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001 10.0 points Consider the head-on collision of two masses, m 1 = 6 kg moving to the right with speed 8 m/s and m 2 = 12 kg moving to the left with speed 4 m/s. Given that they stick to- gether after the collision, what is the increase in internal energy of the system during the collision? Assume no energy is lost to the surroundings and neglect any external forces acting on the two masses.
Correct answer: 288 J.
Explanation: We know that the velocity of the center of mass is given by
vCM =
m 1 v 1 + m 2 v 2 m 1 + m 2
By substituting the values of m 1 , m 2 , v 1 and v 2 we have
vCM =
(m)(v 0 ) + (2m)(−
v 0 2
m + 2m
vCM = 0
Thus, the velocity of the center of mass is zero. We then apply the energy principle to this problem
∆E = ∆K + ∆Eint = 0
∆Eint = − ∆K = Ki − Kf
But, the final kinetic energy is zero since the velocity of the center of mass is zero. Thus, we have ∆Eint = Ki
Ki =
m v^20 +
(2m)
v 0 2
∆Eint =
m v 02
∆Eint =
m v 02
∆Eint =
(6 kg) (8 m/s)^2
∆Eint = 288 J
002 10.0 points As seen from above in the image, a string is wrapped around the edge of a uniform cylin- der of radius R and mass M which is initially resting motionless on a frictionless table. M^ F
ω
R
The end of the string is pulled with a force of F over a total distance l. If the linear speed of the cylinder is found to be v after pulling this distance, what is the angular speed of the cylinder? (Note that v 6 = ωR in this case.)
- ω =
M R^2
F R
- ω =
M R^2
F l −
M v^2
correct
- ω =
M R^2
F R −
M v^2
- ω =
M R^2
F l −
M v^2
- ω =
3 M R^2
F R
- ω =
3 M R^2
F l
- ω =
3 M R^2
F R
- ω =
M R^2
F l
- ω =
M R^2
F R −
M v^2
- ω =
3 M R^2
F l
Explanation: By the energy principle, ∆K = W. The moment of inertia of the disk is I = (1/2)M R^2.
1 2
M v^2 +
Iω^2 = F l
M v^2 +
M R^2 )ω^2 = F l
So,
ω =
M R^2
F l −
M v^2
Note that there is a way to eliminate v from the final answer using the angular momentum principle, but in that case the answer is
ω =
3 M R^2
F l
003 10.0 points Consider the following diagram.
E 0
E 1
E 2
E 3
III
II
I
IV
Say which arrows in the above diagram correspond to the following processes respec- tively: Absorption of a photon from the ground state to the first excited state; Emis- sion of a photon from an excited state to another excited state.
- Transition IV; Transition II
- Transition I; Transition III
- Transition I; Transition IV
- Transition II; Transition III
- Transition IV; Transition III
- Transition II; Transition IV
- Transition III; Transition II
- Transition III; Transition I
- Transition I; Transition II correct
- Transition III; Transition IV Explanation: Absorption sends the atom into a higher energy state, so for the first process the arrow will point upward. Given that the amount of energy is represented by the difference be- tween E 1 and E 0 for the first process, it should be clear that Transition I is the correct choice for the first process. On the other hand, emis- sion sends the atom into a lower energy state, so for the second process the arrow will point downward. The phrase “to another excited state” tells us that the final state should be either E 2 or E 1. Given the possibilities on the diagram, we find that Transition II is the correct choice for the second process.
004 10.0 points Starting from rest, a woman lifts a barbell of mass mbb with a constant force F through a distance h, at which point she is still lifting, and the barbell has acquired a speed v. Let Ewoman stand for the following energy terms associated with the woman:
006 10.0 points
In the figure, a block sitting on a frictionless horizontal surface is attached to a rigid wall on the right through a spring (whose axis is horizontal). A bullet is shot at the block from the left and gets embedded in it, causing the block to move to the right, thus compressing the spring. (Assume the bullet is travelling perfectly horizontally, along the axis of the spring, before hitting the block). Which of the following are true? A. The initial kinetic energy of the bullet is completely converted to spring potential energy when the spring reaches its maximal compression. B. The initial momentum of the bullet is equal to the momentum of the bullet+block system just after the bullet enters the block. C. Part of the momentum of the bul- let+block system is lost during the collision (i.e. before the spring-compression starts). D. Part of the energy of the bullet+block system is ”lost” (no longer present as macro- scopic kinetic energy) during the collision, be- fore the spring-compression starts. E. If we are given the masses of the block and the bullet, the initial speed of the bullet and the spring constant, it is possible to find the maximum compression of the spring.
