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This print-out should have 21 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering.
001 10.0 points Which of the following objects are in uni- form motion? (Remaining at rest is also con- sidered to be uniform motion.) A. an object held motionlessly by two strings B. a planet traveling in a circle at a constant speed C. a skateboard sliding down a hill (ignore friction) D. a bullet shot on the moon by a gun E. an electron traveling at constant velocity F. a proton coming near an electron G. a mass oscillating on a spring H. an object without an applied force in a frictionless environment
- E, F, H
- None
- A, C, E, F
- A, B, D, E, H
- B, C, E, H
- A, B, E, H
- A, C, E
- All
- A, E, H correct
- B, D, E
Explanation: All are experiencing a net force and hence will not be in uniform motion except: A– The net force of the strings and any external forces is zero. E– Constant velocity implies constant momentum and hence uniform motion. H– Again, no forces are present.
002 10.0 points When they are far apart, the momentum of a particle is 〈− 4 × 10 −^22 kg m/s, 0, 0〉, as it ap- proaches another particle that is initially at rest. The two particles repel each other with- out coming close enough to touch. At a later time, the particles are again far apart, but now one has momentum 〈 5. 6 × 10 −^21 kg m/s,
3 × 10 −^21 kg m/s, 0〉. What is the y component of the momentum of the other particle now?
-2.4e-
-4.9e-
-2.8e-
6.5e-
-4.1e-
7.8e-
7.6e-
8.5e-
8.8e-
5.3e-
Correct answer: 5. 3 × 10 −^21 kg m/s. Explanation: Since there is no net external force, the total momentum of the system will be conserved. Initially, only one particle is moving so the initial momentum is ~pi = 〈− 4 × 10 −^22 kg m/s, 0 , 0 〉. The final momen- tum will be the sum of the final momenta of the two particle, and will be equal to the initial momentum
~pf = 〈 5. 6 × 10 −^21 kg m/s, 5. 3 × 10 −^21 kg m/s, 0 〉+~p 2
From here we can see that the y component of the final momentum of the other particle is
py 2 f = − 5. 3 × 10 −^21 kg m/s.
003 10.0 points A tug boat is pulling a barge of mass
- 2 × 105 kg into the harbor by a cable. The barge’s momentum is changing at a rate of
- 278 × 105 kg m/s^2. The water exerts a con- stant resistive (frictional) force of 5. 3 × 105 N. The cable is made of a material with a Young’s modulus of 3. 2 × 1011 N/m^2 and has a diam- eter of 15 cm and a natural length 65 m.
Find the amount by which the cable stretches.
Correct answer: 0.0110095 m.
Explanation: By the momentum principle we know that
F^ ~net = d~p/dt.
F~net is the sum of the resistive force ~Fres, which points in the negative direction, and the force provided by the cable F~cable. Thus we can write
Fcable = d~p/dt + Fres.
The cross-sectional area of the cable is
A =
πd^2 4
= 0.0176714 m^2
. The fractional change in length is then
∆L L
Fcable AY
9. 578 × 105 N
(0.0176714 m^2 )(3. 2 × 1011 N/m^2 )
For an initial length 65 m this gives
∆L = (0.000169376)(65 m) = 0.0110095 m.
004 10.0 points A crazed monkey man is swinging from a vine of natural length r. His mass is m. At the lowest point of his trajectory, he is moving horizontally with momentum ~p = 〈px, 0 , 0 〉. The magnitude of his momentum is changing at a rate d|~p| dt
= ma.
The vine acts like an elastic rope with spring constant k. Find the expression for that amount that the vine stretches.
mg +
mp^2 x r k
−mg + ma +
p^2 x mr
k
−mg +
p^2 x mr k
−mg + ma +
p^2 x mr k
mg + ma +
p^2 x mr k
mg +
p^2 x mr k
correct
mg +
p^2 x mr
k
mg −
mp^2 x r k
mg + a +
p^2 x mr k
mg − ma +
p^2 x mr k
Explanation: The increase in the magnitude of the mo- mentum must be caused by some force which acts parallel to the momentum, and thus does not come from the vine. The tension in the vine must counteract gravity and provide the perpendicular part of the change in momen- tum, mv x^2 /r or equivalently p^2 x/mr. Inserting this tension force into the spring equation F = kx and solving for x gives the correct answer.
005 10.0 points A helium-filled balloon is tied to a 0.3 m long, 0 .067 kg string. When deflated, the balloon has a mass of 0.2 kg. Assume the balloon
Correct answer: 1.11304 N/m.
Explanation: From the given data, we can find the Young’s modulus of the material.
