Download PHYSICS 101 MIDTERM

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PHYSICS 101 MIDTERM
October 27, 2005
2 hours
Please Circle your section
1
9 am
Galbiati
2
10 am
3
11 am
Hasan
4 12:30 am
5 12:30 pm Olsen
Problem
1
2
3
4
5
6
Total
Wang
Hasan
Score
/16
/16
/16
/18
/16
/18
/100
Instructions: When you are told to begin, check that this examination booklet
contains all the numbered pages from 2 through 17. The exam contains 6 problems.
Read each problem carefully. You must show your work. The grade you get depends
on your solution even when you write down the correct answer. BOX your final
answer. Do not panic or be discouraged if you cannot do every problem; there are
both easy and hard parts in this exam. If a part of a problem depends on
a previous answer you have not obtained, assume it and proceed. Keep
moving and finish as much as you can!
Possibly useful constants and equations are on the last page, which you may
want to tear off and keep handy
Rewrite and sign the pledge: I pledge my honor that I have not violated the Honor Code
during this examination.
Signature
2
Problem 1: Grab Bag
(a) [4 pts] A bowling ball and golf ball have the same initial momentum. The same constant
force F is applied to slow down both balls.
Which ball stops faster? Box your answer and detail your reasoning in writing on the
side.
1. The golf ball stops over a shorter time.
2. The bowling ball stops over a shorter time.
3. They stop over the same time.
4. They don’t stop.
The correct answer is answer 3 .
F ∆t = ∆p
and therefore
∆t = ∆p/F
Since F and ∆p are the same for the two balls, also ∆t is the same and they stop
over the same time.
(b) [4 pts] A bowling ball and golf ball have the same initial momentum. The same constant
force F is applied to slow down both balls.
Which ball stops in the shortest distance? Box your answer and detail your reasoning
in writing on the side.
1. The golf ball stops over a shorter distance.
2. The bowling ball stops over a shorter distance.
3. They stop over the same distance.
4. They don’t stop.
The correct answer is answer 2 .
W = −F s = KEf − KEi = −p2i /(2m)
Therefore:
s = p2i /(2mF )
and the distance is shorter for the ball with the larger mass.
3
(c) [4 pts] Massless strings are guided through four massless and frictionless pulleys as
shown. The masses of the pulleys are negligible.
What is the relation between m and M so that the system stays in equilibrium?
Explain your answer in detail. Box your answer.
The tension
The tension
The tension
The tension
Therefore:
in
in
in
in
the
the
the
the
red rope is M g.
green rope is M g/2.
yellow rope is M g/4.
blue rope is M g/8 = mg.
m = M/8
Note: the color code of the ropes is not essential for solving the problem, is just
an easy way to identify the ropes in the solution. If the copy of the solutions you
are handed is black and white, please refer to the file posted on blackboard to
see the color code.
4
(d) [4 pts] A tennis ball of mass m is held just above a basketball of mass M . You can
assume M >> m throughout the problem. With their centers vertically aligned, they are
released from rest at the same time, to fall through a distance h = 6 m. Note: the sizes of
both the tennis ball and the basketball are negligible with respect to h.
To what height does the tennis ball rebound?
Explain your answer in detail. Box your answer.
m
M
h
When the basketball has just hit the floor, it has velocity gt directed upward.
The tennis ball is still traveling downward and has a velocity gt. At this moment,
they collide.
The basketball is much more massive, so the center of mass of the system is,
effectively, the basketball. With respect to the center of mass and just prior to
the collision, the tennis ball is falling with velocity 2gt directed downward. Just
after the collision, the velocity of the tennis ball with respect to the center of
mass flips of sign: it is 2gt directed upward. The velocity of the tennis ball in the
lab frame is obtained by summing the velocity of the center of mass (unchanged
in the collision), i.e. 3gt directed upward.
The tennis ball, after colliding with the basketball, has un upward velocity three
times higher than the downward velocity just before the collisions. So the tennis
ball raises to an height 32 = 9 times higher than the distance it fell from, i.e.
54 m.
5
Problem 2: Golfing
Tiger Woods hits a golf ball with an initial velocity v0 at an angle of 45◦ with respect to the
ground (neglect air resistance in this problem).
