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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 (part 1 of 2) 10.0 points Which of the following choices corresponds to a system of two electrons that start out far apart, moving toward each other (that is, their initial velocities are nonzero and they are heading straight at each other)? Note that the horizontal and vertical axes in each plot are the separation between the particles and energy, respectively.
K + U
U
K
(I)
r
K + U
U
(II) K
r
K + U
U
(III) K
r
U K + U
K
(IV)
r
K + U
U
K
(V)
r K^ +^ U
U
K
(VI)
r
- Figure III
- Figure I correct
- Figure II
- Figure IV
- Figure VI
- Figure V
Explanation: When the two electrons 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 positive energy at r = ∞, which is also equal to the kinetic energy at that location. As the electrons get closer, due to their Coulomb repulsion their kinetic energies drop to 0 while the potential energy rises. Thus the correct answer is Figure (I).
002 (part 2 of 2) 10.0 points Which of the diagrams corresponds to a sys- tem of a proton and an electron that start out far apart, moving toward each other (that is, their initial velocities are nonzero and they are heading straight at each other)?
- Figure I
- Figure V
- Figure II correct
- Figure III
- Figure VI
- Figure IV
Explanation: 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).
003 10.0 points
Two identical objects are moving directly to- ward one another at the same speed v.
~v −~v m m
What is the total kinetic energy of the sys- tem of the two objects?
- Zero since (~v − ~v)^2 = 0
- m v^2 since energy is a scalar correct
- Still
m v^2 from the definition of kinetic
energy
- 4 m v^2 since (2 v)^2 = 4 v^2
- None of these
Explanation: Since kinetic energy is an additive scalar, the total kinetic energy of the system of two objects is
1 2
m v^2 +
m v^2 = m v^2.
004 10.0 points An astronaut in a space shuttle at a height 200 km above the surface of the earth (radius 6380 km) drops a wrench that has mass M. The wrench floats nearby the astronaut. What is the net force on the wrench?
- Approximately Mg (g = 9.8 m/s^2 ) cor- rect
- Impossible to determine without more information
- Much less than Mg (g = 9.8 m/s^2 )
- Zero
- None of the above
Explanation: The wrench experiences approximately the same gravitational force as it would on the surface of the earth Mg, though slightly
smaller because it is at a radius that is 200km larger than it would be at the surface. As it circles around the earth, the gravitational force accelerates the wrench, changing only the direction, not the magnitude of it’s ve- locity. Likewise, the astronaut experiences a gravitational force about the same as he or she would at the surface, and an acceleration also approximately equal to g. Thus, the wrench appears to not accelerate at all with respect to the astronaut, i.e. appears weightless to the astronaut.
005 10.0 points An object of volume V is floating at the in- terface between water and a denser liquid. 3 4
of the object’s volume is displacing water,
while
of the object’s volume is displacing the heavier liquid.
ρw ρ
If the density of the heavier liquid is ρ and the density of water is ρw, what is the total buoyant force on the object?
ρ V g 4
- (ρ + ρw) V g
ρw V g 4
(3 ρ + ρw) V g 4
- (ρ − ρw) V g
3 ρw V g 4
(ρ + 3 ρw) V g 4
correct
Explanation:
gravitational force giving
may = N cosθ − mg = 0
Eliminating N from the above equations, we get v =
gRtanθ.
008 10.0 points A body oscillates with simple harmonic mo- tion along the x-axis. Its displacement varies with time according to the equation
x(t) = A cos(ω t).
Given the maximum acceleration amax = 2 m/s^2 , the period T = 3 s, determine the amplitude of oscillation, A.
Correct answer: 0.455945 m.
Explanation: The acceleration is the rate of change of the velocity and so it’s also the second time derivative of the position of the ob- ject, i.e. a = dv/dt = d^2 x/dt^2. Thus we have a = −ω^2 A cos(ω t), which has a max- imum amplitude of amax = Aω^2. We need only the angular frequency ω which is related to the period by ω = 2π/T so in the end:
amax = A
( 2 π T
giving A = amax
( T
2 π
009 10.0 points A bucket containing a rock of mass m is ro- tated in a vertical circle of radius 0.873 m. What must be the minimum speed of the pail at the top of the circle so that the rock won’t fall out?