Y =
F
A × (∆L/L)
(72 kg)(9.8 m/s^2 ) (9.13 mm)^2
×
3 .4 m 5 .4 mm
= 5. 32969 × 109 N/m^2
We also need to find the interatomic bond length. Note that the volume of the wire is V = (3.4 m)(9.13 mm)^2 , and so given the density, we can find the mass of the wire as
mwire = V ρ = (3.4 m)(9.13 mm)^2 (7.97 g/cm^3 ) = 2258.81 g.
Then, since we know the mass of a mole of the material, and the number of atoms in a mole, we can find the number of atoms in the wire:
N = NA
mwire mmole
= (6. 02 × 1023 ) ×
2258 .81 g 43 .7 g
We visualize atoms as cubes with side length equal to the interatomic bond length d, so the volume of the wire must be equal to the number of atoms times the volume of a single atom, or V = N d^3. Solving for d and using the above numbers gives
d = 2. 08838 × 10 −^10 m.
Finally, we have the following formula for the stiffness of the bond:
k = Y d = (5. 32969 × 109 N/m^2 )(2. 08838 × 10 −^10 m) = 1.11304 N/m.
007 10.0 points Alpha decay is a type of radioactivity in which a nucleus can emit what is known as an alpha particle. Consider a nucleus which is initially moving in the -z direction and which then emits an alpha particle in the +x direc- tion Which of the following statements about the components of the nucleus’s final momen- tum are true?
1a. The x component is positive. 1b. The x component is negative.
1c. The x component is zero. 2a. The y component is positive. 2b. The y component is negative. 2c. The y component is zero. 3a. The z component is positive. 3b. The z component is negative. 3c. The z component is zero.
- 1c, 2c, 3c
- 1a, 2c, 3a
- 1b, 2c, 3b correct
- 1b, 2c, 3c
- 1c, 2c, 3b
- 1a, 2a, 3b
- 1a, 2c, 3b
- 1b, 2c, 3a
- 1a, 2b, 3b
- 1b, 2b, 3b Explanation: This process involves no external forces so momentum will be conserved. The nucleus starts with some momentum in the -z direc- tion and the alpha particle carries no momen- tum in the z direction so the z component of the nucleus’s momentum will continue to be zero. Likewise, the nucleus starts with zero momentum in the x direction, but the alpha particle carries momentum in the +x direc- tion, so the nucleus must be left with an equal amount of momentum in the -x direction.
008 10.0 points A 4 kg metal block slides along the floor. The coefficient of sliding (or kinetic) friction between the block and the floor is 0.26. The initial speed of the block is 6 m/s. How far does the block slide before coming to a stop?
Correct answer: 7.06436. Explanation:
Let : m = 4 kg , g = 9.8 m/s^2 , vi = 6 m/s , and μk = 0. 26.
We know that the frictional force fk is given by −μkFN where FN is the normal force and equal to mg. Thus, we make use of the rela- tion
vf = vi +
fk m
∆t
= vi −
μk (mg) m
∆t
But, the final velocity is zero since the block comes to a stop.
0 = 6 m/s − (0.26)(9.8 m/s^2 )∆t
∆t = 2.35479 s The distance covered is given by
xf = xi + vi∆t +
fk 2 m
∆t^2
= xi + vi∆t −
μk (mg) 2 m
∆t^2
= 0 + (6 m/s)(2.35479 s) −
(0.26)(9.8 m/s^2 ) 2
(2.35479 s)^2 = 7. 06436
009 10.0 points A planet is in a circular orbit of radius r around a massive star. The planet has mo- mentum p and goes once around in a time T. The gravitational force exerted on the planet by the star is equal to
p^2 r
, in the direction of the momentum
p^2 r
, at a right angle to the momentum
2 π T
) p, in the direction of the momentum
- 0
2 π T
) p, at a right angle to the momentum correct
Explanation: The only force acting on the planet is the gravitational force from the massive star, which acts always perpendicular to the path of the planet. Therefore, dp dt
mv^2 R
vp R
2 π T
) p.
010 10.0 points
A mass is hung as shown above. The ten- sion in String 2 is 32.34 N. The mass remains stationary. What is the mass of the block?
Correct answer: 3.3 kg. Explanation:
Correct answer: 7590.05 N.
Explanation: The momentum principle tells us that
∆~p = F~avg∆t.
For Object 2, since it starts at rest, we have
∆~p = 〈71 kg m/s, 89 kg m/s, 0 〉.
And so
F^ ~ = ∆~p ∆t
71 kg m/s 0 .015 s
89 kg m/s 0 .015 s
The magnitude is then
|~F | =
71 kg m/s 0 .015 s
89 kg m/s 0 .015 s
013 10.0 points The graph below represents the relationship between the force applied to a single inter- atomic bond and the change in length of the bond.
Force (N)
Applied Force vs. Compression
Compression (Displacement) (m)
Suppose this is a substance with a natu- ral interatomic bond length of 6. 5 × 10 −^10 m. What is the Young’s modulus of the sub- stance?
