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Linear Motion - multiple choice
What is the cross product A x B of the vectors A = (1,0,1) and B = (-1,0,1)?
(a) (0,0,2) (b) (0,-1,0) (c) (0,2,0) (d) (0,-2,0) (e) none of [a]-[d]
The two vectors each have length √2 and lie in the xz-plane, making angles of +
o and
o with respect to the z-axis, respectively. The magnitude of the cross product is then
2 sin
o = 2. By the right-hand rule, the direction is in the - y direction. Hence, (0,-
Consider two vectors A = (2,2,0) and B (4,-1,0). If a vector C = (1,0,0) is added to B ,
what is the change in the cross-product A x B ( i.e., the new value minus the old value).
[a] (0,0,-2) [b] 2 [c] (0,0,2) [d] (-10,0,0) [e] none of
[a]-[d]
The difference in the cross products is A x( B + C ) - A x B = A x B + A x C - A x B = A x C.
Thus, the difference is (0, 0, 0 - 1•2) = (0, 0, - 2).
Find the vector C in the cross product C = A x B , if A = (0,1,0) and B = (1,0,0).
(a) (1,1,0) (b) (0,0,1) (c) (0,0, - 1) (d) (1, - 1,0) (e) (0,0,1)
Either use the explicit Cartesian representation, or the following logic:
Vectors A and B are unit vectors pointing along the y and x axes, respectively. Thus,
the cross product must be perpendicular to the xy-plane, pointing along the + z or - z
directions. From the right-hand rule, the direction must be negative. Lastly, the
magnitude of C equals the product of the magnitudes of A and B (each of which equals
- and the sine of the angle between them. Here, the sine equals 1, so the resulting
vector is C = (0,0, - 1).
The displacement of an object for a round trip between two locations
(a) is always greater than zero.
(b) is always less than zero.
(c) is zero.
(d) is not zero.
(e) can have any value.
For a round trip, the displacement must vanish because the initial and final positions
coincide.
An object moves with constant speed in a straight line. Which of the following
statements must be true?
(a) No force acts on the object.
(b) A single constant force acts on the object in the direction of motion.
(c) The net force acting on the object depends on the value of the speed.
(d) The net force acting on the object is zero.
(e) The net force acting on the object cannot be determined.
The acceleration is determined by the net force; hence, a vanishing acceleration implies
that the net force vanishes, but not that there is no force.
As a sky diver falls through the air, her terminal speed
(a) depends on her mass.
(b) depends on her body’s orientation.
(c) depends on the local value of the acceleration due to gravity.
(d) depends on the density of air.
(e) is described by all of the above.
All of these features apply to terminal velocity.
A man riding in an elevator has an apparent weight greater than his actual weight.
Which one of the following statements could be true?
(a) The elevator moves upward with constant speed.
(b) The elevator moves downward with constant speed.
(c) The elevator moves upward with decreasing speed.
(d) The elevator moves downward with decreasing speed.
(e) The elevator moves downward with increasing speed.
If his apparent weight is less than his real weight, then his acceleration must be upward.
This could arise in two situations:
v is upward and increasing or v is downward and decreasing.
A moving box receives an impulse directed to the north. As a result, the box
(a) has a velocity pointing north.
(b) is north of the equator.
(c) stops.
(d) has a velocity towards the south.
(e) accelerates towards the north.
The direction of the acceleration must be the same as the net force, but the velocity may
be in any direction.
A lamp of mass m hangs from a spring scale that is attached to the ceiling of an
elevator. When the elevator is stopped at the fortieth floor, the scale reads mg. What
does it read while the elevator decends toward the ground floor at a constant speed?
(a) More than mg.
(b) Less than mg.
(c) mg.
(d) Zero.
(e)This cannot be answered without knowing how fast the elevator is descending.
If the speed of the lamp with respect to the elevator cage is constant, then there is no
relative acceleration, and the scale must still read mg. Hence: (c).
A falling abject experiences the drag force due to air resistance. Which statement is
NOT true?
(a) The drag force depends on the falling speed.
(b) The faster the ball falls, the stronger the air resistance.
(c) The mechanical energy of the object is conserved.
(d) The speed of the object will reach a maximum value and then stop changing.
(e) The net force acting on the object will eventually reach zero.
The position of an object with zero initial velocity and subject to constant gravitational
acceleration - g is z = - gt
2 /2. Thus, if the elapsed time is 3 t, the height fallen must be 9 h.
