Download 1 - Net Start Class

Physics 12 WK DCA Review Key
Forces
1. Construct free-body diagrams for the following physical situations.
a. A ball is dropped from rest from the top
of a building. Assume negligible air
resistance.
b. After being thrown, a football is moving
upwards and rightwards towards the peak
of its trajectory. Assume negligible air
resistance.
c. After reaching a terminal velocity, a
falling skydiver then opens up the
parachute.
In the absence of Fair, the only force acting
upon the ball is gravity. It is a projectile. In the absence of Fair, the only force acting
upon the ball is gravity. It is a projectile.
Note that an upwards moving object does When the terminal velocity was reached, Fair
= Fgrav. Then the parachute was opened,
not need an upwards force. Only an
making Fair even greater than before. This
upwards accelerating object requires an
is represented by the larger arrow.
upwards force.
d. An air track glider (frictionless) is gliding e. A car is skidding to a stop while traveling
to the right at constant velocity.
to the right.
There is no rightwards force. A rightwards There is no rightwards force. A rightwards
force would only be required if there is a
force would only be required if there is a
rightwards acceleration. If the glider is
rightwards acceleration. If the car is
gliding; there is no mention of it being
skidding (wheels are locked), friction acts in
pushed or pulled (Fapp) and if there is a
a direction opposite its motion to slow it
constant velocity, there must be balanced
down.
forces.
2. Which of the following are always true of an object that is at equilibrium (balanced forces)? Include all that apply.
a.
All the forces acting upon the object are equal.
b.
The object is at rest.
c.
The object is moving and moving with a constant velocity.
d.
The object has an acceleration of zero.
e.
There is no change in the object's velocity.
f.
The sum of all the forces is 0 N.
g.
All the forces acting upon an object are balanced.
Answer: DEFG
These (DEFG) statements are always true. Statements ABC might be true but are not always true.
3. Which of the following are never true of an object that is at equilibrium (balanced forces)? Include all that apply.
a.
The object is accelerating.
b.
The object is at rest.
c.
The object is moving in a circle at constant speed.
d.
All the forces acting upon the object are equal.
e.
The object is in free-fall.
f.
The object is falling and has reached its terminal velocity.
g.
There is a net force acting upon the object.
h.
The object is moving and moving with a constant velocity.
i.
The object is at rest.
Answer: ACEG
An object that is at equilibrium can never be accelerating; its acceleration MUST be 0 m/s/s. Thus, A is an answer; and because an
object in free-fall (E) and an object moving in a circle (C) are also accelerating, they must be counted as answers as well. (NOTE: an
object moving in a circle is changing its direction and as such has an acceleration.) If there is a net force (G), then by definition the
object is not at equilibrium. Choices B, D, F and H could be true of an object at equilibrium (though none of them are always true).
4. Big Bubba has a mass of 100 kg on the earth. What is Big Bubba's mass on the moon where the force of gravity is approximately
1/6-th that of Earth's? ________ Explain or show your work.
Answer: 100 kg
Mass is the amount of matter present in the object and is independent of the weight of the object. The weight or force of gravity
acting upon an object depends upon the mass and the acceleration of gravity (F grav = m•g). Because g is different on different
planets and locations in the universe, the force of gravity or weight of an object will not always be the same. If Big Bubba goes to
the moon, he will weigh less due to the decreased acceleration of gravity. However, he will still look the same, still have the same
amount of matter, and still have a mass of 100 kg. KNOW the distinction between mass and weight.
5. How much net force is required to keep a 5-kg object moving rightward with a constant velocity of 2 m/s? ________ Explain or
show your work.
Answer: 0 N
If the velocity is constant, then the acceleration is 0 m/s/s and the net force is zero. A net force is only required in order to
accelerate an object. Yes, this means that an object can be moving to the right and NOT have a rightward's net force.
6. TRUE or FALSE: For an object resting upon a non-accelerating surface, the normal force is equal to the force of gravity.
Answer: FALSE
In this instance, the normal force could be equal to the force of gravity. But all that we can conclusively know is that the all the
vertical forces sum up to 0 N. If the object is upon an incline, then the normal force will not be equal to the force of gravity. Or if
there is another force with an upward or downward component, then the normal force is not equal to the force of gravity.