(a) [5 pts] If the ball lands 400 m away, what is v0 ?
Detail your work and provide both an analytical AND a numerical solutions.
Box your answers.
Let’s call the range of the ball D = 400 m.
t=
D
v0 cos θ
1
2v0 sin θ
∆y = 0 = v0 sin θt − gt2 ⇒ t =
2
g
D
2v0 sin θ
=
v0 cos θ
g
v0 =
h
i1/2
gD
2 sin θ cos θ
(9.8 m/s2 )(400 m)
=
2 sin 45◦ cos 45◦
"
#1/2
= 62.6 m/s
(b) [3 pts] Suppose Tiger decides to spend some of his earnings for a trip to the first golf
tournament on the moon. What is the acceleration of gravity on the moon? (Recall that
Mmoon = 7.35 × 1022 kg and Rmoon = 1.74 × 106 m.)
Detail your work and provide both an analytical AND a numerical solutions.
Box your answers.
F =G
mMmoon
= mgmoon
2
Rmoon
moon
gmoon = G M
= (6.67 × 10−11 m3 kg−1 s−2 )
R2
moon
7.35 × 1022 kg
= 1.6 m/s2
(1.74 × 106 m)2
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(c) [4 pts] If he hits a tee shot with the same initial velocity ~v0 as part (a), how far would
it travel on the moon?
Detail your work and provide both an analytical AND a numerical solutions.
Box your answers.
Rearranging the solution to part (a) of this problem:
D=
2v02 sin θ cos θ
g
Only g is different, so:
Dmoon
gearth
=
Dearth
gmoon
This implies:
Dmoon =
earth
Dearth ggmoon
9.8 m/s2
= (400 m)
= 2450 m
1.6 m/s2
(d) [4 pts] What would be the maximum height of the ball above the lunar surface?
Detail your work and provide both an analytical AND a numerical solutions.
Box your answers.
Time it takes to reach max altitude is:
t=
1 D
= 27.7 s
2 v0 cos θ
Therefore:
∆y = v0 sin θt − 21 gmoon t2 = 612 m
7
Problem 3: Airplane Toy
A toy wooden airplane (M = 0.2 kg) is
hung to the ceiling by a string and moves
in a circular trajectory at constant speed.
The distance h with the horizontal is
0.25 m. The length L of the string is 0.5 m.
θ
L
h
R
(a) [6 pts] What is the period for the circular motion of the toy airplane?
Detail your work and provide both an analytical AND a numerical solutions.
Box your answers.
We call F the tension in the rope and T = 2πR/v the period of the motion. The
conditions for equilibrium are:
M g = F sin θ
M v 2 /R = F cos θ
Taking the ratio of the two equations we get:
gR
h
R
tan θ = 2 =
⇒
=
v
R
v
Therefore:
s
T =
2πR
v
= 2π
q
h
g
= 2π
s
h
g
0.25 m
= 1.0 s
9.8 m/s2
8
(b) [4 pts] What is the velocity of the airplane? If you did not answer to part (a), make
an educated guess for the period of the motion.
Detail your work and provide both an analytical AND a numerical solutions.
Box your answers.
Let’s calculate first the radius of the trajectory:
√
√
R = L2 − h2 = 0.52 − 0.252 m = 0.43 m
Therefore the velocity is:
v=
2πR
T
= 2π
0.43 m
= 2.70 m/s
1s
The airplane is hit from behind in the tail by a bullet of mass 10 g. The speed of the bullet,
relative to the ground, is 20 m/s. The bullet sticks into the airplane.
(c) [6 pts] What is the momentum of the airplane right after the collision? Attention: the
bullet stays into the airplane!
Detail your work and provide both an analytical AND a numerical solutions.
Box your answers.
M is the mass of the toy airplane, v = 3.14 m/s its velocity before the collision,
m is the mass of the bullet and u its velocity before the collision. We will call w
the final velocity of the two bodies after the inelastic collision.
M v + mu = (M + m)w
Therefore:
w=
M v+mu
M +m
=
(2.70 × 0.2 + 20 × 0.01) kg m/s
= 3.52 m/s
0.21 kg
9
Problem 4: Roller Coaster
A roller coaster of mass m = 50 kg travels on the track shown below. At point C the coaster
enters a frictionless loop of radius R1 =20 m. After exiting the loop, the coaster travels the
distance L = 40 m from C to D with friction, and then onto a straight frictionless ramp that
leads to a circular arc of radius R2 =10 m.