An arrow shot vertically into the air rises to a height h before it falls back to Earth. If the
speed of the arrow as it leaves the bow is doubled, how high could the arrow rise?
(a) 8 h (b) h (c) 16 h (d) 2 h (e) 4 h
The position of an object with zero final velocity (when the arrow reaches its maximum
height) and subject to constant gravitational acceleration - g is
h = - ( v f
2
2 )/2 g = v i
2 /2 g.
Thus, if the initial velocity is doubled, then the maximum height increases by a factor of
2 = 4.
A girl is standing in an elevator traveling upwards at a constant speed v. She observes
that it takes a time t for a penny to drop from her hand to the floor of the elevator. If the
elevator were traveling downwards at a constant speed of 2 v, how long would it take for
the penny to drop?
(a) t (b) 2 t (c) 4 t (d) t/2 (e) none of [a]-[d]
Because the velocity of the elevator is constant in both situations, then the time for
relative motion is unchanged. Hence, it takes t whether the elevator is at rest or moving
with constant velocity.
A projectile is fired at an angle of 35
o above the horizontal. At the highest point in its
trajectory, its speed is 200 m/s. If air resistance is neglected, what is the initial
horizontal component of the projectile's velocity (in m/s)?
(a) 0 (b) 200 cos
o (c) 200 sin
o (d) 200/cos
o (e) 200
At the highest point in the trajectory, the velocity is entirely horizontal. Hence, the
horizontal component must be 200 m/s throughout the flight of the projectile.
A golfball is shot horizontally from the top of a vertical cliff of height h. If the initial speed
of the ball is v, how far away from the bottom of the cliff does the ball land?
(a) v(2 h/ g)
2 (b) gt
2 /2 (c) v(2 h/ g)
1/ (d) vg / 2 h (e) none of (a)-(d)
Resolving the motion into components:
horizontal distance = R = vt
vertical distance = h = gt
2 /2.
---> R = v(2 h / g)
1/ .
A car moving with constant speed in a circular path experiences a centripetal
acceleration a c
. If the speed of the car increases by a factor of three, but the circular
path remains the same, what is the new centripetal acceleration in terms of the original
a c
(a) 3 a c
(b) a c
/9 (c) a c
/3 (d) 9 a c
(e) a c
Since the centripetal acceleration is proportional to v
2 / R, then the acceleration increases
by 3
2 =9 if the speed increases by 3.
Two cars are travelling in concentric circles of radius R and 2 R, as in the diagram.
2 R 4 R
Each car completes its own circle in the same time T. What is the centripetal
acceleration of the outer car divided by that of the inner car?
(a) 1 (b) 2 (c) 1/2 (d) 4π
2 (e) 8π
2
In general, a c
= v
2 / r with v = 2π r/ T, so that a c
= (2π r/ T)/ r = 4π
2 r/ T
2
. Hence,
a c
(outer)/ a c
(inner) = 2 R / R = 2.
A Ferris wheel ride at a circus has a diameter of 20 m. What must be the speed of the
perimeter of the wheel to produce a centripetal acceleration (at the perimeter) the same
as that of gravity (use g = 10 m/s
2 ; answers quoted in m/s)?
(a) 10 (b) 14 (c) 200 (d) 20 (e) 100
Centripetal acceleration a c
= v
2 / R. Demanding a c
= 10 m/s
2 ,
v = ( a c
R)
1/ = (10 x 20 / 2)
1/ = 10 m/s.
An object at the end of a string is swung in a circular path at constant speed with a
period T. If the period is shortened to T/2 without changing the radius of the circle, what
is the new centripetal acceleration in terms of the original acceleration a?
(a) 4 a (b) a/2 (c) a/4 (d) a (e) 2 a
The centripetal acceleration a is equal to 2π v/ T. In turn, the velocity is 2π R/ T, so the
acceleration is 4 π
2 R/ T
2
. Thus, if the period is decreased by a factor of 2, the
acceleration is increased by a factor of 4.
A car travels at constant speed on a circular test track of radius R, completing each lap
around the track in time T. The centripetal acceleration of the car, a c
, is at the limit
where the tires start to skid. If the test track were three times as large (i.e., had a radius
of 3 R), what would be the shortest period in which the car could complete a lap without
its acceleration exceeding the same a c
as for the smaller track?