7. Which one(s) of the following force diagrams depict an object moving to the right with constant speed? Write all possible answers
(if any) in the blank: ________________
Explain your answer(s) to this question.
Answer: AC
Only A and C could have a constant speed. Both B and D would have an acceleration that is directed rightwards. Do not confuse
acceleration, motion, and force. A rightwards acceleration requires that there be a rightwards net force, On the other hand, a
rightwards motion could have no net force, a leftward net force or a rightwards net force; it depends upon whether the object is
slowing down, speeding up or maintaining a constant speed. Only A and C show a balance of forces, so only A and C are moving with
constant speed.
Use the approximation that g=~10 m/s2 to fill in the blanks in the following diagrams.
8.
9.
Fgrav = m•g = ~800 N
Since Fgrav = m•g, m can be calculated
to be ~70 kg (m=Fgrav/g).
Fnet = ma = (80 kg)•(2.0 m/s/s)
Fnet = 160 N, down
Since ay = 0 m/s/s, Fnorm must equal
Fgrav; so Fnorm = 700 N.
The Fgrav (down) and the Fair (up) must
add up to 160 N, down. Thus, Fair must
be smaller than Fgrav by 160 N.
Fnet x = m•ax = (70 kg)•(5.0 m/s/s)
Fair = 640 N
(Note that the ·Fx direction is always
the same as the ax direction.)
Fnet x = 350 N, left
With Fapp being the only horizontal force,
its value must be 350 N - the same as
·Fx.
10. A 945-kg car traveling rightward at 22.6 m/s slams on the brakes and skids to a stop (with locked wheels). If the coefficient of
friction between tires and road is 0.972, determine the distance required to stop.
Answer: 26.8 m
Like most problems, this problem begins with a free-body diagram (as shown at right). Note that there is no
rightwards applied force (a common mistake). Note also that the force of friction is the only force responsible for
the acceleration (deceleration) of the car. The Ffrict value is the net force. So determining the acceleration involves
finding Fgrav (Fgrav=m•g = 945•9.8 = 9261 N), Fnorm = Fgrav= 9261 N, and Ffrict (Ffrict = µ•Fnorm = 0.972 • 9261 N =
9002 N). With the Fnet x =Ffric= 9002 N, the acceleration can be calculated (ax = Fnet x/m = 9002 N/945 kg = 9.53
m/s/s.) This is a leftwards acceleration; ax will be assigned the numerical value of -9.53 m/s/s in the next part of
this problem.
Now in the kinematic part of this problem, the distance must be found using the known information (vf = 0 m/s, vi = 22.6 m/s and a
= 9.53 m/s/s). Use the equation:
vf2 = vi2 + 2 a d
(0 m/s)2 = (22.6 m/s)2 + 2•(-9.53 m/s/s)•d
d = [(22.6 m/s)2] / [2•(9.53 m/s/s)] = 26.8 m
Projectiles
11. Which of the following statements are true of projectiles? List all that apply.
a.
A projectile is a free-falling object.
b.
A projectile experiences negligible or no air resistance.
c.
A projectile must be moving in the downward direction.
d.
A projectile must be accelerating in the downward direction.
e.
A projectile does not have to have horizontal motion.
f.
A projectile could begin its projectile motion with a downward velocity.
g.
A projectile does not need to be "falling."
Answer: ABDEF and possibly G
a. TRUE - Free-falling objects, like projectiles, are objects upon which the only significant force is gravity.
b. TRUE - The only force on a projectile is gravity; air resistance must not be present or must not have an influence upon the motion
of the projectile.
c. FALSE - Projectiles can be moving either upward or downward or at an angle to the vertical. They must however be accelerating
downward, consistent with gravity's effect on an object.
d. TRUE - The force of gravity acts directly downwards upon an object, causing a downward acceleration. Any projectile must be
accelerating downwards regardless of other features of its motion.
e. TRUE - A projectile could be moving strictly in a vertical direction with no horizontal motion. A ball thrown straight up in the air
would be such a case.
f. TRUE - There is no rule about which direction a projectile must be moving at the instant it is projected. It could begin its motion
with a initial downward velocity.
g. TRUE - The word "falling" can mean different things to different people. If "falling" involves moving in the downward direction at
all instants in time, then a projectile does not need to be "falling." To many, "falling" means being pulled downward by gravity's
force. In this case, a projectile must be "falling."
