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This print-out should have 31 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.
001 10.0 points A planet with a mass of 1 × 1023 kg travels around a star in a nearly circular obrbit in the xy plane as shown in the diagram. Its speed is constant at 15000 m/s.
Star
C A
D
B
Which arrow best describes the direction of ∆P~ in going from B to C?
a
e
g c
h b
f d
- h
- e
- b correct
- a
- c
- g.
- d
- f Explanation: We know that direction of ∆~P is given by ~PB + ∆~P = P~C
∆~P = P~C − P~B Thus, from the directions of P~C and P~B it follows that the correct answer is b
002 10.0 points A disk of radius 8.4 cm and mass 0.48 kg is pulled along a frictionless surface with a force of 14.3 N by a string wrapped around the edge (Fig. 11.43, displayed below). 24 cm of string has unwound off the disk. At the instant shown the angular velocity is 18 rad/s in the clockwise sense.
8 .4 cm
24 cm 14 .3 N
At time ∆t = 0.2 s later, what is the mag- nitude of the angular momentum about the center of the disk?
Correct answer: 0.270722 kg · m^2 /s. Explanation: Use ~τ = ~r × F~. Therefore, |~τ | = |~r⊥||F~ | = (0.084 m) (14.3 N). Thus, |~τ | = 1.2012 N · m.
~L = I~ω =< 0 , 0 , −Iω >, where I = (1/2)M R^2 for the solid disc and ~ω is directed along the negative z-axis. Using the values given from the beginning of the problem, the equation becomes
~L =< 0 , 0 , −(1/2)M R^2 ω > =< 0 , 0 , − 0 .0304819 kg · m^2 /s > |L~| = 0.0304819 kg · m^2 /s
Since the torque acting on the disc is con- stant over time, we conclude the change in angular momentum is given by
∆~L = ~τ ∆t ∆~L =< 0 , 0 , −RF ∆t >
The final angular momentum is then given by
~Lf = ~L + ∆~L
= ~L+ < 0 , 0 , −RF ∆t > =< 0 , 0 , −(1/2)M R^2 ω − RF ∆t > =< 0 , 0 , − 0. 0304819 − 0. 24024 > kg · m^2 /s =< 0 , 0 , − 0 .270722 kg · m^2 /s >
|~Lf | = 0.270722 kg · m^2 /s
003 10.0 points Planet A and planet B of equal mass m orbit the same star of mass M in circular trajecto- ries of radii rA = R and rB = 2R respectively. Calculate the ratio of the kinetic energy of A to the kinetic energy of B.
- 1/
- 1/
- 1/
- 4
- 2 correct
- 16
- 1/
Explanation: Since A has a circular orbit, m v A^2 R
G m M R^2
which implies
KEA =
m v A^2 =
G m M 2 R
Similarly, for B, m v^2 B 2 R
G M m 4 R^2
which implies
KEB =
m v B^2 =
G M m 4 R
Therefore, KEA KEB
004 10.0 points A rubber ball is dropped from rest onto the floor, and bounces back up to the same height from which it started. Ignore the force of fric- tion due to the air. Which of the following sets of plots most accurately depict this mo- tion? (The force plots depict the force on the ball by the environment.)
- (^) t
y
t
vy
t
Fy
- (^) t
y
t
vy
t
Fy
correct
- (^) t
y
t
vy
t
Fy
Solving the equations in Ib, one obtains
vmin,rope =
gL
vmin,rod = 0
Hence, it is seen that vmin,rope > vmin,rod. Thus, IIa is the correct answer.
006 10.0 points A proton of mass mp is moving along the -ˆz direction and undergoes a change in its speed from 0.994c to 0.998c. What is the magnitude and direction of the impulse acting on the proton?
- 2 mpc along -ˆy
- 2 mpc along +ˆz
- 1 mpc along -ˆz
- 1 mpc along +ˆx
- 004 mpc along +ˆz
- 7 mpc along -ˆz correct
- 7 mpc along +ˆz
- 004 mpc along -ˆz
Explanation:
Let : mp = 1. 7 × 10 −^27 kg , v 1 = 0. 994 c , v 2 = 0. 998 c , and c = 3 × 108 m/s.