- A, B
- A, E
- A, B, D
- A, B, E
- B, D, E correct
- B, D
- D
- A, C
- A, C, E Explanation: Since the bullet gets embedded in the block, the collision is inelastic. Hence, some the the Kinetic energy of the bullet is converted into internal energy, heat, etc during the collision and does not get stored up in the spring. Therefore, A is false and D is true. However, momentum is conserved during the collision, since forces acting on the bullet-block system during the short time period are negligible. Therefore, B is true and C is false. If we are given initial velocity and masses, we can use momentum conservation to find velocity (and hence kinetic energy) of bullet-block system just after collision. We can then use energy principle to get maximum compression of the spring. Hence, E is true.
007 10.0 points A skater pushes straight away from a wall. She pushes on the wall with a force whose magnitude is F , so the wall pushes on her with a force F (in the direction of her motion). As she moves away from the wall, her center of mass moves a distance d. Consider the following statements regarding energy.
I ∆Ktrans + ∆Einternal = F d II. ∆Ktrans + ∆Einternal = −F d III. ∆Ktrans + ∆Einternal = 0 IV. ∆Ktrans = F d V. ∆Ktrans = −F d
What is the correct form of the energy prin- ciple for the skater as a real system and as a point particle (PP) system?
- Real: III, PP: IV correct
- Real: II, PP: V
- Real: I, PP: IV
- Real: II, PP: IV
- Real: I, PP: V
- Real: V, PP: IV
- Real: V, PP: I
- Real: IV, PP: IV
- Real: IV, PP: III
- Real: III, PP: V
Explanation: As a real system, the wall does no work on the skater since the contact point of the skater’s hand and the wall does not move. Also, the wall does not transfer energy to the skater, i.e. by cooling down. The right hand side of the energy principle must therefore be zero for the real system. Furthermore, the left hand side must contain the skater’s change in internal energy for a real system. III is correct. As a point particle system, the skater’s cen- ter of mass moves a distance d while a force F is applied. Thus, the work done is F d. Translational kinetic energy is the only type of energy a point particle system can have. IV is correct.
008 10.0 points Find the moment of inertia of a solid sphere of mass M = 2.5 kg and radius R = 0.6 m about an axis that is tangent to the sphere.
ω
m r
Correct answer: 1.26 kg · m^2. Explanation: The moment of inertia of a solid about a diameter is
Icm =
M R^2.
Using the parallel-axis theorem, the moment of inertia about an axis that is tangent to the sphere is
I = Icm + M R^2
=
M R^2 + M R^2
M R^2
(2.5 kg)(0.6 m)^2
= 1 .26 kg · m^2
009 10.0 points A test car of mass 705 kg is moving at a speed of 7.4 m/s when it crashes into a wall to test its bumper. If the car comes to rest in 0 .35 s, how much average power is expended in the process?
Correct answer: 55151.1 W. Explanation:
- Id, IIa
- Ib, IId correct
- Ia, IIa
- Ia, IIc
Explanation: Since the temperature doesn’t change, the amount of thermal energy cannot change, so ∆Ethermal = 0. Since we have no change in the thermal energy, the energy that is being added by the natural gas and the sunshine must transfer out to the surrounding outside air. Hence, Q = −12000 J, where the sign is negative since the energy flows from the system (the house) to the surroundings (the outside air).
012 10.0 points A spring of stiffness k and relaxed length L stands vertically on a table. A person takes a mass M and very slowly lets the mass down onto the spring. On letting go, he finds that the mass does not move and the spring is compressed by a length d 1. The same experiment is now repeated by letting the mass go all of a sudden when the spring is still unstretched. The spring is found to compress by a length d 2 when the mass momentarily comes to rest. What is the ratio d 1 /d 2?