(a) 3 T (b) 9 T (c) T / 3 (d) √ 3 T (e) none of (a - d)
The speed of the car is 2π R / T, so the centripetal acceleration is proportional to R / T
2 ,
or T
2 ~ R / a c
. If the centripetal acceleration is to remain unchanged in the new track,
then the period must be at least √ 3 T.
The total linear momentum of a system of particles will be conserved if:
(a) the positions of the particles do not change with respect to each other
(b) one particle is at rest
(c) no external force acts on the system
(d) the internal forces equal the external forces
(e) the particles do not rotate about their axes.
Force is the rate of change of linear momentum. Hence, there must be no external
force for the rate of change of momentum to be zero.
the Moon? Use g = 10 m/s
2 on Earth.
(a) 6 kg (b) 1 kg (c) 0.6 kg (d) 60 kg (e) 10 kg
Mass on Earth = w / g = 60 / 10 = 6 kg. The same mass must be measured on the
Moon.
Two masses ( m and 2 m) are attached to one another by a string as illustrated. A force
F acts on mass m to accelerate the whole system. What is the magnitude of the net
force on m?
F
2 m m
(a) F /3 (b) F (c) 2 F /3 (d) F /2 (e) 3 F /
The overall acceleration of the blocks is a = F /3 m. But this acceleration is the result of
the net force on a given block. Hence, the net force required to produce an acceleration
of a on m is F net
= ma = mF /(3 m) = F /3.
Two masses ( m and 2 m) are attached to one another by a string as illustrated. A force
F acts on mass m to accelerate the whole system. What is the magnitude of the force
on mass 2 m?
F
2 m m
(a) F /3 (b) F (c) 2 F /3 (d) F /2 (e) 3 F /
The overall acceleration of the blocks is a = F /3 m. The net force required to produce
an acceleration of a on 2 m is F = 2 ma = 2 mF /(3 m) = 2 F /3.
Two masses ( m 1
and m 2
) are connected by a massless string and accelerated
uniformly on a frictionless surface, as shown. The ratio of the tensions T 1
/ T
2
is given
by:
m 1 T 1
m 2
T 2
(a) m 1
/ m 2
(b) m 2
/ m 1
(c) ( m 1
)/ m 2
(d) m 1
/( m 1
) (e) m 2
/( m 1
m 1 T 1
m 2
T
2
T
1
At m 1
, an unbalanced force results in an acceleration T 1
= m 1
a.
At m 2
, an unbalanced force results in an acceleration T 2
= T
1
a = ( m 1
) a.
Hence, T 1
/ T
2
= m 1
/( m 1
A picture frame of mass 0.5 kg is held in place by a massless wire making an angle of
o with respect to the horizontal. What is the tension in the wire (in Newtons)?
T
o
(a) 7.2 N (b) 4.9 N (c) 2.5 N (d) 14.3 N (e) 0.84 N
Consider the forces at the nail. Taking components in the vertical direction
T
o
T
mg
2 T sin 20
o = mg hence T = = 7.2 N.
A vertical rope is attached to an object that has a mass of 40 kg and is at rest. The
tension in the rope needed to give the object an upward speed of 3.5 m/s in 0.
seconds is
(a) 392 N (b) 192 N (c) 532 N (d) 592 N (e) 140 N
The free-body diagram for the mass is
T
mg
The tension in the string is unbalanced, such that
T - mg = ma or T = m( g - a)
The acceleration of the object is a = ( v f
)/ t = 3.50 / 0.7 = 5 m/s
2
. Hence,
T = 40•(9.8 - 5) = 192 N.
The radius of the planet Xion is three times that of the Earth, although the densities of
the planets are the same. What is the acceleration due to gravity on the surface of
Xion?
(a) 27 g (b) g (c) 3 g (d) g / 3 (e) g / 9
From the gravitational force expression F = G M 1
M
/ R
2 , the acceleration due to gravity
on the surface of a planet is a g
= G M / R
2
. Since the mass of a planet with constant
density grows like R
3 , then the gravitational acceleration must grow like R
3 / R
2 = R.
Hence, the acceleration due to gravity on Xion must be 3 g.
Two identical springs, each with force constant k are attached in series
k k
What is the effective force constant of the springs, taken together?
(a) k (b) 2 k (c) k / 2 (d) 4 k (e) k / 4
If a single spring is stretched by a distance x, then the spring system must stretch by 2 x.
