Concept Questions Exam 1 Solutions
W01_D3 Concept Questions
What is your best estimate for the volume of the earth’s atmosphere?
1)
between 101 and 105 cubic meters
2)
between 105 and 1010 cubic meters
3)
between 1010 and 1015 cubic meters
4)
between 1015 and 1020 cubic meters
5)
between 1020 and 1025 cubic meters
6)
between 1025 and 1030 cubic meters
Answer 4: The volume is a scalar quantity. Approximate the volume of the
atmosphere by a spherical shell of radius r and thickness t with volume
Ve  4 r 2t where the thickness of shell is approximately t  10 km  104 m and
the radius of shell equals the radius of the earth, Re  6  103 km  6  106 m . So
the volume is approximately

Ve  4 r 2t  4 6  106 m
 10 m 4  10
2
4
18
m3
W02D1 Concept Questions Solutions
An object goes from one point in space to another. After it
arrives at its destination, the distance it traveled is:
1) either greater than or equal to
2) always greater than
3) always equal to
4) either smaller than or equal to
5) always smaller than
6) either smaller or larger than
the distance it traveled the magnitude of the displacement.
Answer 1: The magnitude of the displacement is equal to
the magnitude of the length of the straight line connecting
the initial point to the final point. The distance is the length
of the path traveled between the two points. The distance
traveled is always greater or equal to the magnitude of the
displacement since the actual path traveled between the two
points need not be the straight line path.
The graph shows position as a function of time for
two trains running on parallel tracks. For times
greater than t =0, which is true:
1) At time t , both trains have the same velocity.
B
2) Both trains speed up all the time.
3) Both trains have the same velocity at some time
before t .
B
4) Somewhere on the graph, both trains have the
same acceleration.
Answer: 3. The slope of curve B is parallel to line A
at some point t < tB.
You are throwing a ball straight up in the air. At the
highest point, the ball’s
1) velocity and acceleration are zero
2) velocity is nonzero but its acceleration is zero
3) acceleration is nonzero, but its velocity is zero
4) velocity and acceleration are both nonzero.
Answer: 3. The ball reaches its highest point when its
velocity is zero; the acceleration of gravity is never
zero (it is always 9.8 m/s2 downward).
A person standing at the edge of a cliff throws one
ball straight up and another ball straight down, each
at the same initial speed. Neglecting air resistance,
which ball hits the ground below the cliff with the
greater speed:
1) ball initially thrown upward;
2) ball initially thrown downward;
3) neither; they both hit at the same speed.
Answer: 3. (coordinate system: positive x-axis
upwards) Upon its descent, the velocity of an object
thrown straight up with an initial x-component of
velocity vx,0  0 has velocity vx  vx,0  0 v when it
passes the point at which it was first released. This is
exactly the same x-component of velocity has the ball
that was thrown downward, so both balls will hit the
ground with the same x-component of velocity. Let
t f denote the time interval that the ball thrown
downwards takes to hit the ground, then the xcomponent of the velocity of both balls when they hit
the ground is given by
vx (t f )  vx,0  gt f .
W02D2 PRS Solutions
Consider an object released at time t = 0 with an initial x-component of velocity vx,0  0 ,
and accelerating according to
ax 
dvx
 c0  c1vx
dt
After a very long time, the x-component of the velocity is
1. Zero
2. c0  c1
3. c0 / c1
4. c0  c1
5. Not sure
Answer: 3. Terminal velocity is the velocity when the acceleration is zero. So
0
dvx
 c0  c1 (vx )term
dt
implies that terminal velocity is
(vx )term  c0 / c1 .
A person simultaneously throws two objects in the air. The objects leave the person’s
hands at different angles and travel along the parabolic trajectories indicated by A and B
in the figure below. Which of the following statements best describes the motion of the
two objects?
a) The object moving along the trajectory A hits the ground before the object
moving along the trajectory B.
b) The object moving along the higher trajectory A hits the ground after the object
moving along the lower trajectory B.
c) Both objects hit the ground at the same time.
d) There is not enough information is specified in order to determine which object
hits the ground first.