- M g kL
- (^14)
- 4
- kL M g
- 1
- 2
- 12 correct
Explanation: When the mass is very gradually let down
onto the spring, the net work done on the mass is 0 because at every instant while it is being let down, the net force acting on the mass is 0 (since the man always provides just enough force to balance all the forces). The final spring compression is determined by the fact that there is no net force acting on the mass even when the man has let it down, i. e., the spring force and the force of gravity cancel. Hence we can write
kd 1 − M g = 0.
From this, we get
d 1 =
M g k
When the mass is let go all of a sudden, it is easy to see that when the spring has been compressed by d 1 , there will still be a down- ward momentum for the mass due to the fact that it has been pulled down by the force of gravity which was always greater in magni- tude than the spring force (the work done till that instant would be non-zero). Hence, the mass goes down further, until all its Kinetic energy has been converted into potential en- ergy. Eventually, the mass oscillates about the equilibrium position conserving its total energy all the time. By using conservation of energy between the topmost point of os- cillation (the point at which the man just let the mass go) and the bottommost point of os- cillation (when the mass just comes to stop momentarily), we get
KE 1 +GP E 1 +SP E 1 = KE 2 +GP E 2 +SP E 2.
This translates to
0 + 0 + 0 = 0 − M gd 2 +
kd^22.
From this, we get
d 2 =
2 M g k
From these two expressions, we can see that the required ratio is 1/2.
013 10.0 points
The energy levels of a particular quantum object are − 8 .6 eV, − 5 .6 eV, and − 1 .2 eV. If a collection of these objects is bombarded by an electron beam so that there are some objects in each excited state, what are the energies of the photons that will be emitted?
- 3 eV, 7.4 eV, 4.4 eV correct
- 3 eV, 6.8 eV, 14.2 eV
- 14 .2 eV, 9.8 eV, 6.8 eV
- 14 .2 eV, 4.4 eV, 9.8 eV
- 3 eV, 14.2 eV, 4.4 eV
- 8 .6 eV, − 5 .6 eV, − 1 .2 eV
- − 8 .6 eV, − 5 .6 eV, − 1 .2 eV
- −3 eV, − 7 .4 eV, − 4 .4 eV
- 7 .4 eV, − 5 .6 eV, 1.2 eV
- 9 .8 eV, −3 eV, 7.4 eV
Explanation: 3 eV, 7.4 eV, 4.4 eV is the correct answer. The 3 emitted photons correspond to 3 tran- sitions between the states. Label them as E 1 = − 8 .6 eV E 2 = − 5 .6 eV E 3 = − 1 .2 eV
Then the possible transitions are given by
Ephoton = |∆E 1 , 2 | = |E 1 − E 2 | = |− 8 .6 eV + 5.6 eV| = 3 eV Ephoton = |∆E 1 , 3 | = |E 1 − E 3 | = |− 8 .6 eV + 1.2 eV| = 7.4 eV Ephoton = |∆E 2 , 3 | = |E 2 − E 3 | = |− 5 .6 eV + 1.2 eV| = 4.4 eV
014 10.0 points A projectile of mass m 1 moving with a speed v 1 in the +x direction strikes a stationary tar- get of mass m 2 head-on in an elastic collision. Find the final velocity of the projectile m 1. Hint: You can use the energy and momentum principles, OR you can solve the problem in the center of mass reference frame.
m 1 + m 2 m 1 − m 2
v 1
m 1 m 2
v 1
m 2 m 1 + m 2
m 2 2 m 1
v 1
m 1 m 1 + m 2
v 1
2 m 1 m 1 + m 2
v 1
m 1 − m 2 m 1 + m 2
v 1 correct
m 2 m 1
v 1
Explanation: From momentum conservation,
m 1 v 1 = m 1 v 1 ,f + m 2 v 2 ,f
which implies
v 1 ,f = v 1 − m 2 v 2 ,f /m 1.
From energy conservation,
m 1 v 12 = m 1 v 12 ,f + m 2 v 22 ,f.
