1
Your Name:
PHYSICS 101 MIDTERM
October 25, 2007
1
3
5
2 hours
Please circle your section
9 am
Nappi
2 10 am McDonald
10 am
Galbiati
4 11 am McDonald
12:30 pm Pretorius
Problem
1
2
3
4
5
6
Total
Score
/15
/10
/20
/15
/20
/20
/100
Instructions: When you are told to begin, check that this examination booklet
contains all the numbered pages from 2 through 15. 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
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Problem 1: Grab Bag
(a) [6 pts] Three balls, A, B, and C are thrown from the top of a cliff of height h = 100 m
overlooking a flat land. The speed of the the three balls is identical: vA = vB = vC = 10 m/s.
The angle in the figure is φ=π/4.
VC
φ
φ
VA
VB
h
a.1 [3 pts] Which one of the three balls will have the largest speed at the moment it makes
contact with the ground? Box your answer and explain your reasoning on the side.
• Ball A
• Ball B
• Ball C
• They all reach ground with the same speed
The three balls have equal initial kinetic energy. They will also have equal final kinetic energy,
as they drop by the same height. So they all reach ground with the same speed.
a.2 [3 pts] Which of the three balls will reach the ground in the shortest time? Box your
answer and explain your reasoning on the side.
• Ball A
• Ball B
• Ball C
• They all reach ground at the same time
• Ball B and C reach ground at the same time, before ball A.
Ball B reaches the ground first. What matters here is only the motion in the y direction. Since
ball B is the only one with an initial velocity directed towards ground, it will reach the ground
first.
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(b) [9 pts] A uniform plank of length L=2 m and mass m=10 kg is at rest and leaning
against a frictionless wall. The horizontal surface is rough, and the coefficient of static
friction is µs =0.8.
L
θ
b.1 [3 pts] Draw a complete body diagram on the figure, identifying all forces.
N
Fy
mg
Fx
b.2 [2 pts] Find the vertical component Fy of the force exerted by the horizontal surface
on the plank.
The plank is at rest, therefore the sum of the forces in the vertical direction is null. Therefore:
Fy = mg = 10 kg · 9.8 m/s2 = 98 N
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b.3 [2 pts] Find the normal force N exerted by the vertical wall on the plank.
The plank is at rest, therefore the total torque is null. It is convenient to calculate the torque
around the point of contact between the plank and the horizontal surface: in this case, the two
forces Fx and Fy do not bear contributions to the torque because their lever of arm is null.
τ = −mg
Therefore:
N=
L
cos θ + N L sin θ = 0
2
10 kg · 9.8 m/s2
mg
=
= 28.3 N
2 tan θ
2 tan 60◦
b.4 [2 pts] Find the horizontal component Fx of the force exerted by the horizontal surface
on the plank.
The plank is at rest, therefore the sum of the forces in the horizontal direction is null. Therefore:
Fx = N = 28.3 N
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Problem 2: Rock the Cart!
A cart on wheels of mass M =100 kg is standing still on frictionless rails on a horizontal
surface. Billy is playing on the car, and has a mass m of 50 kg. The cart is initially at rest,
and Billy plays while sitting on one end of the cart. All of a sudden, Billy stands up and
starts running towards the other end of the cart. While on the cart, Billy reaches a steady
velocity v = 10 m/s relative to the cart.
Billy
(a) [4 pts] As a result of Billy’s motion, the cart also gets in motion with a velocity u
relative to the ground when Billy has a velocity v relative to the cart. What is the velocity
of Billy relative to the ground? Express your answer as a function of u and v.
vBilly,Ground = vBilly,Cart + vCart,Ground = v + u
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(b) [6 pts] Calculate the velocity u of the cart relative to the ground at the moment when
Billy achieves the maximum velocity v.
The total momentum is conserved. The initial momentum (as seen from the ground reference
frame) is null, so the final momentum must be null as well.