γ =
v^2 c^2
, so
γ 1 =
(0. 994 c)^2 c^2 = 9. 14243 and
γ 2 =
(0. 998 c)^2 c^2 = 15. 8193.
We use the formulae for the impulse and the momentum
Impulse = p~f − p~i
pf − pi
(−ˆz)
M omentum = γ m v
Thus
I =
(γ 2 v 2 − γ 1 v 1 ) c
× mp c
(15.8193) (0. 998 c) c
× mp c
(9.14243) (0. 994 c) c
× mp c = 6. 7 mp · c.
007 10.0 points If you were designing a circular race track for F1 cars that travel 295 km/hr on a track banked at 36 degrees all the way around the circuit, what track radius would you use so that no frictional force would be required for a car to make the turn? Use g = 9.8 m/s^2.
Correct answer: 943.087 m. Explanation: Looking at a free body diagram on which we have drawn the normal and gravitational
forces, we see that the normal force must be decomposed in order to write the equations of motion in the x and y directions. The x direction is a natural choice for writing the momentum principle because it happens to be the centripetal direction in this problem. Writing the equations of motion along both directions,
y dir :
dpy dt
= FN cos θ − m g
0 = FN cos θ − m g ⇒ FN =
m g cos θ
We can also write
x dir :
dpx dt
= FN sin θ
Putting these together, we have
m v^2 R
= m g tan θ
⇒ R =
v^2 g tan θ
=
(295 km/hr)^2 (9.8 m/s^2 ) tan 36◦ = 943 .087 m.
008 10.0 points The energy levels in hydrogen are given by
En =
− 13 .6 eV n^2
If the hydrogen atom goes from its third ex- cited state to the ground state, what would be the energy of the emitted photon?
- − 12 .09 eV
- None of the above.
- 10.2 eV
- − 10 .2 eV
- − 12 .75 eV
- 12.09 eV
- Since energy is conserved, there would be no emitted photon.
- Not enough information.
- 12.75 eV correct Explanation: If the hydrogen atom goes from its third excited state (n = 4) to its ground state (n = 1), the energy difference (positive value) gives the emitted photon energy:
Ephoton(4 → 1) =E 4 − E 1
= − 13. 6
eV
= 12 .75 eV
009 10.0 points Two identical 0.4 kg blocks (labeled 1 and 2 in the diagram) are initially at rest on a nearly frictionless surface, connected by an unstretched spring, as shown in the upper portion of the figure.
140 N
140 N
x = 0 m 0.6 m
x = 0.2 m 2.0 m
Initial
Final
Then a constant force of 140 N to the right is applied to block 2, and at a later time the blocks are in the new positions shown in the lower portion of the figure. At this final time, the system is moving to the right and also vibrating, and the spring is stretched. What is the final vibrational energy (Uspring + Kvib)of the two block and spring system?
Let : L 0 = 20 cm , L 1 = 24 cm , L 2 = 12 cm , and L 3 = 32 cm.
The spring force is given by
= k
| L~s| − L 0
Lˆ
The value of F~ 1 is
= k
| L~ 1 | − L 0
Lˆ
= k (4) Lˆ
The value of F~ 2 is
= k
| L~ 2 | − L 0
Lˆ
= k (−8) Lˆ
The value of F~ 3 is
= k
| L~ 3 | − L 0
Lˆ
= k (12) Lˆ
Thus, the ratio of the three forces is
3 F~ 1 = (− 3 /2) F~ 2 = F~ 3
012 10.0 points A cart on a spring rides along a track which will eventually launch it at an angle over a gap. What’s the minimum amount of spring compression s that will allow the cart to jump the gap?
H
θ
d h
The total mass of the cart is m and the spring constant is ks. Hint: Let the speed
of the cart when it hits the gap be vg. Solve for the range d, from which you can obtain vg in terms of d. Then use this result in an energy equation describing the track part of the problem.