Plugging in the expression for v 1 ,f in terms of v 2 ,f in the above expression, one can solve for v 2 ,f and obtain,
v 2 ,f =
2 m 1 m 1 + m 2
v 1
⇒ v 1 ,f =
m 1 − m 2 m 1 + m 2
v 1.
5. I-D, II-C, III-A, IV-B
6. I-C, II-B, III-D, IV-A
7. I-B, II-C, III-A, IV-D
8. I-C, II-A, III-B, IV-C
- I-A, II-B, III-D, IV-C correct
- I-B, II-D, III-C, IV-A
Explanation: I-A, II-B, III-D, IV-C is the correct answer. Vibrational states of diatomic molecules have the typical potential curve, and energy levels become closely spaced as energy goes up to zero - (A) Quantum oscillator has equally spaced en- ergy levels - (B) Electronic states have more wide separation than vibrational states, and vibrational states have more spacing than rotational states; so they combine to give (D) Electronic states of hydrogen are well known, which is also the only remaining op- tion - (C)
017 10.0 points A beam of high-energy π−^ (negative pions) is shot at a flask of liquid hydrogen, and some- times a pion interacts through the strong in- teraction with a proton in the hydrogen, in the reaction
π−^ + p+^ → π−^ + X+,
where X+^ is a positively charged particle of unknown mass.
Before After
π−^ p+
π−
X+
θ
The incoming pion momentum is 3065 MeV/c. The pion is scattered through 36 ◦, and its momentum is measured to be 1515 MeV/c. The X+^ particle is scattered through an unknown angle θ with an un- known momentum. A pion has a rest energy of 140 MeV, and a proton has a rest energy of 938 MeV. Find the scattering angle θ of the X+^ par- ticle.
Correct answer: 25.8334 Degrees.
Explanation: We conserve momentum in the x and y directions:
pπ,i =pπ,f cos 36◦^ + pX cos θ 0 =pπ,f sin 36◦^ − pX sin θ
We rearrange the second expression to solve for pX , and then plug it into the first expres- sion and solve for θ.
pπ,i = pπ,f cos 36◦^ + pX cos θ
pπ,i = pπ,f cos 36◦^ + pπ,f
sin 36◦ tan θ
tan θ =
pπ,f sin 36◦ pπ,i − pπ,f cos 36◦
=
1515 MeV/c sin 36◦ 3065 MeV/c − 1515 MeV/c cos 36◦ ⇒ θ = 25. 8334 ◦^.
018 10.0 points
Two spheres, A and B, have the same mass and radius. However, sphere B is made of a dense core and a less dense shell around it. How does the moment of inertia of sphere A about its center of mass compare to the moment of inertia of sphere B about its center of mass?
Ia. IA > IB Ib. IA < IB Ic. IA = IB
If the two spheres are rolled down an incline from the same height simultaneously,
IIa. sphere A reaches the bottom first. IIb. sphere B reaches the bottom first. IIc. spheres A and B reach the bottom simul- taneously.
Choose the correct pair of statements.
- Ic,IIc
- Ic,IIa
- Ia,IIc
- Ia,IIa
- Ib,IIc
- Ib,IIb
- Ic,IIb
- Ib,IIa
- Ia,IIb correct
Explanation:
Since the two spheres share the same mass and radius, a comparison of their moment of inertias depends on how their mass is dis- tributed relative to an axis through the center of mass. Sphere B has a dense core, implying that its mass is on average closer to the axis than A’s mass. Thus, A has a greater moment of inertia than B. Let β be the unitless parameter in the mo- ment of inertia formula I = βM R^2. Sphere B has a smaller β from the preceding argument. As a consequence of energy conservation, it can be shown that
vCM =
2 gh 1 + β
Therefore, B reaches the bottom first.
019 10.0 points Which of the following forces are conservative forces?
A) Gravitational force between the Earth and the Sun
B) Spring force
C) Air resistance
D) A constant force
E) Friction force
- A,B,D correct
- A,B
- A,B,C,E
- A
- C,D,E
- A,B,D,E
- A,D,E
- B