In the final state, the cart has a velocity u relative to the ground and Billy has a velocity u+v
relative to the ground. Therefore:
Pf = M u + m(u + v) = 0
u = −v
m
50 kg
= −10 m/s
= −3.3 m/s
m+M
50 kg + 100 kg
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Problem 3. Roller Coaster.
A cart of mass m is released with zero velocity at the top of the roller coaster and then runs
in the loop of radius R shown in the picture.
R
H
θ
a. [7 pts] If the cart has to stay on track in the loop of radius R, at what minimal height
must it be released? I.e., find the minimum height h that will allow the cart to complete the
loop of radius R.
At the top of the loop:
m
v2
= N + mg.
r
For zero normal force, one gets v 2 = Rg. Using this equation and the conservation of energy:
mgh =
1
1
mv 2 + mg2R = mRg + mg2R
2
2
The solution of this equation is:
h=
5
2R
.
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b. [6 pts] Assume that the cart starts at an height H > h that allows it to complete the
loop. Find the expression of the normal force as function of H and of the angle θ from the
vertical.
At each angle θ the equation of motion is:
m
v2
= N + mg cos θ.
r
The conservation of energy gives:
mgH =
1
mv 2 + mgR(1 + cos θ).
2
Eliminating v 2 , one gets the result for the normal force:
N = mg
2H
R
− 2 − 3 cos θ .
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c. [7 pts] Compute the apparent weight of the cart at the bottom and at the top of the
loop. Show that the difference between the two is always 6 mg, where m is the mass and g is
the acceleration of gravity.(To answer this question you do not necessarily need to use your
answer to part b.)
From the expression derived in part b, one finds:
N (θ = π) − N (θ = 0) = 6mg.
Alternatively, one can compute the normal forces at the top and at the bottom of the loop and
compare them:
2
vtop
N top = m
− mg,
r
N bottom = m
2
vbottom
+ mg.
r
From energy conservation one derives:
2
vtop
= 2g(H − 2R)
2
vbottom
= 2gH.
Substituting these velocities in the expressions for N top and N bottom :
N top = 2mg
H
− 5mg,
R
N bottom = 2mg
H
+ mg.
R
Hence:
N bottom − N top = 6mg .
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Problem 4. Target Practice.
l = 100 m
d = 1.52 m
An archer is shooting arrows, each
with a mass m of 165 g, at a target that is a distance l=100 m away.
When the bullseye is in the archer’s
line of sight, i.e., looking directly
down the shaft of the arrow with
the bow drawn as shown in the figure, the arrow makes an angle of 0◦
relative to the horizontal.
a. [6 pts] The archer shoots an arrow and it leaves the bow with a speed v0 of 48.5 m/s. Ignoring air resistance, at what angle relative to the horizontal must the archer loose the
arrow so that it hits the bullseye? Hint: you might find the following identity useful:
2 sin θ cos θ = sin(2θ)
Equations of constant acceleration applicable here:
1
= y0 + v0 sin θ · t − gt2
2
x = x0 + v0 cos θ · t
y
To hit the bullseye, y − y0 = 0 and x − x0 = l = 100 m; v0 = 48.5 m/s. Solving for θ gives:
−2
1
g(x − x0 )
· 100 m
1
−1 9.8 m s
θ = sin−1
sin
= 12.3◦
=
2
v02
2
(48.5 m/s)2
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b. [3 pts] For how much time was the arrow in flight? (If you did not get an answer to part
a, use an angle θ=15.00◦ )
t=
x
= 2.11 s
v0 cos 12.3◦
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c. [6 pts] A steady wind starts blowing. Assume that the effect of the wind on an arrow
during its flight is as if a constant force F~ =−0.118N x̂ was applied to the arrow, where the x̂
direction is along the horizontal as illustrated in the figure. The archer now shoots a second
arrow with exactly the same speed as before (v0 =48.5 m/s) but, anticipating the wind, he
adjusts the angle to θ=12.8◦ . Where, relative to the bullseye, does the second arrow hit?