12. Which of the following statements are true of the time of flight for a projectile? List all that apply.
a.
The time that a projectile is in the air is dependent upon the horizontal component of the initial velocity.
b.
The time that a projectile is in the air is dependent upon the vertical component of the initial velocity.
c.
For a projectile which lands at the same height that it is projected from, the time to rise to the peak is equal to
the time to fall from its peak to the original height.
d.
For the same upward launch angles, projectiles will stay in the air longer if the initial velocity is increased.
e.
Assume that a kicked ball in football is a projectile. If the ball takes 3 seconds to rise to the peak of its
trajectory, then it will take 6 seconds to fall from the peak of its trajectory to the ground.
Answer: BCD
a. FALSE - The time for a projectile to rise vertically to its peak (and subsequently fall back to the ground) is dependent upon the
initial vertical velocity. Alteration in the horizontal velocity will only cause the projectile to have a greater horizontal displacement
(x).
b. TRUE - Absolutely true. Projectiles with a greater vertical component of initial velocity will be in the air for longer amount of times
(assuming that the direction of viy is upward). An alteration in the viy value will alter the time of flight of the projectile, regardless of
the direction of viy.
c. TRUE - For projectiles launched at upward angles and landing at the original height, the time to the rise to the peak equals the
time to fall from the peak. If it takes 3 seconds to rise upward, it will take 3 seconds to fall.
d. TRUE - For a constant launch angle, an increase in the initial velocity (vi) will increase the vertical velocity (viy). This results in an
increased time for the projectile to decelerate to 0 m/s as it rises towards its peak. So the projectile takes longer to get to the peak,
longer to fall from the peak and overall is in the air for a longer time.
e. FALSE - Close, but very false. If it takes 3 seconds to rise to the peak, then it takes 3 seconds to fall from the peak; The 6
seconds is the total time of flight of the projectile.
13. Vector A is directed northward and vector B is directed eastward. Which of the following vector addition diagrams best represent
the addition of vectors A and B and the subsequent resultant?
Answer: E
If the head-to-tail method (sometimes referred to as the triangle method) is being used, then the tail of B should be drawn starting
at the head of A. Both diagrams D and E show this. Then the resultant should be drawn from the tail of A to the head of B (which is
not shown in diagram D). There is also a parallelogram method for adding vectors. In this method, the tails of the two vectors are
placed together (as in diagrams A and B). Then a parallelogram should be drawn with the two vectors forming the adjacent sides of
the parallelogram. The resultant is drawn from the tails of the vectors to the opposite corner of the parallelogram. This is not done
correctly on diagrams A and B. Diagrams C and F do not resemble any (accurate) vector addition method known to humankind.
14. When adding vector B to vector A geometrically (or graphically) using the head to tail method, the resultant is drawn from ____
to the ____.
a. head of A, tail of B
b. tail of A, head of B
c. head of B, tail of A
d. tail of B, head of A
Answer: B
Adding vector B to vector A is equivalent to saying A + B. In such an instance, A is drawn first, then B is drawn with its tail at the
head of A. Finally, the resultant is drawn from the tail of the first vector (A) to the head of the last vector (B).
15. The velocity vector of a projectile with a vertical velocity of 25.0 m/s and a horizontal velocity of 18.0 m/s is ___ m/s.
a. 7.00
b. 21.5
c. 30.8
d. 35.8
e. 43.0
f. 54.2
g. 949
h. None of these
Answer: C
These two vectors are directed at right angles to each other. In such instances, the vector sum can be determined using the
Pythagorean Theorem. The resultant (R) is equal to the square root of the sum of the squares of the two vectors being added. That
is, V2 = Vx2 + Vy2, thus V2 = (25.0 m/s)2 + (18.0 m/s)2
V = 30.805 m/s = ~30.8 m/s
16. An object is undergoing free fall motion. As it falls, the object's ____.
a. speed increases
b. acceleration increases
c. both of these
d. none of these
Answer: A
As an object free-falls, its velocity (and also its speed) changes by approximately 9.8 m/s every second. This means that the
acceleration is a constant value of 9.8 m/s/s. An object has a changing speed (or velocity) and a constant acceleration if the speed
changes by the same amount (a "constant amount") in each consecutive second of its motion.
