Instructor: Hacker Course: Physics 121

Instructor: Hacker
Course: Physics 121
Sample Exam 5 Solutions (Work, Energy, and Momentum)
Print your name neatly. If you forget to write your name, or if the grader can’t read your
writing, you can lose up to 100 points. Answer all the questions that you can.
This exam will consist of 21 multiple-choice problems. You may not use calculators or
other electronic devices on this exam. The use of such a device will be regarded as an
attempt to cheat, and will be pursued accordingly. All diagrams and figures on this exam
are rough sketches: they are not generally drawn to scale.
No partial credit will be given for these problems. However, you can miss one of the 21
problems without penalty. Your grade will be based on your best 20 problems. You will
not receive extra credit for getting all 21 right.
Your grade on the exam will be based entirely on the answers that you circle on this
sheet. If you have no answer or a wrong answer there, the grader will not look at the
page with the problem to see if the right answer appears there. Illegible or ambiguous
answers will be graded as wrong. You are responsible for copying your answers clearly,
correctly, and in the right place.
Although there is no partial credit on this exam, you must show your work in the space
provided on the exam. There is additional scratch paper at the end of the exam: do
not use it unless you have filled all the scratch space provided on the page with the
problem. If you answer a difficult problem without doing any written work, the grader
will assume that you got the answer by guessing or by copying from someone else, and
will not give you credit for the problem even though you’ve indicated the correct solution
on the answer sheet.
Circle your answers here. Do not detach this sheet from the test.
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Physics 121, sample exam 5 solns
Copyright ©Wayne Hacker 2010. All rights reserved. 2
Work
Problem 1. An ice skater is gliding across the frictionless surface of a frozen pond. As
he crosses the pond, the work done by gravity on the skater is:
(a) positive
(b) negative
*(c) zero
Solution: The force of gravity is pulling straight down. The skater is moving horizontally. Since the force is perpendicular to the motion, the work done by the force is
zero.
Problem 2. A sidewalk runs parallel to a street. A skateboarder is being towed down
the sidewalk by a rope attached to the back of a car on the street. The rope makes an
angle of 30◦ to the direction in which the car and the skateboarder are moving. The
tension in the rope is 200 N. How much work does the car do in towing the skateboarder
100 m down the sidewalk?
√
(a) 2000 J√
*(b) 10,000 3 J
(c) 20,000 3 J
(d) 10,000 J
(e) None of these
Solution: The component of the tension in the direction of the sidewalk is T cos 30◦ .
The work is this parallel component of force, times the distance through which it acts:
√
W = (T cos 30◦ )s = (200 N)(100 m) cos 30◦ = 10, 000 3 J
Kinetic energy
Problem 3. A skier with a mass of 80 kg is sliding down a slope at 20 m/s. What is
the skier’s kinetic energy?
(a) 3.2 × 103 J
(b) 6.4 × 103 J
*(c) 1.6 × 104 J
(d) 3.2 × 104 J
(e) None of these
Solution: K = 12 mv 2 = 12 (80 kg)(20 m/s)2 = 1.6 × 104 J
Potential energy
Problem 4. A case of rat food has a mass of 10 kg. To keep rats from eating it, you
move it from the floor to a shelf 2 m high. By how much did you increase the potential
energy of the case? Take g = 10 m/s2 .
(a) 160 J
*(c) 200 J
(e) None of these
(b) 180 J
(d) 220 J
Solution: Let y = 0 be the level of the floor. Then the initial potential energy of the
case is zero. The potential energy of the case on the shelf is
Ugrav = mgy = (10 kg)(10m/s2 )(2 m) = 200 J
Physics 121, sample exam 5 solns
Copyright ©Wayne Hacker 2010. All rights reserved. 3
Work-Mechanical-Energy theorem
Problem 5. With no force applied to it, a spring hanging from the ceiling is 24 cm long.
You hang a weight of 300 N from it, stretching it to a length of 30 cm. How much work
was done by gravity in stretching the spring?
*(a) 9 J
(b) 18 J
(c)
900 J
(d) 1800 J
(e)
None of these
Solution: By applying a force (the weight of the object) through a distance x (the
displacement) gravity has done work on the system. The work done on the springmass system is equal to the potential energy stored in the spring. We can find that
using the formula: U = 12 kx2 . Here x is the change in length from equilibrium: x =
Ls − Lu = 30 cm − 24 cm = 6 cm = 0.06 m. We can calculate k using the spring formula:
F = kx ⇒ k = F/x. Then the potential energy is:
U=
kx2
F x2
Fx
(300 N)(0.06 m)
=
=
=
=9J
2
2x
2
2
Problem 6. A bartender slides a glass of beer with a mass of 500 g along the top of a
level bar. The coefficient of kinetic friction between the glass and the bar is µk = 0.1.