Again, equations of constant acceleration, but now we have a net acceleration in the x̂ direction:
ax =
Fx
0.118 N
=−
= −0.715 m s−2
m
0.165 kg
Therefore:
1
= y0 + vy0 t − gt2
2
1
x = x0 + vx0 t + ax t2
2
y
Now we want to solve for y − y0 , given x − x0 = l = 100 m, v0 = 48.5 m/s, θ = 12.8◦ ,
vx0 = v0 cos θ, and vy0 = v0 sin θ. From the second equation we find the time of flight:
p
−vx0 ± vx0 2 + 2ax (x − x0 )
t=
.
ax
We get two positive solutions, the smaller one represents the initial impact: t = 2.15 s. Plugging this solution into the first equation, we obtain (y − y0 ) = 0.45 m. I.e., the arrow hits
0.45 m above the bullseye .
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Problem 5. The Velodrome.
A velodrome has the form of a circular
bowl whose flat bottom is a circle of radius
r=25 m. The side of the velodrome has
the same radius of curvature r=25 m, as
shown in the figure below which represents
a vertical cross section of the velodrome,
through its central axis.
r
r r
r
R
r
a. [7 pts] When on the edge of the flat bottom of the velodrome, riders are moving in a
circle with radius r=25 m. If the coefficient of static friction is µs =0.5, what is the largest
speed vA for uniform circular motion (rolling without slipping) around this circle?
For rolling without slipping, the friction is static friction, and the frictional force is perpendcular
to the velocity vector ~vA . For motion in a circle, the frictional force FS provides the needed
2
centripetal acceleration: FS = mvA
/r, where m is the combined mass of the bicycle and the
rider.
For motion in a horizontal plane, the normal force has magnitude N = mg. Hence, the
2
maximum speed vA of the circular motion is related by mvA
/r = FS = µs N = µs mg, and:
vA =
√
µs gr =
p
0.5 · 9.8 m/s2 · 25 m = 11.1 m/s = 39.8 km/hr .
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b. [3 pt] What is the angular velocity ωA for the motion in part a?
ωA =
vA
11.1 m/s
=
= 0.44 rad/s .
r
25 m
c. [7 pts] Now suppose a rider goes on the banked curve of radius R that passes through
point B on the figure, corresponding to an angle θ=30◦ , and that he rides at a speed such
that no friction is required to maintain uniform circular motion. What is the speed vB for
this motion?
The only forces on the bicycle are its weight m~g and the
~ . There is no vertical acceleration, so:
normal force N
mg = N cos θ.
The equation for uniform circular motion in the horizontal
circle of radius R = r(1 + sin θ) is:
v2
m B = N sin θ = mg tan θ.
R
Hence:
tan θ(1 + sin θ) =
2
vB
rg
r
r
r r
R
Therefore:
p
p
vB = gr tan θ(1 + sin θ) = 9.8 m/s2 · 25 m tan 30◦ (1 + sin 30◦ ) = 14.6 m/s ,
or:
vB = 52.6 km/hr .
r
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d. [3 pt] What is the angular velocity ωB for the motion in part c?
ωB =
vB
vB
14.6 m/s
=
=
= 0.39 rad/s .
R
r(1 + sin θ)
25 m(1 + sin 30◦ )
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Problem 6. The Rube Goldberg Orange Juice Squeezing Machine.
Milkman takes empty milk bottle (A), pulling string (B) which causes sword (C) to sever
cord (D) and allow guillotine blade (E) to drop and cut rope (F) which releases battering
ram (G). Ram bumps against open door (H), causing it to close. Grass sickle (I) cuts a slice
off end of orange (J), at the same time spike (K) stabs “prune hawk” (L) who opens his
mouth to yell in agony, thereby releasing prune and allowing diver’s boot (M) to drop and
step on sleeping octopus (N). Octopus awakens in a rage and, seeing diver’s face which is
painted on orange, attacks it and crushes it with tentacles, thereby causing all the juice in
the orange to run into glass (O)...
a. [4 pt] If the guillotine blade (E) falls by a distance h=65 cm before cutting rope (F),
what is the speed u of the blade just before it cuts the rope?
u=
p
√
2gh =
2 · 9.8 m s−2 · 0.65 m = 3.6 m/s .