- s =
k m
[
2 g(h − H) +
gd 2 sin θ cos θ
]
- s =
k m
[2g(h + H) + 2gd sin θ cos θ]
- s =
k m
[2g(h − H) + 2gd sin θ cos θ]
- s =
m k
[2g(H − h) + 2gd sin θ cos θ]
- s =
m k
[2g(h − H) + 2gd sin θ cos θ]
- s =
m k
[
2 g(h − H) +
gd 2 sin θ cos θ
]
correct
- s =
k m
[2g(H − h) + 2gd sin θ cos θ]
- s =
m k
[2g(h + H) + 2gd sin θ cos θ]
- s =
k m
[
2 g(H − h) +
gd 2 sin θ cos θ
]
- s =
m k
[
2 g(H − h) +
gd 2 sin θ cos θ
]
Explanation:
y − dir : 0 = vg sin θ − gtf / 2
tf =
2 vg sin θ g
x − dir : d = vg cos θtf =
2 v g^2 sin θ cos θ g
Ef = Ei
mv^2 + mgh = mgH +
kss^2
s =
m k
[
2 g(h − H) +
gd 2 sin θ cos θ
]
013 10.0 points You know a collision must be “elastic” if:
I. The colliding objects stick together. II. The colliding objects are stretchy or squishy. III. The sum of the final kinetic energies equals the sum of the initial kinetic ener- gies. IV. There is no change in the internal ener- gies of the objects (thermal energy, vi- brational energy, etc.) V. The momentum of the two-object sys- tem doesn’t change and the total energy decreases.
- I and III
- I and IV
- IV
- III and IV correct
- II and V
- III
Explanation: If there is no change in the internal ener- gies, assuming that the total energy doesn’t change, then there will be no change in the kinetic energy, and this defines an “elastic” collision. Thus III and IV are correct.
014 10.0 points A horse whose mass is M gallops at constant speed v up a long hill whose vertical height is h, taking an amount of time t to reach the top. Assume the horse’s hooves do not slip on the rocky ground and that air resistance is neg- ligible. When the horse started running, its
temperature rose quickly to a point at which from then on, heat transferred from the horse to the air keeps the horse’s temperature con- stant. Once the temperature of the horse is constant, which equation is the correct state- ment of the energy principle, ∆Esys = W +Q, taking the horse as the system? (Leave out any terms that are equal to zero.)
- ∆Echemical = −M gh − |Q| correct
- ∆Erest+∆K+∆Echemical = +M gh−|Q|
- ∆Erest + ∆K + ∆Echemical + ∆Ethermal + ∆Ugrav = 0
- ∆Echemical + ∆Ugrav = −M gh
- ∆Echemical = +M gh + |Q|
- ∆Erest+∆K+∆Echemical = −M gh+|Q|
- ∆Echemical = −M gh
- ∆Echemical + ∆Ethermal = 0
- ∆Echemical + ∆Ethermal = −M gh − |Q|
- ∆Erest + ∆K + ∆Echemical + ∆Ethermal + ∆Ugrav = −M gh − |Q|
Explanation: We will consider each possible energy term on the left hand side of the equation, which represents the change of the horse’s en- ergy. The horse’s mass does not change, so ∆Erest = 0. The horse’s velocity is constant, so ∆K = 0. The horse’s chemical energy decreases, so ∆Echem 6 = 0. The horse’s tem- perature is constant, so ∆Ethermal = 0. The horse is one object, so it has no gravitational potential energy: ∆Ugrav = 0. On the right hand side, the work done on the horse is due to gravity of the Earth and is given by WEarth = −M gh. Heat is leaving the horse, so Q < 0.
015 10.0 points
Correct answer: 1.04533 cm. Explanation:
Let : ka = 48 N/m , d = 2 × 10 −^10 m , L = 2 m , m = 128 kg , A = 1 × 10 −^6 m^2 , and g = 9.8 m/s^2.
We know that the Young’s modulus is given by
Y =
ka d
48 N/m 2 × 10 −^10 m
= 2. 4 × 1011 N/m^2
By using the formula below, we can determine ∆L and we use the fact that F = m g.