When the glass leaves the bartender’s hand, it is moving at a speed of v0 = 2 m/s. How
far does the glass travel before coming to a stop? Take g=10 m/s2 .
(a) 100 cm
(b) 180 cm
*(c) 200 cm
(d) 220 cm
(e) None of these
Solution: We know the mass m, the coefficient of friction µk , the initial velocity vi , and
the final velocity vf = 0. We want the stopping distance ∆x.
Apply the work-energy principle: Wnet = ∆K = Kf − Ki . The net force after the glass
is released is friction, so
Wnet = Fnet · (distance) = −µk N ∆x
Since the top of the bar is level, N = mg. Hence Wnet = −µk mg∆x.
Since vf = 0, Kf = 21 mvf2 = 0. Hence Wnet = −Ki = − 12 mv02 . Hence
1
− µk mg∆x = − mv02
2
v02
÷(−m)
−−−−−−−→ µk g∆x =
2
2
v
(2 m/s)2
÷µk g
−−−−−
−→ ∆x = 0 =
= 2 m = 200 cm
2µk g
2(0.1)(10 m/s2 )
Physics 121, sample exam 5 solns
Copyright ©Wayne Hacker 2010. All rights reserved. 4
Problem 7. (Similarity Problem) You are driving at a speed of V on a straight and
level road when you see a deer in front of you and lock up the brakes. You travel a
distance D before coming to a complete stop. If your original speed had been 2V when
you hit the brakes, what distance would you have travelled before stopping?
√
2D
(a) D
(b)
(c) 2D
*(d) 4D
(e) None of these
Solution: We will use the work-energy principle. Your car will move until the negative
work done by friction equals its initial kinetic energy. If m is the mass of the car, v is its
initial speed, Fk is the force of kinetic friction, and d is the stopping distance, then
1 2
mv = Fk d
2
⇒
d=
mv 2
2Fk
The mass of the car and the force of friction don’t change between trials. In the first
trial, your initial speed is V , so
mV 2
D=
2Fk
In the second trial, your initial speed is 2V , so
d=
m(2V )2
4mV 2
=
= 4D
2Fk
2Fk
Problem 8. A box has a mass of 22 kg. It is sitting on an icy sidewalk, which constitutes
a horizontal frictionless surface. A spring with a spring constant of 88 N/m is attached
to the box and pulled horizontally, so that the box accelerates at 2 m/s2 . How much
does the spring stretch?
*(a)
(c)
(e)
0.5 m
2m
None of these
(b) 1.5 m
(d) 2.5 m
Solution: The spring is pulled horizontally and there is no friction, so the force exerted
by the spring is the mass of the box times its acceleration: F = ma. By Hooke’s law,
F = kx. Hence
ma = kx
⇒
x=
(22 kg)(2 m/s2 )
ma
=
= 0.5 m
k
88 N/m
Physics 121, sample exam 5 solns
Copyright ©Wayne Hacker 2010. All rights reserved. 5
Conservation of mechanical energy
Problem 9. Two frictionless inclined planes have the same vertical height Y . Plane 1
makes an angle of 60◦ to the horizontal; plane 2 makes an angle of 30◦ to the horizontal.
Two objects with the same mass M are released from the tops of the planes to slide
down. The object that slides down plane 1 has speed v1 when it reaches the bottom; the
object that slides down plane 2 has speed v2 when it reaches the bottom. What is the
relationship between v1 and v2 ?
*(a) v1 = v2
(b) v1 = 2v2
◦
◦
(c)
v1 sin 30 = v2 sin 60
(d) v1 cos 60◦ = v2 cos 30◦
(e)
None of these
Solution: The key word is “frictionless”. Since there are no nonconservative forces,
mechanical energy must be conserved. Initially, both objects have zero kinetic energy
and the same potential energy (since they have the same mass M and they’re at the
same height Y ). When they reach the bottom of their planes, they still have the same
mechanical energy; since they have height zero, their potential energy is zero; so their
mechanical energy is all kinetic, and it’s the same for both. Since they have the same
kinetic energy and the same mass, their speeds must be the same.
