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Physics 125-1 Sample Final Questions, Fall 2010
With Solutions
1) Suppose we have a traffic light hanging from the
end of a boom of mass = 40 kg, length = 7.5 m. The
boom is of uniform width and density, and is free to
pivot around a frictionless pin at the bottom. A 5.2 m
cord is tied to the boom at a height of 3.9 m above the
pivot. The traffic light is connected to the very end
of the boom, and has a mass of 60 kg. What is the
tension in the cord?
Solution: The torques acting to rotate the boom
around its pivot must add to zero. There are two
clockwise torques: the weight of the boom and the
weight of the stop light. The boom’s weight acts at
its center of mass, (7.5 m)/2 = 3.75 m from the pivot.
The stop light’s weight acts at the end, 7.5 m from the pivot. The component of the gravitational force
perpendicular to the boom can be obtained by using the triangle made by the boom, cord, and pillar, and a little
trig. Pythagoras gives the hypotenuse of the triangle as
3.92 + 5.22 = 6.52 , which is a 3:4:5 triangle, and this tells us that the angle in
r x F is sinθ = 4/5 for both weights. The combined torque due to the weight of the boom and the stop light is τ =
(4/5)(40 kg)g(3.75 m) + (4/5)(60 kg)g(7.5 m) = (4g/5)(600 kg-m). This must equal the torque supplied by the
cord in the counterclockwise direction. The cord is attached to the boom 6.5 m from the pivot, and once again
using a bit of trig, we find that sinθ = 3/5 for the tension T in the cord acting on the boom. We have (3/5)T(6.5
m) = (4g/5)(600 kg-m), or T = (4g/3)(600/6.5) = 1206 N.
2) Mars has two tiny moons, Deimos and Phobos, aka Panic and Fear. Deimos orbits 23,436 km from Mars with
a period of 1.262 Earth days. Phobos orbits 9377 km from Mars. How long does Phobos take to make one orbit?
Solution: We can use Kepler's Third Law, T2 = (4π2/GM)R3. Taking the ratio, (TP/TD)2 = (RP/RD)3 gives us
(TP/1.262 day)2 = (9377 / 23,436)3, or TP = (1.262 day)(9377 / 23,436)3/2 = 0.319 day.
3) The "large" Martian moon, Phobos, has a radius of about 11 km and a mass of 1.08 x 1016 kg. If you stood on
the surface of Phobos, how fast would you have to throw a baseball to place it into orbit around Phobos (assuming
the orbit is barely above the ground)?
Solution: For a stable orbit, the gravitational force must equal the centrifugal force acting on the baseball. We
have GMm/r2 = mv2/r, or v = (GM/r)1/2 = [(6.67 x 10-11)(1.08 x 1016)/(11,000)]1/2 = 8.1 m/s, or 18 miles per hour!
4) We have a 500-gm mass and two springs with constants k1 and k2. If only the first spring (k1) is connected to
the mass, then its oscillation frequency is 5 Hz. If only the second spring (k2) is connected to the mass, then its
oscillation frequency is 10 Hz. What is the oscillation frequency when both springs are connected to the mass?
Solution: With both springs attached to the mass, F = –k1x – k2x = –(k1 + k2)x. This means ω = [(k1 + k2)/m]1/2
with both springs attached. Since ω1 = 2πf1 = (k1/m)1/2, we have k1 = 4π2f12m, and likewise for k2. Doing the
algebra: ω = [(k1 + k2)/m]1/2 = [(4π2f12m + 4π2f22m)/m]1/2 = 2π( f12 + f22)1/2. This gives f = (52 + 102)1/2 = 11.2 Hz.
(Note that you never need to use or even know the value of the mass!)
5) The tube shown at left contains a mineral oil with a density of 3.4 gm/cm3.
