. What is the smallest amount of vibrational energy that can be added to this system? Use the values
¯h = 6. 582 × 10 −^16 eV · s 1 eV = 1. 602 × 10 −^19 J c = 2. 998 × 108 m/s.
Correct answer: 0.0212524 eV.
Explanation:
ω 0 =
k m
=
49 N/m
- 7 × 10 −^26 kg = 3. 22886 × 1013 rad/s. For a quantum oscillator, the difference in successive energy levels is
∆E = ¯h ω 0 = (6. 582 × 10 −^16 eV · s) × (3. 22886 × 1013 rad/s) = 0 .0212524 eV.
Note that this is in the IR range of the electromagnetic spectrum.
019 10.0 points A triangular wedge 5 m high, 13 m base length, and with a 11 kg mass is placed on a frictionless table. A small block with a 6 kg mass (and negligible size) is placed on top of the wedge as shown in the figure below.
11 kg
13 m
5 m
6 kg
∆Xwedge
11 kg
13 m
5 m 6 kg
All surfaces are frictionless, so the block slides down the wedge while the wedge slides sidewise on the table. By the time the block slides all the way down to the bottom of the wedge, how far ∆Xwedge does the wedge slide to the right?
Correct answer: 4.58824 m. Explanation:
Let : M = 11 kg , m = 6 kg , L = 13 m , and H = 5 m.
Consider the wedge and the block as a two- body system. The external forces acting on this system — the weight of the wedge, the weight of the block and the normal force from the table — are all vertical, hence the net hor- izontal momentum of the system is conserved,
P (^) xwedge + P (^) xblock = constant.
Furthermore, we start from rest =⇒ center- of-mass is not moving, and therefore the X coordinate of the center-of-mass will remain constant while the wedge slides to the right and the block slides down and to the left,
Xcm =
m Xblock + M Xwedge m + M
= constant.
Note: Only the X coordinate of the center- of-mass is a constant of motion; i.e., the Ycm accelerates downward because the Py compo- nent of the net momentum is not conserved. Constant Xcm means ∆Xcm = 0 and there- fore
m ∆Xblock + M ∆Xwedge = 0.
Note that this formula does not depend on where the wedge has its own center-of-mass; as long as the wedge is rigid, its overall dis- placement ∆Xwedge is all we need to know. Finally, consider the geometry of the prob- lem: By the time the block slides all the way
What mass of water is used to cool the engine? The specific heat of water is 4180 J/kg ·◦C and of iron 450 J/kg ·◦C.
Correct answer: 17.9813 kg.
Explanation:
Let : Qlost = 4. 3 × 106 J , Th = 36◦C , Tl = 15◦C , ce = 450 J/kg ·◦C , cw = 4180 J/kg ·◦C , and me = 288 kg.
∆T = ∆Tw = ∆Te = Th − Tl , so
Qlost = mw cw ∆ Tw + me ce ∆ Te mw cw∆T = Qlost − me ce ∆T
mw =
Qlost cw ∆T
me ce cw
=
4. 3 × 106 J
(4180 J/kg ·◦C) (21◦C)
−
(288 kg) (450 J/kg ·◦C) 4180 J/kg ·◦C = 17 .9813 kg.
023 10.0 points A mass of 11 kg is fixed at the origin, and a mass of 42 kg is fixed at
〈 45 , 0 , 0 〉 m.
Where should a mass of 2 kg be placed so that there is no net gravitational force on it? Use G = 6. 7 × 10 −^11 N · m^2 /kg^2.
- 〈 18. 3738 , 0 , 0 〉 m
- 〈 18. 6396 , 0 , 0 〉 m
- 〈 16. 1652 , 0 , 0 〉 m
- 〈 15. 2335 , 0 , 0 〉 m correct
- 〈 17. 3783 , 0 , 0 〉 m
- 〈 17. 6871 , 0 , 0 〉 m
- 〈 14. 8971 , 0 , 0 〉 m
- 〈 16. 7092 , 0 , 0 〉 m
- 〈 20. 0446 , 0 , 0 〉 m
- 〈 15. 8233 , 0 , 0 〉 m Explanation: Because both the m 1 = 11 kg and m 2 = 42 kg masses rest on the x-axis, any nonzero y or z coordinate of the m = 2 kg mass would mean it would feel a nonzero force in the y or z direction. Thus the mass must also rest on the x-axis to be in equilibrium. Furthermore, as the gravitational force is attractive, mass m must rest between masses m 1 and m 2 in order to cancel the forces on it. Letting mass m rest at x-coordinate x, we equate the magnitude of the forces on it:
G
m 1 m x^2
= G
m 2 m (d − x)^2
where d is the x coordinate of m 2. Using the quadratic formula to solve for x, we choose the solution such that 0 < x < d.
x = d
m 1 −
m 1 m 2 m 1 − m 2
= 15.2335 m.