Physics 121, sample exam 5 solns
Copyright ©Wayne Hacker 2010. All rights reserved. 6
Problem 10. Ricky, the 10 kg snowboarding raccoon, is
on his snowboard atop a frictionless ice-covered quarterpipe with radius 2 meters (see figure at right). If he
starts from rest at the top of the quarter-pipe, what is
his speed at the bottom of the quarter-pipe? For ease of
calculation, assume that g = 10 m/s2 .
√
(a) 20√
m/s
*(b) 2√ 10 m/s
(c) 10 2 m/s
(d)
20 m/s
(e) None of these
x
Raccoon: 10 kg
2m
2m
Solution:
We are given: m = 10 Kg, R = 2 m, vi = 0 m/s, and g = 10 m/s2 .
We want: vf =?
We assume: no friction, hence the mechanical energy of the system is conserved.
Since the system, Ricky and the earth, has no external forces on it, such as friction, we
can apply the conservation of mechanical energy: MEi = MEf . Now, the initial potential
due to gravity is Ugrav = mgh = mgR and the initial kinetic energy is Ki = 0 since Ricky
is at rest. The final P.E. is zero, since Ricky is now at ground level h = 0. The final K.E.
is Kf = 12 mvf2 . Using the definition of mechanical energy: ME = PE + KE gives
1
mgR = PEi + KEi = MEi = MEf = PEf + KEf = mvf2
2
÷1m
2
−−−−
−−→ vf2 = 2Rg
√
p
√
1/2
−−−−−−→ vf = 2Rg = (2)(2 m)(10 m/s2 )
= 2 10 m/s
Notice that since R = h, this is the same speed that Ricky would have if he had jumped
straight down (go look back at our 1-D kinematic problems). The only difference is the
direction of his speed would have been different. If he had jumped straight down, then
his velocity would also have been straight down, and into the ground would go Ricky; but
the quarter pipe redirected his motion to being tangent to the ground, while preserving
his speed. So instead of busting his head, and winding up dead; Ricky sped off to work,
where he was a clerk. Were it not for conservation of mechanical energy he would have
been late for his job, and replaced by his competitor Rob.
Physics 121, sample exam 5 solns
Copyright ©Wayne Hacker 2010. All rights reserved. 7
Problem 11. A frictionless horizontal air track has a spring at either end. The spring
on the left has a force constant of kL ; the one on the right has a force constant of kR .
A glider with a mass of m is pressed against the left-hand spring, compressing it by
xL . The glider is then released, so that the spring propels it rightward. It slides along
the track and into the right-hand spring. What is the maximum compression of the
right-hand spring? (The diagram following the answers shows the situation before the
glider is released, when the left-hand spring is compressed and the glider is not moving.)
r
kL
kL
xL
(b)
xL
(a) m
kR
kR
r
r
kL
kR
xL
(d) m
xL
*(c)
kR
kL
(e) None of these
m
E E E E
C C C
C C C
Solution: We know the mass m of the glider, the spring constants KL and KR of the
left- and right-hand springs, and the compression xL of the left-hand spring. We want to
know the compression xR of the right-hand spring.
There is no friction, so we can assume that mechanical energy is conserved: Ui + Ki =
Uf + Kf . We will take the initial situation to be that in which the left-hand spring is
fully compressed, before the glider is released to move; and the final situation to be that
in which the glider has pushed the right-hand spring to its maximum compression. Since
the glider isn’t moving in either case, Ki = Kf = 0; so Ui = Uf .
In both situations, the potential energy is all in a spring: Ui = 12 kL x2L = 12 kR x2R = Uf .
We can solve this for xR :
·2
−−−−→ kR x2R = kL x2L
kL x2L
÷kR
−−−−−
−→ x2R =
k
sR
r
√
kL x2L
kL
−−−−−→ xR =
=
xL
kR
kR
Notice that we never needed the mass of the glider. Its only function in the problem was
to carry all the energy from one spring to the other.
Physics 121, sample exam 5 solns
Copyright ©Wayne Hacker 2010. All rights reserved. 8
Problem 12. (Lab Problem) A spring cannon shoots a ball with a mass of m vertically
into the air from ground level. In this situation, air resistance can be ignored. If the
spring is compressed by a distance of d, the ball reaches a maximum height of ymax . What
is the spring constant of the spring in the cannon?
mgymax
mgymax
(b)
(a)
2
d
d
mgymax
2mgymax
(c)
*(d)
2d2
d2
(e) None of these
Solution: We know the mass of the ball m, the compression distance d of the spring,
the maximum height ymax of the ball, and the value of g. We want to know the spring
constant k.