The end of the tube is open to the local atmosphere, which happens to be the
atmosphere on the planet Zargon. The other end of the tube is connected to a
bulb of gas imported from Earth which is at a pressure of exactly one standard
Earth atmosphere (atm). Zargon has the same mass as the Earth, but its radius
is only 85% that of Earth. What is the atmospheric pressure on the planet
Zargon?
Solution: We have P = P0 + ρgh, where P = 1 atm = 1.01 x 105 N/m2,
3
3
ρ = 3.4 gm/cm = 3400 kg/m , and h = 0.25 m. Since Zargon has the same
mass as Earth but a radius 85% as large, the ratio of the two gravitational
forces is: FZ/FE = (GMm/rZ2 )(rE2/GMm) = (rE/rZ)2 = 1 ÷ 0.852 = 1.384, so "g"
on Zargon is (1.384)(9.8) = 13.6 m/s2. Inserting all the numbers: P0 = P – ρgh
= 1.01 x 105 – (3400)(13.6)(0.25) = 8.94 x 104 N/m2. Dividing this by
1.01 x 105 N/m2 gives us 0.89 atm. (The higher gravity on Zargon does not
imply higher atmospheric pressure. It may have less air than Earth.)
6) Prior to the end of the 18th century, most firearms had smooth bores – i.e., their barrels were perfectly smooth
cylinders on the inside. After that time, arms makers increasingly moved to “rifled” barrels, which were barrels
that had fine lines or grooves engraved on the inside in a shallow spiral (see illustration at right). The purpose of
the grooves was to put a lot of spin on the bullet. The question is, why would you want to put a lot of spin on a
bullet?
A) Spin makes the bullet give off a frightening whine.
B) The added rotational kinetic energy makes the bullet more
destructive.
C) The angular momentum vector keeps it pointed ahead for better
accuracy.
D) The centrifugal force helps the bullet maintain its shape.
E) A spinning bullet experiences less air friction.
F) All of the above
G) None of the above
Answer: C. The spinning bullet acts like a tiny gyroscope, so the angular
momentum vector directed away from its nose helps to keep it pointed
straight forward rather than tumbling. This stabilizes it against erratic flight
due to the severe buffeting it receives as it pushes its way through the air at high speed.
7) A bullet with a mass of 50 gm is travelling at 250 m/s. In front of it is a thin
rod with a length of 1 meter that has a frictionless pivot point at its center. The
bullet slams into the thin rod at a distance of 0.3 meter from the pivot, and sticks
to the rod. After the collision, the bullet is travelling at 160 m/s. What is the
mass of the rod?
Solution: We can use conservation of angular momentum. At the instant of
collision the bullet is travelling at right angles to the rod, so the angular
momentum of the bullet around the pivot point is L = r × p = r mv =
(0.3 m)(0.05 kg)(250 m/s) = 3.75 kg m2 /s. After the collision, the angular
momentum of the bullet is (0.3)(0.05)(160) = 2.4 kg m2 /s. So, to conserve
angular momentum, the rod must be rotating after the collision such that:
(3.75 – 2.4) kg m2/s = Iω = [1/12 M(1 m)2][(160 m/s)/(0.3 m)] = 44.4 M m2/s. (We have used the fact that
I = 1/12 ML2 for a thin rod, v = rω for the bullet, and ω for the bullet and the rod must be the same.) So,
M = (3.75 – 2.4)(kg m2/s) / (44.4 m2/ s) = 0.03 kg = 30 g.
8) You take a lead cube that is 1 cm on a side and a wooden cube that is 2 cm on a side, and tie them together
with a silk string. You toss them into a lake. What is their acceleration as they sink? (Neglect water friction.)
Density of lead = 11.35 g/cm3
Density of wood = 0.7 g/cm3
Solution: The net force acting on the cubes is their weight minus the buoyancy force. Buoyancy equals the
weight of the liquid displaced, so F = –(11.35 g/cm3)(1 cm3)g – (0.7 g/cm3)(2 cm)3g + (1.0 g/cm3)(9 cm3)g =
–(7.95 g)(9.8 m/s2). We have a = F / m = –(7.95)(9.8 m/s2) / (11.35 + 5.6) = –4.6 m/s2.