17. A football is kicked into the air at an angle of 45 degrees with the horizontal. At the very top of the ball's path, its velocity is
_______.
a. entirely vertical
b. entirely horizontal
c. both vertical and horizontal
d. not enough information given to know.
Answer: B
As a projectile rises towards its peak, its horizontal velocity remains constant while its vertical velocity decreases. At its peak, its
vertical velocity becomes 0 m/s. At this instant in time, the velocity is entirely horizontal; there is no vertical component to the
velocity.
18. A football is kicked into the air at an angle of 45 degrees with the horizontal. At the very top of the ball's path, its acceleration is
_______. (Neglect the effects of air resistance.)
a. entirely vertical
b. entirely horizontal
c. both vertical and horizontal
d. not enough information given to know.
Answer: A
As a projectile rises towards its peak, its horizontal velocity remains constant while its vertical velocity decreases. This is to say that
the acceleration of the object is vertical, not horizontal. At the peak and everywhere throughout the trajectory, there is a vertical
(downward) acceleration. In fact, a projectile is an object upon which the only force is gravity. This force causes an acceleration
which is in the same direction as the force - downward.
19. A football is kicked into the air at an angle of 45 degrees with the horizontal. At the very top of the ball's path, the net force
acting upon it is _______. (Neglect the effects of air resistance.)
a. entirely vertical
b. entirely horizontal
c. both vertical and horizontal
d. not enough information given to know.
Answer: A
A projectile is an object upon which the only force is gravity. Since no other forces act upon the object, the net force would be
downward.
20. At what point in its path is the horizontal component of the velocity (v x) of a projectile the smallest?
a. The instant it is thrown.
b. Halfway to the top.
d. As it nears the top.
e. It is the same throughout the path.
c. At the top.
Answer: E
As a projectile rises towards its peak, its horizontal velocity remains constant while its vertical velocity decreases. This is to say that
the acceleration of the object is vertical, not horizontal. Having a constant horizontal velocity, there is no point along the trajectory
where the vx value is smaller than at other points.
21. At what point in its path is the vertical component of the velocity (v y) of a projectile the smallest?
a. The instant it is thrown.
b. Halfway to the top.
d. As it nears the top.
e. It is the same throughout the path.
c. At the top.
Answer: C
As a projectile rises towards its peak, its horizontal velocity remains constant while its vertical velocity decreases. During the upward
portion of its trajectory, the vy continuously decreases until it becomes 0 m/s at the peak. Thus, the v y is as small as it will ever be
when it is at the peak of the trajectory.
22. Roll a bowling ball off the edge of a table. As it falls, its horizontal component of velocity ___.
a. decreases
b. remains constant
c. increases
Answer: B
Once the ball leaves the table's edge, it becomes a projectile. As it falls, its horizontal velocity remains constant while its vertical
velocity decreases. This is to say that the acceleration of the object is vertical, not horizontal. In fact, a projectile is an object upon
which the only force is gravity. This force causes an acceleration which is in the same direction as the force - downward.
23. A bullet is fired horizontally and hits the ground in 0.5 seconds. If it had been fired with twice the speed in the same direction, it
would have hit the ground in ____. (Assume no air resistance.)
a. less than 0.5 s.
b. more than 0.5 s.
c. 0.5 s.
Answer: C
Once the bullet leaves the muzzle, it becomes a projectile (assuming no air resistance). As it falls, its horizontal velocity remains
constant while its vertical velocity decreases. The force of gravity acts upon the bullet to cause its downward acceleration. The
motion of the bullet in the downward direction is independent of the motion in the horizontal direction. That is to say, any alteration
in a horizontal aspect of its motion will not affect the motion in the vertical direction. The time to fall vertically to the ground is not
effected by the horizontal speed of the projectile. It would still take 0.5 seconds to fall to the ground from this height regardless of
the horizontal speed.
24. A projectile is launched at an angle of 15 degrees above the horizontal and lands down range. For the same speed, what other
projection angle would produce the same downrange distance?
a. 30 degrees.
b. 45 degrees.
c. 50 degrees.
d. 75 degrees
e. 90 degrees.