Answer: 3. The time it takes a shell to reach the ground again depends on the speed and
angle at which it was fired. Not knowing these, it may seem impossible to tell which ship
gets hit first. But the time a projectile spends in the air is equal to twice the time it takes
to fall from the maximum height the projectile reaches. Since the shell fired at ship A
reaches a higher altitude than the other shell, it will take longer to return to sea level.
Consider the path of a ball moving along a path through the air under the action of the
gravitational force. You may neglect the effects of air friction. As it reaches the highest
point in its arc, which of the following statement is true?
1) The magnitudes of the velocity and acceleration are zero.
2) The magnitude of the velocity is at a minimum but not equal to zero.
3) The magnitude of the velocity is equal to zero, and the magnitude of the
acceleration is constant and not equal to zero.
4) The magnitude of the velocity is at a minimum but not equal to zero and the
magnitude of the acceleration is zero.
5) Neither the magnitudes of acceleration or velocity has yet attained its minimum
value.
Answer: 2. The velocity has a non-zero horizontal component which remains constant
throughout the flight. The vertical component of the velocity is zero at the top of the orbit
so the magnitude of the velocity is minimum at the top of the orbit. The acceleration is
constant throughout the orbit and points downwards.
For a parabolic orbit under the influence of gravitation,
1) the magnitude of the velocity can be determined from the slope of the tangent line
to the graph of y vs. x but not the direction
2) the magnitude and direction of the velocity can be determined from the slope of
the tangent line to the graph of y vs. x
3) the magnitude and direction of the velocity cannot be determined from the slope
of the tangent line to the graph of y vs. x
4) the direction of the velocity can be determined from the slope of the tangent line
to the graph of y vs. x but not the magnitude.
Answer: 4. The direction is the velocity is always tangent to the spatial orbit but the
magnitude cannot be determined. The magnitude of the slope is dy / dx which is not the
magnitude of the velocity v  vx2  vy2  (dy / dt)2  (dx / dt)2 .
Consider the situation depicted here. A stone is accurately aimed at a person hanging
from the gutter of a building. The target is well within the stone’s range, but the instant
the stone is thrown, the person lets go and drops to the ground. The stone moves with a
speed v0, just as it is released. What happens? The stone
1) hits the person, regardless of the value of v0;
2) hits the person only if v0 is large enough;
3) misses the person.
Answer: 2. Let’s consider the stone’s motion without the effect of gravity. If you aim it at
the ledge, in the time interval t that it takes the stone to reach the target, it would be at
the same height as the ledge. However during this time interval, the effect of gravity is to
decrease the height of the stone by a factor (1 / 2)g(t)2 . This is precisely the same
distance that the person has fallen downwards during the same time interval t . So in
principle the stone will strike the person. However if the stone does not have enough
initial speed, it will hit the ground before it reaches the person.
W03D1 Concept Questions Solutions
Concept Question: Equivalent Spring Constant
Two identical springs with spring constant k are attached to
each other. A block of mass m is suspended from the lower
spring. What is the equivalent spring constant keq of the
system of two springs? Note the block displaces the same
amount in the two figures.
1)
keq = k
2)
keq = k/2
3)
keq = 2k
4) an unknown relation; there is not enough information.
Solution: Each spring undergoes an equal stretch x . The total stretch for both springs
is
xtotal  2x
(0.1)
The magnitude of the spring force is the same in each spring, i.e. the spring force is
transmitted uniformly through the two springs because the two springs can be considered
as one and the spring force is constant inside (a massless) spring, thus
r
F  kx
(0.2)
r
Note that this last equation implies that x  F / k . Since the total stretch adds Eq. (0.1)
becomes
r
xtotal  2x  2 F / k
(0.3)
r
Since the equivalent spring is under the same tension (same spring force), F  keq xtotal ,
where keq is the equivalent spring constant when the two springs are replaced by one.
The total displacement is therefore
r
xtotal  F / keq
(0.4)
Substitute Eq. (0.4) for the total displacement into Eq. (0.3) yielding
r
r
2 F / k  F / keq
(0.5)
The forces cancel from the above equation, and the equivalent spring constant satisfies
1 / keq  2 / k
Thus
keq  k / 2
(0.6)
Concept Question: Car-Earth Interaction
Consider a car at rest. We can conclude that the downward
gravitational pull of Earth on the car and the upward
contact force of Earth on it are equal and opposite because
1. the two forces form an interaction pair.