9) If you take a spinning gyroscope and drop it off a tall building (you can assume that it falls without air
friction), it will:
a) precess all the way down at constant ω
b) precess all the way down with ω gradually decreasing
c) not precess, but tumble randomly
d) precess all the way down with ω gradually increasing
e) not fall at all
f) neither precess nor tumble, but point in one direction
Answer: F. Angular momentum conservation keeps the axis pointed in one direction, and it cannot precess
because there is no torque on a freely falling object.
10) A weight attached to a spring is set into oscillation. As the weight gradually loses energy (due to friction in
the spring), which of the following happens?
a) amplitude decreases, frequency increases
b) amplitude increases, frequency increases
c) amplitude decreases, frequency is constant
d) amplitude increases, frequency is constant
e) amplitude decreases, frequency decreases
f) amplitude increases, frequency decreases
Answer: C.
11) If you take a ball of radius R that is rotating very rapidly at a rate ωS and toss it onto a flat surface with
nonzero friction, what will the ball do before it begins to roll without slipping?
a) skid until it has zero rotational kinetic energy
b) skid until its linear velocity equals what it would be if the surface was frictionless
c) skid until its linear velocity is v = RωS
d) skid until the frictional torque equals zero
e) skid until its angular momentum equals its linear momentum
f) none of the above
Answer: D.
12) A block of 5 kg is tied to a cord wrapped around the
equator of a uniform sphere of mass 3 kg and radius 110 cm,
and over a wheel of mass 5 kg and radius 60 cm. You can
assume the wheel is a uniform disk. The cord moves
without slipping or losing contact with the sphere and wheel,
but there is no other friction present. Both the sphere and
wheel can rotate freely. If the system starts at rest, what will
be the angular velocity ω of the sphere when the block has
fallen 4 m?
Solution: From conservation of energy, the gravitational energy released by the falling block will go into the
kinetic energy of the block, the rotating sphere, and the rotating wheel. Using S for the sphere and W for the
wheel: mgh = ½ Isωs2 + ½ Iwωw2 + ½ mv2. (m & v denote the block’s mass & velocity.) With Is = 2/5 MsRs2 and
Iw = ½ MwRw2, we have mgh = ½(2/5 MsRs2)ωs2 + ½(½ MwRw2)ωw2 + ½ mv2. Since the cord is moving at the same
speed for all three objects, we know v = Rsωs and v = Rwωw. Substituting: mgh = 1/5 Ms v2 + ¼ Mw v2 + ½ mv2 =
(1/5Ms + ¼ Mw + ½ m)v2. Inserting numbers: v2 = (5 kg)(9.8 m/s2)(4 m)/[(0.2)(3 kg) + (0.25)(5 kg) + (0.5)(5 kg)],
or v = 6.71 m/s. This gives ωs = v/Rs = 6.71 / 1.1 = 6.1 rad/sec.
13) If the Sun were to suddenly lose 20% of its mass, and the Earth were to remain at the same orbital radius as it
has now, how many days would there be in a year?
Solution: Using Gm1m2/r2 = mv2/r, the velocity for the Earth is v2 = GMSun/r. The ratio between the proposed
situation and the current one is v/vnow = [G (0.8 MSun) RE / G MSun RE]1/2 = 0.81/2 = 0.894, so the year would be
365.24/0.894 = 408.35 days.
14) A disk at the top of a ramp has a coiled spring inside it such
that a torque of –kR2θ is applied to the disk if it is rotated away
from θ = 0. (k = 20 N/m. The disk has M = 10 kg and R = 0.2 m.)
A mass of 5 kg is attached to the disk as shown. If the ramp is
frictionless, and you tap the mass, what will be its frequency of
oscillation?