Answer: D
For projectiles launched at angles, a launch angle of 45 degrees will provide the largest horizontal displacement. Any two launch
angles which are separated from 45 degrees by the same amount (for example, 40 degrees and 50 degrees, 30 degrees and 60
degrees and 15 degrees and 75 degrees) will provide the same horizontal displacement.
25. Two projectiles are fired at equal speeds but different angles. One is fired at angle of 30 degrees and the other at 60 degrees.
The projectile to hit the ground first will be the one fired at (neglect air resistance) ____.
a. 30 degrees
b. 60 degrees
c. both hit at the same time
Answer: A
For projectiles launched at angles, a launch angle of 45 degrees will provide the largest horizontal displacement. Launch angles
greater than 45 degrees result in large vertical components of velocity; these stay in the air longer and rise to higher heights.
Launch angles less than 45 degrees result in small vertical components of velocity; these do not rise as high and end up falling to the
ground in shorter times.
26. Consider the trajectory diagram shown below for a horizontally launched projectile. On the diagram, draw vector arrows
representing the vx and vy velocity components during the course of the motion. The length of the arrows should represent the
magnitude of the velocity components. Label each component. (Note that the velocity components are already shown for the first
position.)
Answer: See diagram above.
A projectile has a downward acceleration. There is no horizontal acceleration; rather, the horizontal velocity remains constant. Thus,
the vx vector is of constant length (i.e., magnitude) throughout the trajectory. The downward acceleration means that the v y vector
will be changing. As the object falls, it speeds up. Thus, the vy vector increases its length (i.e., magnitude) throughout the trajectory.
This is shown in the diagram above.
27. Consider the trajectory diagram shown below for a non-horizontally launched projectile. On the diagram, draw vector arrows
representing the vx and vy velocity components during the course of the motion. The length of the arrows should represent the
magnitude of the velocity components. Label each component. (Note that the velocity components are already shown for the first
position.)
Answer: See diagram above.
A projectile has a constant horizontal velocity and a changing vertical velocity. The changing vertical component of velocity is
consistent with the fact that there is a vertical acceleration. As the projectile rises towards the peak of its trajectory, the vertical
velocity decreases until it reaches 0 m/s at the very peak of the trajectory. As the projectile rises towards the peak of its trajectory,
the vertical velocity increases. The vertical velocity upon falling is of the same magnitude and the opposite direction as any
corresponding location of the same height during the rising motion. The horizontal velocity remains constant throughout the
trajectory. These principles are shown in the diagram above.
28. The launch velocity and angle is given for three different projectiles. Use trigonometric functions to resolve the velocity vectors
into horizontal and vertical velocity components. Then use kinematic equations to determine the time that the projectile is in the air,
the height to which it travels (when it is at its peak), and the horizontal distance that it travels.
29. If a projectile is launched horizontally with a speed of 12.0 m/s from the top of a 24.6-meter high building. Determine the
horizontal displacement of the projectile.
Answer: x = 27.0 m
It is strongly recommended that you begin by listing known values for each of the variables in the kinematic equations. It is helpful
to organize the information into two columns - a column of known horizontal information and a column of known vertical information.
Horizontal Motion
d = ???
vx = 12.0 m/s
a = 0 m/s/s (true for all projectiles)
Vertical Motion
d = -24.6 m (- means moving down)
vi = 0.0 m/s (its launched horizontally)
a = -9.8 m/s/s (true for all projectiles)
ty = ty = 2.25 seconds
d = v i*t + 1/2at2 , solve for ty = 2.25 sec
d = v*t = 27.0 m
For a horizontal projectile ty = tx
31. A projectile is launched horizontally from the top of a 45.2-meter high cliff and lands a distance of 17.6 meters from the base of
the cliff. Determine the magnitude of the launch velocity.
Answer: 5.79 m/s
The best means of starting this problem is to list the known values for each of the variables in the kinematic equations. It is helpful
to organize the information into two columns - a column of known horizontal information and a column of known vertical information.
Horizontal Motion
d = 17.6 m (the distance horizontally from cliff base)
Vertical Motion
d = -45.2 m (it falls down from the cliff to the ground)
vx = ??? m/s
vi = 0 m/s (it is horizontally launched)
a = 0 m/s/s (true for all projectiles)
a = -9.8 m/s/s (true for all projectiles)
v = d/t = 5.79 m/s
d = vit + 1/2at2 , solve ty = 3.0372 s =tx