Correct answer: 2.92496 m/s.
Explanation: The system is the rock. In order for the rock to move in a circle of radius 0.873 m, it ex- periences a net force equal to the centripetal force necessary to maintain the circular mo- tion. A free body diagram of the system includes only the force of gravity and the nor- mal force exerted by the bottom of the bucket, so Fnet = FN + Fg. When moving with the minimum velocity, the rock is just about to fall out, so the normal force is zero and the centripetal acceleration is supplied by gravity alone.
FN + Fg = mg = = m acentripetal
= m
v^2 r
Then,
m g = m
v^2 r
and so,
v =
g r
=
(9.8 m/s^2 ) (0.873 m)
= 2 .92496 m/s.
010 10.0 points A force acting on a particle has the potential energy function U (x), shown by the graph. The particle is moving in one dimension under the influence of this force and has kinetic energy 1.0 Joule when it is at position x 1.
x 0 x 1 x 2 x 3
Position
Potential Energy (J)
Potential Energy vs Position
Which of the following is a correct state- ment about the motion of the particle?
- It comes to rest at either x 0 or x 2 and remains at rest.
- It moves to the left of x 0 and does not return.
- It oscillates with maximum position x 2 and minimum position x 0. correct
- It moves to the right of x 3 and does not return.
- It cannot reach either x 0 or x 2.
Explanation: In this case, the total energy of the particle is conserved so at any point on the axis,
V (x) + U (x) = V (x 1 ) + U (x 1 ) = 1.0 J + (− 1 .0 J) = 0 V (x) = −U (x) ,
where V is the kinetic energy of the particle. Since kinetic energy ≥ 0 ,
V (x) = −U (x) ≥ 0 U (x) ≤ 0 J ,
so the particle oscillates between position x 0 and x 2.
011 10.0 points Two marbles, one twice as massive as the other, are dropped from the same height. When they strike the ground, how does the kinetic energy of the more massive marble compare to that of the other marble?
- It has 4 times the KE.
- It has one-half the KE.
- It has the same KE.
- It has twice the KE. correct Explanation: KE = 1/ 2 mv^2 The two marbles experience the same ac- celeration while falling and so hit the ground with the same velocity. Therefore the mar- ble with twice the mass has twice the kinetic energy.
012 10.0 points Object A with mass 4 kg moves at an initial speed of 2 m/s along a frictionless horizontal plane. A horizontal force F is applied oppo- site to the direction of the motion and brings object A to a stop in a distance ∆xA. Object B with mass 2 kg moves at an initial speed of 4 m/s along the frictionless horizontal plane. The same horizontal force F is applied oppo- site to the direction of the motion of B and brings object B to a stop in a distance ∆xB. What is the relation between ∆xA and ∆xB?
- ∆xA=4∆xB
- ∆xB =4∆xA
- ∆xB =8∆xA
- ∆xA=∆xB
- ∆xA=6∆xB
- ∆xB =2∆xA correct
- ∆xB =6∆xA
force between them. Then, energy conserva- tion gives:
Ei = Ef Ki + Ui = Kf + Uf
0 +
k q^2 d
m v f^2 + 0
vf =
k q^2 d m
015 10.0 points Sally applies a total force of 115 N with a rope to drag a wooden crate of mass 100 kg across a floor with a constant acceleration in the x-direction of a = 0.1 m/s^2. The rope tied to the crate is pulled at an angle of θ = 54◦ relative to the floor.
m
F
Fx
◦
Calculate the total work done by friction as the block moves a distance 45.4 m over the horizontal surface. The acceleration due to gravity is 9.8 m/s^2.
- -10932.
- -2614.
- -4498.
- -14178.
- -17580.
- -4803.
- -2334.
- -8505.
- -7851.
- -12064.
Correct answer: − 2614 .83 J.
Explanation: The system is the block. We know that Fnet,x = ma because the block is accelerating in the x-direction. A free body diagram gives
Ff + Fx = ma
so that the frictional force is given by
Ff = ma − F cos θ
The total work done by friction in moving the block from xi to xf is given by
Wf =
∫ (^) xf
xi
Ff dx =
∫ (^) xf
xi
(ma − F cos θ)dx
Wf = (ma − F cos θ)d.