So the final coordinate of mass m is
〈 15. 2335 , 0 , 0 〉 m.
024 10.0 points Consider a loop-the-loop system where the radius of the loop is R. A small block (of mass m and negligible size) is released from rest at the point P , which is at a height of h.
h (^) R
P C
m
If the mass presses on the track at C with
a force of magnitude
m g , find the initial
height h of the block.
- h =
R
- h =
R
- h =
R
- h =
R
- h =
R
- h =
R
- h =
R
- h =
R
- h =
R
- h =
R correct
Explanation: Consider the free-body diagram at C:
m g
N
The centripetal acceleration acts toward the center of motion, so applying Newton’s 2nd law for the mass at C, m aC = NC + m g
m
v C^2 R
m g + m g =
m g 1 2
m v^2 C =
m g R.
From conservation of energy, UC + KC = UC + KC
m g h + 0 = 2 m g R +
m v C^2
m g (h − 2 R) =
m v^2 C =
m g R
h − 2 R =
R
h =
R.
025 10.0 points A projectile of mass m 1 moving with a speed v 1 in the +x direction strikes a stationary target of mass m 2 = 2 m 1 head-on in an elastic collision. Find the final velocity of the projectile m 1. Hint: You can use the energy and momentum principles.
v 1
v 1
- v 1
- − 5 v 1
v 1 correct
v 1
v 1
where α is the angular acceleration. If the interaction occurs over a time t, then
ω = αt, v = at ,
and v ω
a α
R
i.e.
v =
R
ω.
Finally, we can substitute this into the work- energy equation above and get
F l =
M
R
ω
M R^2
ω^2 ,
or
ω =
F l M R^2
027 10.0 points A certain material is kept at very low temper- ature. It is observed that when photons with energies between 0.2 and 0.9 eV strike the ma- terial, only photons of 0.3 eV and 0.7 eV are absorbed. Next the material is warmed up so that it starts to emit photons. When it has been warmed up enough that 0.7 eV photons begin to be emitted, what other photon ener- gies (in eV) are also observed to be emitted by the material?
- 0.4 eV
- 0.2 eV, 0.4 eV
- No other energy
- 0.3 eV
- 0.2 eV, 0.3 eV, 0.4 eV
- 0.3 eV, 0.4 eV correct
Explanation: 0.3 eV, 0.4 eV is the correct answer. Absorption spectra correspond to absorp- tion from ground state, hence we know that the 3 states can be represented by energies
0, 0.3 and 0.7 eV respectively (the ground state energy is taken as 0 eV). When emis- sion of 0.7eV photons is seen, there must be a substantial number of electrons in all these 3 states; hence, the photons emitted correspond to all 3 transitions - 0.4, 0.3 and 0.7 eV.
028 10.0 points A car of weight 3250 N operating at a rate of 175 kW develops a maximum speed of 43 m/s on a level, horizontal road. Assuming that the resistive force (due to friction and air resistance) remains constant, what is the car’s maximum speed on an incline of 1 in 20; i.e., if θ is the angle of the incline with the horizontal, sin θ =
Correct answer: 41.349 m/s. Explanation: If f is the resisting force on a horizontal road, then the power is
P = f vhorizontal
f =
P
vh
On the incline, the resisting force is
F = f + m g sin θ = f +
W
P
vh
W
, so
v =
P
F
P
P
vh
W
1. 75 × 105 W
1. 75 × 105 W
43 m/s
3250 N
= 41 .349 m/s.
029 10.0 points A mouse hangs on the rim of a spinning turntable. The rotational inertia of the mouse about the center of the turntable is compara- ble to rotational inertia of the turntable by itself. Mouse and turntable are freely spin- ning with constant angular velocity ω. The mouse starts to crawl toward the center of the turntable. At a particular position the angular velocity of mouse and turntable has increased to 3 ω.
Before After
b b
ω 3 ω At this instant, what is the total angular momentum of the system in terms of its orig- inal value Li?
- Lf =
Li 3
- Lf = 9 Li
- Lf =
Li 9
- Lf = 3 Li
- Lf = Li correct
Explanation: No external torques act on the system of mouse and turntable, so as long as the mouse stays on the turntable, the total angular mo- mentum cannot change.
d L dt
τi.
030 10.0 points A projectile fired into the air suddenly ex- plodes into several fragments. What can be said about the motion of the center of mass of the system made up of all the fragments after the explosion?
- Center of mass of the system follows the
same parabolic path the projectile would have followed if there had been no explosion. cor- rect
- Center of mass does not move.
- Center of mass of the system moves in the direction of a fragment with the biggest mass.
- There is not enough information given.
- Center of mass of the system moves in the direction opposite to the direction of a fragment with the biggest mass.
- Center of mass of the system moves in the direction opposite to the direction of a projectile just before it exploded. Explanation: Since the explosion is due to the internal forces within the system, the center of mass will continue with the same motion as if there were no explosion.