But the force exerted by the individual spring is kx , and must be constant along the
(e) none of (a)-(d)
The frictional force reduces the applied force to produce a constant acceleration:
ma = Tcosθ - f ----> f = T cosθ - ma.
A spring under a compression x has a potential energy V o
. When the compression is
doubled 2 x, the potential energy stored in the spring is
(a) V o
(b) 2 V o
(c) 3 V o
(d) 8 V o
(e) None of [a-d] is correct.
The potential energy of a spring is quadratic in the displacement from equilibrium.
Thus, if the compression is doubled, the potential energy must increase by a factor of
four.
A car of mass 1000 kg rolls without friction down Burnaby Mountain for a total vertical
distance of 300 m. If the car started from rest at the top of the hill, what is its speed at
the bottom of the hill in km/hr?
(a) 195 (b) 54.2 (c) 76.7 (d) 2940 (e) 276
Using conservation of energy,
mgh = mv
2 /2 -----> v = √(2 gh).
= √(2 x 9.8 x 300) = √5880 = 76.7 m/s.
Converting to km/hr, v = 76.7 x 3600 / 1000 = 276 km/hr.
A 50 gram piece of cake is dragged across a 2 meter long strip of sandpaper, losing
about half of its mass by the time it reaches the end of the strip. Only just enough force
is applied to the cake to keep it moving. How much work was done in moving the cake?
(Take g = 10 m/s
2 and the coefficient of friction μ for cake on sandpaper to be 0.8.)
(a) 60 J (b) 0.3 J (c) 0.6 J (d) 0.4 J (e) 0.8 J
The average weight of the cake is ( mg + mg /2) / 2 = 3 mg /4. Hence, the average
frictional force is just f = μ(3 mg /4). Multiplying the average force to obtain the work
gives
W = fD = μ(3 mg /4) D = 0.8 • 0.75 • 0.050 • 10 • 2 = 0.60 J.
The potential energy experienced by a ball at the bottom of a particular wine glass is
ax
4 , where a = 1 J/m
. What force is experienced by the ball at x = - 2 m?
(a) 16 N (b) 32 N (c) - 32 N (d) - 16 N (e) none of (a) - (d)
The relation between force and potential is F = - d U / d x. In this problem, F = - d( ax
dx, or
F = - 4 ax
3
. At x = - 2, then F = - 4 • 1 • (-2)
3 = + 32 kg-m/s
2 .
A particle is subject to a conservative force that does not depend on distance x. How
does its potential energy change with x?
(a) 1/ x
2 (b) x (c) 1/ x (d) constant (e) x
2
Starting with the definition of work W = F • Δ x , the work must be proportional to x if F is
constant. The work done on a system raises its potential energy, so we expect Δ U ~ x.
Using a cable, an engine applies a constant force to move a mass m across a
frictionless table in time t, starting from rest. The average power generated by the
engine during this process is P. If the same engine delivered the same force to a mass
2 m in moving it across the table, what would be the average power expended by the
engine (in terms of the original P)?
(a) P (b) √ 2 P (c) P / 2 (d) P / √ 2 (e) 2 P
The force and the distance, and hence the work done by the engine, is the same in both
situations. The average power then depends upon the time t required for the motion,
via
d = at
2 /2 --> t = (2 d /a)
1/ .
Since the force is constant, the acceleration experienced by 2 m is half of that
experienced by m, and hence t 2m
= √ 2 t m
. Therefore,
P
2m
= P
m
The power required to keep a car moving at a constant speed v against turbulent drag is
P. What is the power required to move the car at 2 v?
(a) P (b) 2 P(c) 4 P (d) 8 P (e) none of (a) - (d)
The drag force in the presence of turbulence grows quadratically with the speed
F
drag
~ v
2 .
But power is given by P = F • v , so the power grows like v
3
. Thus, if the velocity is
doubled, the power increases by 2
3 = 8.
What is the exponent n in the expression [ power] ∝ [ speed]
n for a drag force that varies
with speed as F drag
∝ v
2 ?
(a) 0 (b) 1 (c) 2 (d) 3 (e) 4
The power P associated with a force F acting on an object moving with velocity v is P =
F•v. In this problem, F ∝ v
2 , so
P ∝ v
2
3 .
Four squares of equal mass and
dimension L x L are arranged as shown.
What is the y-component of their
centre-of-mass?