Since there’s no air resistance and no friction, we can assume that mechanical energy is
conserved: Ki + Ui = Kf + Uf . We will take the initial position (subscript i) as that
of the cannon with its spring compressed, just before it is fired; and the final position
(subscript f ) as the system when the ball has reached its maximum height. Since the
ball isn’t moving in either case, Ki = Kf = 0. Hence Ui = Uf .
The initial potential energy is all in the spring: Ui = 12 kd2 . The final potential energy is
all gravitational: Uf = mgymax . Hence
1 2
· (2/d2 )
kd = mgymax −−−−−−−→
2
k=
2mgymax
.
d2
Computations with momentum
Problem 13. Your potato gun launches a potato with a mass of 500 g at a speed of 20
m/s. What is the magnitude of the potato’s momentum?
(a) 1000 kg m/s
(c) 25 kg m/s
(e) None of these
*(b) 10 kg m/s
(d) 100 kg m/s
Solution: p = mv = (0.5 kg)(20 m/s) = 10 kg m/s. Don’t forget to convert grams to
kilograms.
Problem 14. An artillery shell has a mass of 5 kg and a momentum of 4000 kg m/s.
What is the shell’s speed?
(a) 40 m/s
(c) 160 m/s
(e) None of these
Solution: p = mv
(b)
*(d)
⇒
v=
80 m/s
800 m/s
p
4000 kg m/s
=
= 800 m/s
m
5 kg
Physics 121, sample exam 5 solns
Copyright ©Wayne Hacker 2010. All rights reserved. 9
Conservation of momentum
Problem 15. You drop an object from rest on a planet with no atmosphere. As the
object falls, which of the quantities of the object listed below is conserved?
(a) Momentum
*(b) Mechanical energy
(c) Potential energy
(d) Kinetic energy
Solution: The object is accelerating in a straight line due to the force of gravity. Its
speed increases as it falls; so its momentum and its kinetic energy are both increasing.
Thus they are not conserved. The potential energy decreases as the object falls, so it
is not conserved. Since the planet has no atmosphere, there is no air drag or other
dissipative force. Thus the object-planet system is an isolated system and conserves
mechanical energy.
Problem 16. An astronaut with a mass of 80 kg is adrift in space outside of his space
station. Fortunately, he is holding a space wrench with a mass of 2 kg. How fast must
he throw the wrench to give himself a recoil speed of 20 cm/s?
*(a)
(c)
(e)
8 m/s
20 m/s
None of these
(b) 16 m/s
(d) 40 m/s
Solution: The initial momentum of the system is zero; so the backward momentum of
the astronaut must equal the forward momentum of the wrench. Don’t forget to convert
centimeters to meters.
ma va = mw vw
⇒
vw =
(80 kg)(0.2 m/s)
ma va
=
= 8 m/s
mw
2 kg
Impulse
Problem 17. A ball weighing 300 g and moving horizontally at 14 m/s strikes a wall
and rebounds horizontally at 10 m/s. What is the magnitude of the impulse that the
wall gives to the ball? Round your answer to two significant digits.
(a) 1.2 kg m/s
(c) 14 kg m/s
(e) None of these
*(b)
(d)
7.2 kg m/s
43 kg m/s
Solution: The impulse is the ball’s change in momentum. If we take the direction away
from the wall as positive, then the ball’s initial momentum is pi = (0.29 kg)(−14 m/s);
its final momentum is pf = (0.29 kg)(10 m/s). The change in momentum is
J = ∆p = pf − pi = (0.3 kg)(10 m/s) − (0.3 kg)(−14 m/s)
= (0.3 kg)(24 m/s) = 7.2 kg m/s
Physics 121, sample exam 5 solns
Copyright ©Wayne Hacker 2010. All rights reserved.10
Collisions
Problem 18. Two gliders collide on a horizontal frictionless air track. Which of
the following statements must be true regardless of whether the collision is elastic or
inelastic?
(a) Neither momentum nor kinetic energy is necessarily conserved.
(b) Kinetic energy is conserved; momentum is not necessarily conserved.
*(c) Momentum is conserved; kinetic energy is not necessarily conserved.
(d) Momentum and kinetic energy are both conserved.
Solution: In all collisions in an isolated system (no external forces), momentum is
conserved. In elastic collisions, kinetic energy is conserved; in inelastic collisions, it is not.
Physics 121, sample exam 5 solns
Copyright ©Wayne Hacker 2010. All rights reserved.11
Problem 19. A block of wood with a mass of mw is sitting on a horizontal tabletop.