A
E
h
F
R1
θ=60o
B
R2
c
D
x
L
(a) [6 pts] Assume no friction for the A-B-C segment of the track, from what height h1 must
the coaster start to stay barely on the track at point E? (Use this value of h1 in part (b).)
Detail your work and provide both an analytical AND a numerical solutions.
Box your answers.
When the coaster barely stays on track at E, the normal force N is zero. Then
the centripetal force is:
m
vE2
= mg => vE2 = gR1
R1
Applying the principle of energy conservation:
1
1
mg(h1 − 2R1 ) = mvE2 = mgR1
2
2
Which can be simplified into:
h1 − 2R1 =
R1
2
Therefore:
h1 = 52 R1 = 50 m
10
(b) [6 pts] Determine the kinetic friction coefficient µk of segment C–D if the roller coaster
stays barely on the track at point F. If you did not complete part (a), assume that h1 = 40 m.
Detail your work and provide both an analytical AND a numerical solutions.
Box your answers.
When the coaster barely stays on track at F, the normal force N is zero. Then
the centripetal force is:
m
vF2
= mg => vF2 = gR2
R2
The work done by friction is:
WNC = Ef − Ei
Therefore:
1
1
−µk mgL = mgR2 + mvF2 − mgh1 = mgR2 + mgR2 − mgh1 =
2
2
µk =
−3R2 +2h1
2L
=
−3 × 10 m + 2 × 50 m
= 0.88
80 m
(c) [4 pts] If the coefficient of kinetic friction is 0.1 from A to B, from what height h2 must
the coaster start to stay barely on the track at point E.
Detail your work and provide both an analytical AND a numerical solutions.
Box your answers.
The work done by friction is:
WNC = Ef − Ei
Therefore:
−mgµk cos θ
h2
1
1
= 2mgR1 + mvE2 − mgh2 = 2mgR1 + mgR1 − mgh2
sin θ
2
2
Taking the first and last member of the above expression, and eliminating the
common factors, we obtain:
5
h2 (1 − µk cot θ) = R1
2
h2 =
5R1
2(1−µk cot θ)
=
5(20 m)
= 53 m
2 (1 − 0.1 cot 60◦ )
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(d) [2 pts] What must be the new µk of segment C–D if the the coaster stays barely on the
track at point F?
Provide a detailed answer. Box your answer.
Since the coaster has the same mechanical energy at C as in part b, µk stays the same
(µk =0.88) for the coaster to stays barely on the track at point F.
12
Problem 5: Platforms on the ice
A man of mass M = 60 kg stands in the center of a platform one of mass m = 40 kg that
rests on ice. Another identical platform floats at a distance L = 1.5 m away. The contact of
the two platforms with the ice is frictionless. The man jumps from the first to the second
platform, and lands in the center of the second platform.
1
2
0
x
(a) [6 pts] Locate the center of mass of the two-platform and man system before the man
jumps. Use the center of mass of platform one as origin.
Detail your work and provide both an analytical AND a numerical solutions.
Box your answers.
xcm =
x1,i (m+M )+x2,i m
2m+M
=
0 × (40 kg + 60 kg) + 1.5 m × 40 kg
= 0.43 m
2 × 40 kg + 60 kg
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(b) [4 pts] What is the velocity vcm of the center of mass of the two-platform and man
system when the man is midway between the two platforms? Detail your work and
provide both an analytical AND a numerical solutions. Box your answers.
The answer is zero . The two-platform and man system is an isolated system.
Since there is no net external force acting on the system, the velocity of the center
of mass is conserved. The initial value of vcm is zero. Therefore vcm stays zero at
any time.
(c) [6 pts] How far apart are the two platforms when the man lands?
Detail your work and provide both an analytical AND a numerical solutions.
Box your answers.
We will use the fact that the center of mass does not move. Even in the final
instant, xcm = 0.43 m.