F A
= Y
∆L
L
By making use of the fact that
∆L =
F
A
L
Y
(128 kg)(9.8 m/s^2 ) 1 × 10 −^6 m^2
2 m
- 4 × 1011 N/m^2
= 1.04533 cm
018 10.0 points Consider following statements which compare U 235 medium with U 238 medium based on the ball-spring model. Which of the following statements are correct? On the length of the interatomic bonds: Ia. d 235 < d 238 Ib. d 235 = d 238 On the stiffness of the interactomic bonds: IIa. k 235 > k 238 IIb. k 235 = k 238
On the speeds of sound: Define the ratio R =
v 235 v 238
IIIa. R =
IIIb. R =
- Ia, IIb, IIIb
- Ib, IIb, IIIb
- Ib, IIb, IIIa correct
- Ia, IIa, IIIa
- Ib, IIa, IIIa
- Ib, IIa, IIIb
- Ia, IIb, IIIa
- Ia, IIa, IIIb Explanation: U 238 contains 3 more neutrons than U 235. So far as the electric content of the two atoms is concerned, the two atoms are identical. This implies that the two interatomic bonds, i.e. their length and their stiffness, are identi- cal. Since the speed of sound is
v = d
ks ma
so
R =
v(235) v(238)
d(235)
k(235) m(235) d(238)
k(238) m(238)
=
m 238 m 235
019 10.0 points How close would another Earth have to be to our Earth (center-to-center distance) so that you would weigh half of your current weight? Neglect rotation. (In the figure, FE 1 is the gravitational force due to the Earth you are standing on, FE 2 is the gravitational force due to the other Earth, and Fscale is
the normal force which measures your weight. The condition of the problem is that Fscale = 1 2
FE 1 .)
D
D − RE
RE (^) F E 1
FE 2 Fscale
1. D = (2 +
2)RE
2. D = 3RE
3. D = RE
4. D =
GME
RE
5. D =
2 GME
R^2 E
6. D = (1 +
2)RE correct
7. D =
2 GME
R^2 E
8. D =
GME
R^2 E
9. D = 2RE
10. D = 2
2 RE
Explanation: For you to stand stationary on the surface of the Earth,
FE 2 + Fscale − FE 1 = 0. We are solving for your scale weight to be half of the force you usually feel due to the Earth.
FE 2 +
FE 1
− FE 1 = 0
⇒ FE 2 =
FE 1
We know that
GME 2 m (D − RE )^2
GME 1 m 2 R^2 E
⇒ 2 R E^2 = (D − RE)^2
⇒ D = RE +
2 RE > 2 RE
We see that our result is greater than 2RE, so it is sensible.
020 (part 1 of 2) 10.0 points What is the minimum speed with which a person of mass 70 kg riding the Barrel of Fun of radius 9 m must spin so he will held in place vertically? The coefficient of static friction between the person and the wall is 0 .1. Use g = 9.8 m/s^2.
ω
Correct answer: 29.6985 m/s. Explanation: Writing the momentum principle in the ra- dial and vertical directions,
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.
023 10.0 points If all three collisions in the figure are totally inelastic, which cause(s) the most damage (deformation of objects, thermal energy in- crease, etc.)? Assume that the wall is station- ary and the car is completely stopped by it in the first diagram.
m
v
brick
wall
I
m^2 m v 0.5 v II
III^ m^ v^^2 v 0.5 m
- all three
- III correct
- II
4. II, III
5. I, II
6. I
7. I, III
Explanation: By momentum conservation the final veloc- ity in all three cases is zero, so in each case all the kinetic energy is lost, causing damage by crumpling the vehicles. The kinetic energy lost in case I is
KI =
mv^2
. The kinetic energy lost in case II is
KII =
mv^2 +
(2m)(0. 5 v)^2
mv^2.
The kinetic energy lost in case III is
KIII =
mv^2 +
m 2
(2v)^2
mv^2.
Since K 3 > K 2 > K 1 , case III causes the most damage.
024 10.0 points
For a satellite and Earth system, which set of curves shown above correspond to the the case when the satellite is in a circular orbit? Note that some sets of graphs may be physically impossible.
- Diagram D
- Diagram F
- Diagram C
- Diagram A
- Diagram B correct
- Diagram E
- Diagram G
Explanation: For a circular orbit, K, U and E are ex- pected to be constant, and at the same time E < 0. Diagram B fits this criteria.