2. the net force on the car is zero.
3. neither of the above.
4. unsure
Answer: 2. These two forces cannot be an interaction pair
because they act on the same object. Because the car is at
rest, however, its momentum is constant (and zero).
Because net force equals the time rate of change in
momentum, the net force on the car must be zero. This
means that the two forces must be equal and opposite.
Concept Question: Tension in a Rope
You are trying to pull a rock resting on the ground with a heavy
rope (the rope has non-zero mass). Just before the rock slips and
starts to move, the magnitude of the tension in the rope is
1)
greater than the magnitude of the pulling force?
2)
equal to the magnitude of the pulling force?
3)
less than the magnitude of the pulling force?
4)
Not enough information is given to answer.
Answer 2: Suppose you are pulling the rope from the rope side. If
you make an imaginary slice anywhere inrthe rope, the forces
acting on the rope are your pulling force Fr,pull
p , and the tension
r
exerted by the left side of the rope on the right side, Ftension
. Since
r,l
the rope is not accelerating, the sum of the forces on the rope are
zero. Therefore,
r
r
tension  0
Fr,pull
p  Fr,l
From this, we can conclude that the tension exerted by the left side
of the rope on the right side is equal in magnitude but pointing in
the opposite direction from your pulling force.
r tension r pull
Fr,l  Fr, p .
Note that this is not a Third Law action reaction pair. Since the
above argument is independent of the location of the imaginary
slice, we can conclude that the magnitude of the tension in the rope
is equal to the magnitude of the pulling force.
W03D2 Concept Questions Solutions
Consider a person standing in an elevator that is
accelerating upward. The upward normal force
N exerted by the elevator floor on the person is
1. larger than
2. identical to
3. smaller than
the downward force of gravity on the person.
Answer: The normal force on the person is
greater than the gravitational force on the person
because the normal force must also accelerate
the person, as well as oppose the gravitational
force. Thus
N - mg = ma
implies that
N = m(g+a).
A force sensor on a cart is attached via a string
to a hanging weight. The cart is initially held.
When the cart is allowed to move does the
tension in the string
1. increase?
2. stay the same?
3. decrease?
4. cannot determine. Need more information
about friction acting on the system.
Answer: The tension must decrease. This is
analogous to an elevator accelerating downward,
where the normal force is less than the
gravitational force (since mg-N =ma). Here mgT=ma.
W04_D2 Concept Questions Solutions
A 1-kg rock is suspended by a massless string from
one end of a 1-m measuring stick. What is the weight
of the measuring stick if it is balanced by a support
force at the 0.25-m mark?
1)
2)
3)
4)
5)
6)
0.25 kg
0.5 kg
1.0 kg
2.0 kg
4.0 kg
impossible to determine
Answer: 3. Because the stick is a uniform,
symmetric body, we can consider all its weight
as being concentrated at the center of mass at
the 0.5-m mark. Therefore the point of support
lies midway between the two weights, and the
system is balanced only if the total weight on the
right is also 1 kg.
You are using a wrench and trying to tighten a
nut. Which of the arrangements is most effective
in tightening the nut?
Answer: 2. (Please note that 1 and 4 have the same
torque). To increase a torque, you can increase either
the applied force or the moment arm. Here the force
is the same in all four situations, and so this question
boils down to comparing moment arms.
A box, with its center-of-mass off-center as
indicated by the dot, is placed on an inclined
plane. In which of the four orientations
shown, if any, does the box tip over?
Answer: 3. In order to tip over, the box must
pivot about its bottom left corner. Only in
case 3 does the torque about this pivot (due
to gravity) rotate the box in such a way that
it tips over.
W5D1 Concept Questions Solutions
A sports car drives along the coastal highway at
a constant speed. The acceleration of the car
is
1. zero
2. sometimes zero
3. never zero
4. constant
Answer 2: Whenever the car is moving in a
straight line at constant speed the
acceleration is zero. On the photo this may
occur in a few places.