016 10.0 points
The buoyant force exerted on a solid object immersed in a fluid can be: A. less than the weight of the object; B. greater than the weight of the object; C. the same as the weight of the object.
- B only
- C only
- A and C only
- A only
- All are true. correct
- A and B only
- None is true.
- B and C only
Explanation: Imagine an object made of the fluid with the same shape as the solid object. When immersed into the fluid, the net force on this fluid object would be zero, which means the pressure of the surrounding liquid must just balance the gravitational force. Now replace the fluid object with the solid object that has the same geometry; the pressure of the surrounding fluid would be the same. Hence the force upward on the object (the buoyant force) equals the weight of the displaced fluid.
That force depends only on the geometric shape of the object and not its weight, i.e. gravitational force it experiences. Therefore all three statements A, B, and C could be true depending on the density of the object relative to the density of the fluid.
017 10.0 points The escape speed from a very small asteroid is only 16 m/s. If you throw a rock away from the asteroid at a speed of 42 m/s, what will
be its final speed? G = 6. 7 × 10 −^11
N · m^2 kg^2
Correct answer: 38.833 m/s.
Explanation:
vesc =
2 GM
R
v^2 esc =
2 GM
R
Use the Energy Principle.
Ei = Ef Ui + Ki = Uf + Kf −GM m ri
mv i^2 = 0 +
mv f^2
Where Uf is zero.
v f^2 = v^2 i −
2 GM
R
v f^2 = v^2 i − v^2 esc
vf =
v^2 i − v^2 esc
=
(42 m/s)^2 − (16 m/s)^2
vf = 38 .833 m/s
018 10.0 points The following graph represents a hypothetical potential energy curve for a particle of mass m.
r
U (r)
O
3 U 0
2 U 0
U 0
r (^0 2) r 0 If the particle is released from rest at posi- tion r 0 , its speed ‖~v‖ at position 2 r 0 is most nearly
- ‖~v‖ =
U 0
2 m
- ‖~v‖ =
U 0
6 m
- ‖~v‖ =
U 0
m
- ‖~v‖ =
U 0
8 m
- ‖~v‖ =
4 U 0
m
. correct
- ‖~v‖ =
8 U 0
m
- ‖~v‖ =
6 U 0
m
- ‖~v‖ =
2 U 0
m
- ‖~v‖ =
U 0
4 m
Explanation: The total energy of the particle is con- served. So the change of the potential en- ergy is converted into the kinetic energy of the particle, which gives
m v^2 = 3 U 0 − U 0
Explanation: The normal force of the wall on the rider provides the centripetal acceleration neces- sary to keep her going around in a circle. The downward force of gravity is equal and oppo- site to the upward frictional force on her. Note: Since this problem states that it is viewed by a bystander, we assume that the free-body diagrams are in an inertial frame.
021 10.0 points
Three balls with masses: m, 2 m and
m, re-
spectively are thrown from the top of a build- ing, all with the same initial speed vi. The first ball is thrown horizontally, the second at some angle θ 1 above the horizontal, and the third with some angle θ 2 < θ 1 below the hor- izontal. Neglecting air resistance, rank the speeds of the balls as they reach the ground, from the slowest to the fastest:
- 2, 3, 1
- 2, 1, 3
- 3, 1, 2
- 3, 2, 1
- 1, 3, 2
- 1, 2, 3
- All three balls strike the ground with the same speed. correct
Explanation: Consider each ball and the earth as three separate systems. There is no work done on these systems by the surroundings so ∆Esys = Wsurr = 0. Next, consider the system that consists of the ball with mass m and the earth. Ini- tially, it has a total energy (ignoring rest mass energy) of
Esys,i = Ki + Ui
=
m v i^2 + m g yi
and a final energy of
Esys,f = Kf + Uf
=
m v f^2 + m g yf
so
∆Esys =
m (v^2 f − v^2 i ) + m g (yf − yi) = 0
Solving this equation for vf gives
vf =
g (yf − yi) + v^2 i
The result does not depend on the mass of the ball so the same expression can be derived for each one. All three balls begin with the same speed vi and experience the same change in height (yf − yi) so the final speed for all three is the same.