025 10.0 points A ladder of mass m = 10 kg leans against a frictionless vertical wall so that its base makes an angle of 60◦^ with the floor. The coefficient of static friction between the ladder and the floor is 0.24. If a man of mass M = 2m walks up the ladder, what fraction of its length can he climb before it begins to slip? Use g = 9.8 m/s^2.
xL
L
b
m
M = 2m
b 0 .6 m
A^ ×
Clay
A lump of clay with mass m = 0.5 kg falls and sticks to the outer edge of the wheel at the location A = 〈− 0. 5 , 0. 331662 , 0 〉 m (choice of axes: x to the right, y up, z out of the page; origin at the center of the axle). Just before the impact, the clay has speed v 0 = 8 m/s, and the disk is rotating clockwise with angular speed ω 0 = 0.5 rad/s. Just after the impact, what is the angular speed of the combined system of wheel plus clay?
Correct answer: 1.15873 rad/s.
Explanation: To calculate the angular speed of the com- bined system, one needs to compute the mo- ment of inertia of the system.
Isys =
M R^2 + m R^2. Therefore,
ωsys =
∣~Ltot,f
Isys
m v 0 x 0 −
M R^2
ω 0
1 2
M R^2 + m R^2
= 1 .15873 rad/s.
028 10.0 points Compared to an empty ship, would a ship loaded with a cargo of Styrofoam sink deeper or rise in the water?
- The ship will float at the same level be- cause its density hasn’t changed.
- It depends on the weight of the Styro- foam.
- Since Styrofoam is less dense than water, the ship rises.
- Since Styrofoam pushes down on the ship with its weight, the ship sinks deeper. cor- rect
Explanation: Anything loaded onto the ship (regardless of its density) will cause the ship to sink lower.
029 10.0 points A disc with a light string would around it has a mass of 1.5 kg. With your hand, you pull the string straight up with some constant force F such that the center of the disc does not move up or down, but the disc spins faster and faster. Nothing but the vertical string touches the disc.
y 0
yf
ω 0 = 0 ω
Floor Floor
F
M, I
Your hand begins at a height 0.5 m above the floor and the disc is at rest and not rotat- ing. How long does it take for your hand to rise to a height yf = 1.3 m above the floor?
Use g = 9.8 m/s^2.
Correct answer: 0.285714 s.
Explanation: We begin by using the position update equation for the hand and rewriting the hand velocity in terms of the disc’s angular velocity, given by vhand = ω R.
∆y = vavg,y ∆t
∆t =
∆y vavg,y
2∆y vhand,f + vhand,i
2∆y/R ωf + ωi
Now we can apply the angular momentum principle to solve for ωf. Note that F = M g since the disc’s center of mass is stationary.
ωf = ωi +
τCM I
∆t
=
RF
I
∆t
Substituting this back into the position up- date equation, we obtain
∆t =
2∆y/R RF I
∆t
2∆y R^2 M g/( 12 M R^2 )
∆y g = 0 .285714 s.
030 10.0 points A rod of length L and negligible mass is at- tached to a uniform disk of mass M and radius R (see the figure below). A string is wrapped around the disk, and you pull on the string with a constant force F. Two small balls each of mass m slide along the rod with negligible friction. The apparatus starts from rest, and when the center of hte disk has moved a dis- tance d, a length of string s has come off the disk, and the balls have collided with the ends of the rod and stuck there. The apparatus slides on a nearly frictionless table. Here is a view from above:
b
b
b
b
F L
m
m
M, R
d
ω
F
d + s
What is the change in internal energy of the system due to the balls colliding with the ends of the rod?
- ∆Eint = F s −
M R^2 + m L^2
ω^2 cor- rect
When the two particles are very far away, their potential energy is 0, and since they have nonzero initial velocities, this means that they are unbounded and thus have an overall pos- itive energy at r = ∞, which is also equal to the kinetic energy at that location. As the electron and proton get closer, due to their Coulomb attraction their kinetic energies in- crease while the negative potential energy de- creases even further. Thus the correct answer is Figure (II).