Let’s call with x1,f and x2,f the final positions of the platforms, at the moment
the man lands on the second platform. The position of the second platform has
not changed yet and therefore x2,f = x2,i = 1.5 m. Our unknown is x1,f . The
distance between the platforms is x2,f − x1,f .
xcm =
x1,f =
x1,f =
x1,f m + x2,f (m + M )
2m + M
xcm (2m+M )−x2,f (m+M )
m
0.43 m × (2 × 40 kg + 60 kg) − 1.5 m(40 kg + 60 kg)
= −2.25 m
40 kg
x2,f − x1,f = (1.5 + 2.25) m = 3.75 m
14
Problem 6: Well and Bucket
d = 0.2m
r = 0.4m
A farmer raises a water-filled bucket (total mass m =
5 kg) from the bottom of a well by turning a hand crack,
as the figure illustrates. The bucket is attached to a
massless rope that is wound around a solid cylinder with
mass M of 10 kg and a diameter d of 20 cm. The massless
hand crank has a turning radius r of 40 cm.
h = 5m
(a) [2 pts] How many turns of the crank does it take to raise the bucket a height h = 5 m?
Provide a detailed answer. Box your answer.
One turn raises the bucket by:
2π(d/2) = 0.62 m
So it takes:
N = 5 m/0.62 m ≈ 8 turns
15
(b) [4 pts] What is the magnitude of the torque on the cylinder due to the weight of the
bucket?
Detail your work and provide both an analytical AND a numerical solutions.
Box your answers.
τ = F d/2 = mgd/2 =
(5 kg)(9.8 m/s2 )(0.2 m)
= 4.9 N m
2
(c) [4 pts] How much force does the farmer have to apply to the end of the crank to raise
the bucket at constant speed? (Assume the force is applied perpendicular to the lever arm
of the crank.)
Detail your work and provide both an analytical AND a numerical solutions.
Box your answers.
X
τ = 0 ⇒ τW = τhand = F r
F =
τW
r
=
4.9 Nm
= 12.3 N
0.4 m
16
(d) [4 pts] While the bucket is suspended at rest from a height of 5 m, the farmer loses his
grip on the crank. As the bucket falls, the cylinder rotates and the rope unwinds without
slipping. What is the angular acceleration of the cylinder while the bucket is falling? (Hint:
recall that the moment of inertia of a solid cylinder is I = 21 M R2 , where M is the mass and
R is the radius.)
Detail your work and provide both an analytical AND a numerical solutions.
Box your answers.
Sum the forces in the y direction (down is positive):
mg − T = ma
Using a = d2 α, we have T = m g − d2 α . Sum the torques around the axis of the
cylinder:
d
τ = T = Iα
2
Substitute for T :
!
!2
d
1
d
d
= M
α
m g− α
2
2
2
2
α=
mg
( 21 M +m) d2
= 1
2
(5 kg)(9.8 m/s2 )
10 kg + 5 kg
(0.1 m)2
= 49 rad/s2
(e) [4 pts] What is the ratio KErot /KEbucket of the rotational kinetic energy of the cylinder
and the linear kinetic energy of the bucket while the bucket is falling?
Detail your work and provide both an analytical AND a numerical solutions.
Box your answers.
1
1
Iω 2
M r2
KErot
Iω 2
I
2
= 12 2 =
=
=
=
KEbucket
m(rω)2
mr2
mr2
mv
2
M
2m
=
20 kg
= 1
2(10 kg)
17
POSSIBLY USEFUL CONSTANTS AND EQUATIONS
You may want to tear this out to keep at your side
L = Iω
PE = mgh
ω = ω0 + αt
v = v0 + at
F = µN
Στ = Iα
ac = v 2 /r
Wnc = ∆KE + ∆PE
I = 25 mr2 [sphere]
I = Σmi ri2
KE = 12 Iω 2
ω 2 = ω02 + 2α∆θ
F∆t = ∆p
s = Rθ
v = Rω
W = F s cos θ
a = Rα
x = x0 + v0 t + at2 /2
KE = 12 mv 2
∆θ = ω0 t + 12 αt2
F = −GM m/r2
τ = F ` sin θ
p = mv
v 2 = v02 + 2a∆x
I = 12 mr2 [disk]
REarth = 6400 km
MEarth = 6.0 × 1024 kg
G = 6.67 × 10−11 Nm2 /kg2