SOLUTION
The mass on the ramp pulling on the disk will apply a constant
torque of τ = r x F = R mg sin(30°). But, if you were listening
carefully when I discussed a very similar problem in class, then you know that stretching a spring by a constant
amount only shifts its zero point so that it obeys –k(x – x0) rather than –kx. This is a distinction without a
difference, because I can always shift my coordinates to make x0 = 0, so there is NO effect on the spring’s
frequency. We can ignore both the mass and the ramp!
To find ω, we need to write down a differential equation of the form –AΨ
Ψ = B(d2Ψ/dΨ
Ψ2), because this is the form
of a harmonic oscillator. We do not need to solve this equation, we just need to write it down. Once we have, we
know the oscillation frequency will be given by ω2 = A/B.
In this case, the torque on the disk is –kR2θ. This must equal Iα for the disk, and since I = ½
MR2 for a disk, we can substitute to get –kR2θ = ½ MR2 α, or –kθ
θ = ½ M (d2θ/dt2), where
I’ve written out α to remind ourselves that it is, in fact, the double time derivative of θ.
Looking at the two Wildcat bold-purple equations above, we can see that they are identical if
Ψ = θ, A = k and B = ½ M. We have a harmonic oscillator. So ω = (2k / M)1/2 =
[2(20 N / m)/(10 kg)]1/2 = 2 rad/sec.
15) You own a Physicsland® test tube whose walls have no thickness and no mass. Its radius
is 1.2 cm. The test tube contains 6 cm of a tar-like sludge with a density ρ = 1.4 g/cm3.
Above that it contains 17 cm of a light oil with a density ρ = 0.9 g/cm3. The top 5 cm of the
test tube are empty. If I float the test tube in water, how far below the top lip of the test tube
will the waterline be?
Solution: For the test tube to float, its weight must equal the buoyancy force. Since m = ρV,
the mass of the test tube is m = ρtarVtar + ρoilVoil = ρtar πr2(6 cm) + ρoil πr2(17 cm), where we
used V = πr2h for the volume of a cylinder. The given densities yield m = πr2[(1.4)(6) +
(0.9)(17)] = 23.7 πr2. The weight is W = mg = 23.7 πr2g. Buoyancy equals the weight of
displaced fluid. The displaced water’s mass is m = ρH2OVH2O = (1 g/cm3)πr2d, where d is the
length of the tube under water. The buoyancy is then FB = W = mg = πr2dg. Equating this force to the weight of
the test tube yields d = 23.7 cm. The waterline will be (5 + 17 + 6 – 23.7) = 4.3 cm below the lip. (Note that we
did not need to use the values of πr2g).
16) The figure at right shows three spheres connected by two stiff,
massless rods. The large sphere has M = 60 kg and R = 0.35 m. Two
identical smaller spheres have M = 8 kg and R = 0.2 m. The center-tocenter length of the rods is 1 m and they are 100° apart. What is the
moment of inertia of this system around the y-axis?
Solution: The larger sphere is centered on the rotation axis, so it has
I = 2/5 MR2 = (0.4)(60 kg)(0.35 m)2 = 2.94 kg m2. For the moment of
inertia of either of the smaller spheres, we must use the parallel axis
theorem. We have I = ICM + md2 = 2/5(8 kg)(0.2 m)2 +
(8 kg)[(1.0 m)sin(50°)]2 = 4.82 kg m2, where we recall that d refers to
the perpendicular distance from the CM of a small sphere to the rotation axis. The total moment of inertia is
I = 2.94 + 2 x 4.82 = 12.58 kg m2.
17) For the same spheres as in Problem 16, suppose I place a point mass of 1 kg on a line directly below the large
sphere at a distance of 2 m as measured from the large sphere’s center, i.e., at (x, y) = (0, –2) if the coordinate
origin is at the large sphere’s center. What is the direction and magnitude of the gravitational force acting on the
point mass, due to all three spheres?