As the object speeds up along the circular path in a
counterclockwise direction, shown below, its acceleration
points:
1. toward the center of the circular path.
2. in a direction tangential to the circular path.
3. outward.
4. none of the above.
Answer 4: The object always has a component of the
acceleration pointing inward. When it is speeding up, it has
a component of the tangential acceleration in the direction
of motion (counterclockwise). The vector sum of these two
components points somewhere between the arrow 1 and 2.
An object moves counter-clockwise along the circular path
shown below. As it moves along the path its acceleration
vector continuously points toward point S. The object
1. speeds up at P, Q, and R.
2. slows down at P, Q, and R.
3. speeds up at P and slows down at R.
4. slows down at P and speeds up at R.
5. speeds up at Q.
6. slows down at Q.
7. No object can execute such a motion.
Answer 3: A the point P the acceleration has a
positive tangential component so it is speeding
up. A the point S the acceleration has a zero
tangential component so it is moving at a constant
speed. At the Point R the acceleration has a
negative tangential component so it is slowing
down.
You are a passenger in a racecar approaching a turn
after a straight-away. As the car turns left on the
circular arc at constant speed, you are pressed against
the car door. Which of the following is true during the
turn (assume the car doesn't slip on the roadway)?
1. A force pushes you away from the door.
2. A force pushes you against the door.
3. There is no force that pushes you against the door.
4. The frictional force of the between you and the seat
pushes you against the door.
5. There is no force acting on you.
6. You cannot analyze this situation in terms of the
forces on you since you are accelerating.
7. Two of the above.
8. None of the above.
Answer: 1
The force acting on you pushing you away from the door in
this example is the normal component of the contact force
between you and the door and pushing toward the center of
the circle. Although you may feel a tendency to move
outward (continue in your linear motion), there is no force
pushing you outward.
W05D2 Concept Questions
Which of the following types of forces can produce a
centripetal acceleration?
1. Normal force
2. Frictional force
3. Tension force
4. Gravitational force
5. All of the above
Answer: 5. All of the above forces can produce a
centripetal acceleration.
The "barrel of fun" must spin at a certain minimum
angular speed in order for a rider of mass m1 to stick
to the wall. Does this minimum angular speed change
for a rider of mass m2 > m1?
1. yes
2. no
3. Not enough information to determine.
Answer: 2. The normal component of the contact
force pushes the rider toward the center of the orbit
and is equal to mr2, where  is the angular velocity
and r is the radius of the barrel. The vertical
component of the contact force (static friction)
supports the rider and hence is equal to fs = mg . If
the barrel spins at a minimum angular speed then the
static friction has its maximum value and is equal to
(fs)max = sN = mr2, hence proportional to the mass.
Thus mg = mr2 and  = √g/r independent of the
mass.
A stone attached to a string is whirled in a vertical
plane. Let T1, T2, T3, and T4 be the tensions at
locations 1, 2, 3, and 4 required for the stone to have
a given speed v0 at these four locations.
1. T3 > T2 >T1 = T4
2. T1 = T2 = T3 = T4
3. T1 > T2 = T4 > T3
4. none of the above
Answer: 3. When the string is in position 2 and 4, the
tension points radially inward and is equal to T2 = T4
= mv02/r. When the string is in position 3 the tension
and the gravitational force points downward
(inward), hence T3 + mg = mv02/r. Thus T3 = -mg +
mv02/r < T2 = T4. When the string is in position 1, the
tension points upward and the gravitational force
points downward. Thus T1 - mg = mv02/r. Hence T1 =
mg + mv02/r > T2 = T4 > T3.
A puck of inertia M is moving in a circle at uniform
speed on a frictionless table as shown above. It is
held by a string which holds a suspended bob, also of
inertia M, at rest below the table. Half of the length
of the string is above the tabletop and half below.
What is the centripetal acceleration of the moving
puck?
1. less than g
2. g
3. greater than g
4. zero
5. insufficient information
Answer: 2. Since the suspended bob is static, the
tension in string is equal to the gravitational force
acting on the bob, T = Mg. The puck is accelerating
inward due to the tensile force of the string acting on
the puck, hence T = ma . Thus Mg = Ma , or g = a .