The coefficient of kinetic friction between the block and the table is µk . A rifle bullet
with a mass of mb is fired horizontally into the block at a speed of vbi and stops inside
the block. How far does the block with the bullet embedded in it slide before coming to
a stop? Justify your answer.
(a)
(c)
(e)
mb vbi
2µk g(mb + mw )2
2
m2b vbi
µk
g(mb + mw )2
None of these
*(b)
(d)
2
m2b vbi
2µk g(mb + mw )2
2
m2b vbi
µk g(mb + mw )2
Solution: There are two processes going on here: an inelastic collision between the
bullet and the block, followed by the block’s sliding to a halt as friction dissipates its
kinetic energy.
Since the bullet-wood collision is inelastic, we can’t use conservation of energy there. We
have to use conservation of momentum. We know the masses and initial velocities of
the bullet and the block: mb , vbi , mw , and vwi = 0. (We’ll use the subscript “w” for
“wood”, since “block” and “bullet” both begin with the same letter.) We want to know
the kinetic energy Ks of the block-bullet system immediately after the collision, and we
can calculate that if we know the velocity vs at that time. By conservation of momentum,
mb vbi + mw vwi = mb vbi = (mb + mw )vs
mb vbi
÷(mb +mw )
−−−−−−−→ vs =
mb + mw
Now we can calculate the energy of the system immediately after the collision. There is
no potential energy involved.
2
1
(mb + mw )(mb vbi )2
m2b vbi
2
Ks = ms vx =
=
2
2(mb + mw )2
2(mb + mw )
As the block-bullet system slides to a halt, the only force working on it is friction. It will
move until the work done against the frictional force equals its initial energy. Since the
tabletop is horizontal, the normal force equals the weight; so the force of kinetic friction
is
Fk = µk N = µk (mb + mw )g
The work done against friction is the force times the distance x that the block slides:
W = Fk x. We want to know x; and we can find it by setting W = Ks and then solving
for x.
W = Fk x = µk (mb + mw )gx =
÷µk (mb +mw )g
−−−−−−−−−→ x =
2
m2b vbi
2µk g(mb + mw )2
2
m2b vbi
= Ks
2(mb + mw )
Physics 121, sample exam 5 solns
Copyright ©Wayne Hacker 2010. All rights reserved.12
Conservation of angular momentum
Problem 20. A physics instructor has a mass of M ; his wife has a mass of M/2. The
two are on a merry-go-round that is rotating frictionlessly. Initially, the instructor is at
the center and his wife is at the outer edge. The two trade places, passing one another
halfway from the center to the edge. At which of the following three points is the merrygo-round’s angular speed the greatest?
(a) When the instructor is at the center and the wife is at the edge.
(b) When the wife is at the center and the instructor is at the edge.
*(c) When the two are both halfway from the center to the edge.
(d) The merry-go-round’s speed doesn’t change.
Solution: There are no external torques on the system, so angular momentum is conserved: L = Iω ⇒ ω = L/I. The angular speed ω is the greatest when the moment of
inertia I is the smallest. The contribution of each person to the total moment of inertia
is mr2 , where m is their mass and r is their distance from the center. Let R be the
radius of the merry-go-round, so that someone at the edge is at r = R. Look at the three
situations:
Instructor at center, wife at edge
I = M (0)2 +
Wife at center, instructor at edge
I=
Both halfway from center to edge
M 2 M R2
R =
2
2
M 2
(0) + M R2 = M R2
2
2
2
R
M R
3M R2
I=M
+
=
2
2
2
8
Since I changes, the speed of the merry-go-round changes; so we can rule out (d). Since
I is the smallest in situation (c), the merry-go-round’s speed is the greatest in that
situation.
Problem 21. A door’s moment of inertia is 10 kg·m2 . A bullet with a mass of 50 g
moving at 300 m/s strikes the door perpendicularly at a distance of 1 m from the hinges.
The bullet remains embedded in the door. What is the door’s angular speed after the
collision? Assume that the embedded bullet does not change the door’s moment of
inertia.
(a) 0.5 rad/s
(b) 1 rad/s
*(c) 1.5 rad/s
(d) 2 rad/s
(e) None of these
Solution: Use conservation of angular momentum. Initially, the door is at rest, so the
angular momentum of the system is that of the bullet. After the collision, the angular
momentum of the system is that of the door (with the embedded bullet, which we can
ignore). We’ll use subscripts “b” for “bullet” and “d” for “door”.
⇒
Li = Lf = mb vb r = Id ωd
mb vb r
(0.05 kg)(300 m/s)(1 m)
ωd =
=
= 1.5 rad/s
Id
10 kg·m2