(a) L/2 (b) 3 L/2 (c) L (d) 3 L/4 (e) L/
The centre-of-mass of each block lies at its centre, so the cm of the whole system is
y = (1/4 M) • {3 • (1/2) ML + (3/2) ML}
= ( L/4) • (3/2 + 3/2)
= 3/4 L.
L
x
y
L
energy K. If the stationary object had a mass of 2 m, instead of m, what would be the
kinetic energy of the pair?
(a) K/3 (b) 2 K (c) 3 K/2 (d) 2 K/3 (e) K/
By conservation of momentum, the momentum of the pair of objects is p, and this does
not depend on the mass of the struck object. After the collision, the kinetic energy K of
the first case is
K = p
2 /2(2 m) = p
2 /4 m.
In the second case, the kinetic energy K 2
is
K
2
= p
2 /2(3 m) = p
2 /6 m.
Thus, K 2
/ K = 4/6 = 2/3.
An object of mass m, moving on a frictionless surface, collides with, and sticks to, a
second object of mass m, initially at rest. Together, they recoil with a total kinetic
energy K. If the stationary object had a mass of 2 m, instead of m, what would be the
kinetic energy of the pair?
(a) K/3 (b) 2 K (c) 3 K/2 (d) 2 K/3 (e) K/
By conservation of momentum, the momentum of the pair of objects is p, and this does
not depend on the mass of the struck object. After the collision, the kinetic energy K of
the first case is
K = p
2 /2(2 m) = p
2 /4 m.
In the second case, the kinetic energy K 2 is
K
2
= p
2 /2(3 m) = p
2 /6 m.
Thus, K 2
/ K = 4/6 = 2/3.
Object A with mass m is travelling in the positive x direction when it collides inelastically
with a body of mass 2 m and comes to a complete stop. If object A has an initial velocity
v, what is the kinetic energy of object B?
(a) mv
2 /2 (b) mv
2 /4 (c) mv
2 (d) 0 (e) 2 mv
2
Initial momentum = mv = final momentum = p B
Kinetic energy of B is therefore K = p B
2 / 2 m B
= ( mv)
2 /4 m = mv
2 /4.
Two masses M and 5 M are at rest on a frictionless horizontal table with a compressed
(massless) spring between them. When the spring and masses are released, what
share of the spring's potential energy is carried off as kinetic energy by mass M?
(a) 4/5 (b) 1/5 (c) 1/6 (d) 1/2 (e) 5/
Each mass recoils with the same momentum p. Hence, the kinetic energies of the two
masses are p
2 / 2 M and p
2 / 10 M, respectively, for a total of 6 p
2 / 10 M = 3 p
2 / 5 M. The
fraction of this energy carried by mass M is
( p
2 / 2 M) / (3 p
2 / 5 M) = 5/6.
Two identical masses are hung from massless strings of the same length. One mass is
released from a height h above its free-hanging position, and strikes the second mass
which is initially at rest. The two masses stick and move off together. To what height H
do they rise?
h
H
(a) 3 h/4 (b) h/4 (c) h (d) h/2 (e) none of (a)-(d)
Before collision, equate energies to find momentum p
p
2 / 2 m = mgh --> p
2 = 2 m
2 gh.
After the collision, the momentum of the pair is still p, but the height is found from
p
2 / 2(2 m) = 2 mgH
H = p
2 / (8 m
2 g) = (2 m
2 gh) / (8 m
2 g) = h / 4.
A steady stream of 0.02 kg balls roll off a table and bounce elastically on a weigh scale
as shown in the diagram. If the balls drop through a height of 0.1 m in reaching the
scale, and they arrive at 0.5 s intervals, what is the average force measured by the
scales (in kg-m/s
2 )?
0.1 m
(a) 5.6 (b) 1.4 (c) 0.056 (d) 0.028 (e) 0.
The kinetic energy of each ball as it hits the scales is mgh, from which the velocity can
be found from
(1/2) mv
2 = mgh or v
2 = 2 gh or v = (2 • 9.8 • 0.1)
1/ = 1.4 m/s.
Thus, the momentum of the ball is mv = 0.02 • 1.4 = 0.028 kg-m/s. The change in
momentum at the scale is double the incident momentum, or 0.056 kg-m/s. Since this
change occurs every 0.5 seconds, then the force (rate of change of momentum) is
0.056 / 0.5 = 0.11 N.