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b. [4 pts] The battering ram (G) has a mass m of 52 kg. Each of the massless chains that
support it have a length l of 1.73 m and make an initial angle θ of 23◦ to the vertical. What
is the momentum Pram of the battering ram just before it bumps the door (H), if the chains
are vertical at the moment of the collision?
The battering ram falls through a vertical height h1 of:
h1 = l(1 − cos θ) = 1.73 m (1 − cos 23◦ ) = 0.138 m.
The speed v of the ram when the chains are vertical is given by:
√
p
v = 2gh1 = 2 · 9.8 ms−2 · 0.138 m = 1.64 m/s.
The momentum of the ram just before bumping the door is:
Pram = mv = 52 kg · 1.64 m/s = 85.4 kg-m/s .
The ram is moving horizontally at this time, so the momentum vector is horizontal.
c. [3 pt] The battering ram travels parallel to the wall and bumps into the door at a distance
d=67 cm out from the wall. What is the angular momentum of the battering ram about the
axis of rotation of the door? The door is initially at rest and is open by 90◦ with respect to
the wall. For this question, you may consider the battering ram as a point-like object with
mass m and velocity v.
If you did not get the solution of part b, assume that the velocity of the ram v is 2 m/s.
Lram = mvr sin 90◦ = 52 kg · 1.64 m/s · 0.67 m · 1 = 57.1 kg-m2 /s
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d. [3 pt] The door has a mass M of 33 kg and a width w of 75 cm. What is the moment
of inertia of the door about its axis of rotation?
I = M w2 /3 = 33 kg · (0.75 m)2 /3 = 6.2 kg-m2 .
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e. [6 pts] What is the angular velocity of the door about its axis of rotation just after it
has been bumped by the battering ram, assuming that the collision is elastic?
Note: the reaction force of the hinge acts in a direction perpendicular to the axis of rotation
of the door. The external torque is therefore null, and the angular momentum of the system
(door + ram) is conserved. Due to the presence of the reaction force of the hinge, linear
momentum is not conserved.
During the collision there can be an impulsive force on the hinges of the door. So, momentum
is NOT conserved during the collision. But, since the force on the hinges exerts no torque on
the door (about the axis of rotation of the door), angular momentum about the axis of rotation
of the door is conserved:
mvr sin 90◦ = mv 0 r sin 90◦ + Iω,
where v 0 = speed of ram after the collision, I = moment of inertia of the door (found in part
d), and ω = desired angular velocity of the door after the collision.
Conservation of energy during the elastic collision implies that:
1
1
1
2
mv 2 = mv 0 + Iω 2 .
2
2
2
From the first equation, we find:
v0 = v −
Iω
,
mr
and hence
2
v0 = v2 − 2
Iv
I2
ω + 2 2 ω2 ,
mr
m r
and from the second equation we have:
2
v0 = v2 −
Iω 2
.
m
Thus,
v2 − 2
The quadratic equation for ω is:
Iv
I2
Iω 2
ω + 2 2 ω2 = v2 −
.
mr
m r
m
I2
I
+
2
2
m r
m
ω2 − 2
Iv
ω = 0.
mr
The solution ω = 0 is not relevant. The nontrivial solution is:
ω=
v
2
1.64 m/s
=
2
r 1 + I/mr
0.67 m 1 +
2
6.2 kg m2
52 kg·(0.67 m)2
= 3.7 rad/s.
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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]
REarth = 6400 km
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
I = 31 M w2 [thin sheet]
MEarth = 6.0 × 1024 kg
x = x0 + v0 t + at2 /2
KE = 12 mv 2
∆θ = ω0 t + 12 αt2
F = −GM m/r2
τ = F ` sin θ
p~ = m~v
v 2 = v02 + 2a∆x
I = 12 mr2 [disk]
L = mvr
G = 6.67 × 10−11 Nm2 /kg2