Solution: By symmetry, the gravitational force will act upward along the y-axis. The large sphere will contribute
F = GMm/r2 = G (60 kg)(1 kg)/(2 m)2 = 15 G, because its center is on the y-axis. Either of the small spheres will
contribute FY = GMm/r2, where FY means only the y-axis force component is used. (The x-components from the
small spheres cancel each other.) We can use the x-y coordinates of the right sphere to calculate r:
x = (1.0)sin(50°) = 0.766, and y = –(1.0)cos(50°) = –0.6428. This means ∆x = 0.766 – 0 and ∆y = –0.6428 – (–2),
or r = 0.766 i + 1.357 j. The magnitude of this vector is r2 = (0.766)2 + (1.357)2, or r = 1.56 m. The unit vector is
r / 1.56, or r = 0.492 i + 0.871 j. (We don’t need the x-component, but I include it anyway for esthetics.) This
yields a force of FY = G(0.871)(8 kg)(1 kg)/(1.56 m)2 = 2.87 G. By symmetry, the force from the other small
sphere is the same, so the total gravitational force is F = 15 G + (2)(2.87 G) = (20.74)(6.67 X 10–11) =
1.38 X 10–9 N = 1.38 nano-newtons.
18) A small ball of mass m = 0.40 kg is thrown against a thin disk as
shown in the “top down” view at right. The ball has v = 12 m/s oriented at
37° relative to a tangent line at the point of contact. The disk has no
friction and is free to rotate around a fixed axis at its center. It has a radius
of 1.8 m, a mass of 2 kg, and was not moving before the ball hit it. If the
ball sticks to the disk at the point of contact, what will be the angular
rotation rate (rad/sec) of the disk afterwards?
Solution: This is an exercise in conservation of angular momentum. The
angular momentum at the time of collision is L = r x p =
(1.8 m)(0.4 kg)(12 m/s)sin(53°) = 6.9 kg m2/s. This quantity is conserved.
To calculate ω “afterwards”, we need to know the moment of inertia of the
system after the collision. This is I = disk + stuck ball = ½(2 kg)(1.8 m)2 + (0.4 kg)(1.8)2 = 4.536 kg m2. We
have L = Iω, so ω = (6.9 kg m2/s) / (4.536 kg m2) = 1.52 rad/s.
19) A child’s bedroom has a mobile of cute plastic fishes
hanging from the ceiling. The fishes are in perfect balance.
The distances between the strings holding the bars, and the
ends of the bars where the fish are attached, are shown at
right. The bars have no mass. If fish “C” has a mass of
100 g, what are the masses of fishes “A” and “B”?
Solution: Using the lever law (distance times force are
equal at each end of a lever), fish “B” must be three times as
massive as fish “C”, so fish “B” = 300 g. “B” and “C”
together have a mass of 400 g, so fish “A” must have a mass
that is (7.5 / 30) = ¼ of this, or fish “A” = 100 g.
20) Suppose you have a sealed cylinder 1.5 cm in radius and 20 cm long. One end of it is sealed by a cork that
you know will not move until 220 N is applied to it. Estimate how hot you would have to make the cylinder for
the air inside to blow out the cork.
Solution: Pressure is F/A, so the net pressure needed to pop the cork is P = (220 N)/[π(0.015 m)2] = 311,000 Pa.
However, there is atmosphere pressure = 1.01 x 105 Pa operating on the other side of the cork, so the total internal
pressure needed is 311,000 + 101,000 = 412,000 Pa. Since both the volume of the cylinder and the number of
atoms trapped inside it are constant, we can take the ratio of PV = NkT before and after to get PA/PB = TA/TB. We
assume that the “before” temperature is around 300 K° (~ room temperature) and the “before” pressure is one
atmosphere. We have TA = (412 kPa / 101 kPa)(300 K°) = 1220 K°.