The geometry of the "shoot-the-can" demonstration done in class is approximately
tin can
blow tube
10 m
line of sight
o
The projectile is aimed at an angle of 3 0
o with respect to the horizontal, and the
horizontal distance between the end of the blow tube and the can is 10 m. How far will
the can fall before the projectile strikes it, assuming that the can starts to fall as soon as
the projectile leaves the tube? The speed of the projectile when it leaves the tube is 24
m/s (neglect air resistance).
The horizontal velocity, v x
, of the projectile as it leaves the tube is
v x
= 24 cos 30
o = 24 • 0.866 = 20.8 m/s.
Thus, the projectile covers the horizontal distance to the falling can in a time t, given by
t = d / v x
= 10 / 20.8 = 0.48 s.
Since the can starts off with zero velocity, then during time t, the can has fallen a height
given by
h = - gt
2 / 2,
where g is the acceleration due to gravity. Thus,
h = - 9.8 • (0.48)
2 / 2 = - 1.13 m.
The space shuttle orbiter is moving in a circular orbit with a speed of 7 km/s and a
period of 80 minutes. In order to return to Earth, the orbiter fires its engines opposite to
its direction of motion. The engines provide a deceleration in this direction of 20 m/s
2 .
What is the magnitude and direction of the total acceleration of the orbiter when the
engines are first ignited?
The centripetal acceleration is given by
a c
= 2π v / T
= 2π • 7x
3 / (80 • 60)
= 9.16 m/s
2
Because a c
is perpendicular to the direction of motion, the total acceleration can be
found from pythagoras theorem to be
a tot
= ( a c
2
2 )
1/
2
2 )
1/
= 22.0 m/s
2 .
The angle of the acceleration vector is away from the direction of motion, in the general
direction of the Earth, making an angle with respect to the vertical obtained from
tan θ = 20.0 / 9.16 = 2.
or
θ = 65.
o .
In a binary star system, two stars with masses m 1
and m 2
rotate about their common
R
1
R
2
centre of mass. Assume that the orbits are circular, with radii R 1
and R 2
, such that the
distance D = R 1
+ R
2
between the stars is constant.
(i) Establish that R 1
= Gm 2
T
2 / (4π
2 D
2 ), where T is the period of the orbit.
(ii) Find the sum of the masses m 1
in terms of D and T.
(i) Each star experiences an acceleration due to gravity of F = GMm / D
2
. Equating this
force to ma c
, with a c
= v
2 / R, gives
mv
2 / R = GMm / D
2 or v
2 / R = GM / D
2
But the velocity of each star is given by
v = 2π R / T,
so that
(2π R 1
/ T)
2 / R 1
= GM
2
/ D
2 so that
or
4 π
2 R 1
/ T
2 = GM 2
/ D
2 or R 1
= Gm 2
T
2 / (4π
2 D
2 ).
(ii) Using the result from part (i), we can write each mass as
m 2
= 4π
2 D
2 R 1
/ GT
2
which gives a total mass of
m 1
= 4π
2 D
2 R 1
/ GT
2
2 D
2 R 2
/ GT
2
= 4π
2 D
3 / GT
2 .
The viscous drag force exerted by a stationary fluid on a spherical object of radius R is
F = 6π ηRv at low speed (Stokes' law)
F = ( ρ/2) AC D
v
2 at higher speeds,
where the symbols are defined in Lec. 9. Let’s apply this to a spherical cell one micron
in radius, moving in water with η = 10
3 kg/m
3
. Take the cell to have
the same density as water, and let its drag coefficient be 0.5.
(a) Plot the two forms of the drag force as a function of cell speed up to 100 μm/s.
Quote the forces in pN.
(b) Find the speed at which the linear and quadratic drag terms are the same.
For the linear drag term, the constants are numerically equal to
6 π ηR = 6π • 10
= 6π • 10
For the quadratic drag term, we have
( ρ/2) AC D
3
= (π/4) • 10
(a) Both expressions vanish when v = 0. By definition, the linear term increases fastest
at small v. However, even when v = 100 μm/s, the linear term still dominates:
linear drag force = 6π • 10
= 6π • 10
= 6π • 10
quadratic drag term = (π/4) • 10
2
= (π/4) • 10
= (π/4) • 10
(b) Equating the two drag forces gives
6 π • 10
2
or
v = 6π / (π/4) = 24 m/s.