- 84615 × 109 N/m^2 correct
- 07692 × 109 N/m^2
- 15385 × 108 N/m^2
- 53846 × 109 N/m^2
- 07692 × 108 N/m^2
Explanation: By definition the spring constant ks is the slope of the F vs compression plot. From the figure, one finds: ks =
= 2.5 N/m. We may then use Y = ks/d to find
Y =
2 .5 N/m
- 5 × 10 −^10 m
= 3. 84615 × 109 N/m^2.
014 10.0 points
The diagram above shows a top-down view of a ball with a mass of 0.85 kg attached to a 0.70 m long string, which is attached to a post. The ball moves moves in a circular path on a low-friction surface with a constant speed of 2.3 m/s. At the instant shown in the diagram,which of the following statements about the rate of change of the ball’s momentum,
d~p dt
, is TRUE?
d~p dt
points toward the post (the center of the circle). correct
d~p dt
is in the same direction as the ball’s momentum.
d~p dt
is zero, because the ball’s speed is not
changing.
- Both the component of
d~p dt
parallel to the
momentum and the component perpendicular to the momentum are non-zero.
- It is impossible to tell the direction of
d~p dt
because the component of
d~p dt
parallel to the
momentum is unknown.
Explanation: d~p dt
is the sum of the component along (
d~p dt
and perpendicular (
d~p dt
)⊥ to the path. The
ball moves with constant speed and therefore, the component along the path has to be zero. However, there is a change in direction of
the momentum and therefore
d~p dt
is not zero;
(
d~p dt
)⊥ points along the tension force towards
the post.
015 10.0 points One mole of a particular substance has a mass Mmole = 0.0665 kg. This substance has an interatomic bond length of d = 0.226 nm. One mole contains NA = 6. 02 × 1023 atoms. Calculate the volume of 2.91935 kg of this substance.
- 7.45121e-
- 7.61231e-
Correct answer: 0.000305061 m^3.
Explanation: We may find the number of atoms N in a
mass m = 2.91935 kg from
N =
NAm Mmole
We visualize the atoms as cubes with side length d, so the total volume will be
V = N d^3 = (6. 02 × 1023 )×
2 .91935 kg 0 .0665 kg
×(0.226 nm)^3 =
016 10.0 points Astronomical observations show that young stars are surrounded by disks of dust and rocky debris. Imagine a planet in a circular orbit around one of these young stars. It will of course experience a gravitational force, but it will also experience a drag force opposite its direction of motion due to collisions with the debris. Which of the following statements are true?
A. The planet’s momentum is changing. B. The planet’s speed is changing.
C. The gravitational force contributes to the change in momentum. D. The gravitational force contributes to the change in speed.
E. The drag force contributes to the change in momentum. F. The drag force contributes to the change in speed.
1. A, B, C, D, E, F
- None are true
- A, B, C, D, F
- B, D, F
- A, C, E
- A, B, C, D, E
- A, B, D, E, F
- A, B, C, E, F correct
Explanation: The system consists of only of the block and the surroundings of the Earth, the spring and the string. The forces that result from the interactions of the objects with the block are Fgrav (Earth), FT (string) and Fspring (spring).
019 (part 2 of 2) 10.0 points What is the magnitude of the force exerted on the block by the spring?
- 59 N
- 84 N
- 41 N correct
- 28 N
- 34 N
Explanation:
020 10.0 points At how many points on an elliptical (but
non-circular) orbit is
d~p dt
||
, the component
of the change in momentum which is parallel to the momentum, equal to zero?
- 3
- 6
- 0
- more information needed
- at every point
- cannot be predicted
- 4
- 8
- 1
- 2 correct
Explanation: At the points closest to and farthest from the central object, the momentum will be completely perpendicular to the force, so that( d~p dt
||
= 0. That is, at two points.
021 10.0 points You are moving to a new apartment, so you have to slide your couch over to the door and carry it outside. You exert a force F on the couch, which has mass mc, but it does not move. Your friend, who has a mass mf , is a jerk and is sitting on the couch while you try to move it. The coefficient of kinetic (or sliding) fric- tion between the couch and the floor is μk, and the coefficient of static friction is μs. The acceleration of gravity is g. What is the magnitude of the frictional force between the couch and the floor?
- (mc + mf )^2 gμs
- (mc + mf )gμk
- (mc + mf )gμ^2 s
- (mc + mf )gμs
- mcgμs
- (mc + mf )g/μs
- mcgμk
- (mc + mf )g/μk
- F correct
- mcg(μk + μs)
Explanation: Static friction provides enough force to keep an object stationary, up to the limit (mc + mf )gμs. Since the couch is not yet moving, static friction is providing just F.
