Physics Final
STUDY-GUIDE PROBLEMS FOR THE FINAL
I’ve tried to work these the best I can --Animations and Spreadsheets Help Me a Lot.
I’ll be incorporating these better in the future …
True / False Review
True
1. The rate at which velocity changes with time is called acceleration.
a
v change in velocity

t
change in time
2. The SI unit of acceleration is meters per second.
False
a
3. When a car rounds a corner at a constant speed, its acceleration is zero.
4. A ball is thrown into the air. At the highest point of its path, the ball has
zero velocity and zero acceleration.
5. As a ball falls freely, the distance it falls each second is the same.
ms m
 2
s
s
False
Anytime velocity or direction
changes, acceleration must be
changing.
False
For the same reason as #4:
gravity is still acting on the ball.
False
The distance increases
(remember the feather problem).
6. The amount of matter in an object is called the weight of the object.
False
The amount of matter is the
mass. WEIGHT is a force:
Fg  mg
7. The force due to gravity acting on an object is called the mass of the object.
8. The SI unit of force is called the kilogram.
False
See #6.
False
Force is in NEWTONS:
1N 
1kg m
s2
False
9. If a hockey puck slides on a perfectly frictionless surface, it will eventually
Newton’s Laws say an object
slow down because of its inertia.
keeps going unless something
acts on the object.
10. Inertia is the resistance any material object has to a change in its state of
motion.
True
11. The combination of all the forces that act on an object is called the net force.
True
False
12. The acceleration of an object is inversely proportional to the net force acting
Fnet  ma
on the object.
If Force goes up, Acceleration goes up.
If Force goes down, Acceleration goes down.
Therefore, F and a are DIRECTLY
proportional.
13. Air resistance is caused by friction between the air and an object moving
through the air.
True
False
14. The speed of an object dropped in air will continue to increase without limit When the upward force of air
until it strikes the ground.
friction equals weight, you stop
accelerating.
15. When one object exerts a force on another object, the second object always
exerts a force back on the first object.
True
16. A rocketship is pushed forward by gases that are forced out the back of the
ship.
17. In order to make a cart move forward, a horse must pull
harder on the cart than the cart pulls on the horse.
True
Almost True But False
If the horse was on ice, “pulling harder” couldn’t
happen. It’s the ground (and friction) allowing
pulling. The horse pushes on the ground, and the
ground pushes back to move the horse and cart.
18. If a bicycle and a parked car have a head-on collision, the force of impact is
False
greater on the bicycle.
Newton’s 3rd Law –
EQUAL BUT OPPOSITE FORCES.
19. A quantity that has both magnitude and direction is called a scalar.
20. A single vector can be replaced by two vectors in the X and Y directions.
These X and Y vectors are called the resultant of the original vector.
False
A scalar is just a number: 6.
A vector is magnitude AND
direction.
False
Just the opposite – the two
vectors can be replaced by the
resultant vector.
21. When a woman pushes a lawnmower along the handle, she pushes down as
well as forward.
22. Mass is a vector quantity.
23. Wind velocity can be represented as a vector quantity.
True
False
Mass is just a number (no
direction).
True
Weathermen report “wind is 10
mph out of the north” --- both
magnitude and direction.
INCLINED PLANE:
F
F
g
A Quick Overview
Gravity is trying to
pull straight down,
with a force:
cos

Fg  mg

Ff


Fg

Fg  mg
F
F
g
The incline doesn't
allow the full affect of
gravity.
The part of gravity
pulling down the
incline is:
F  Fg cos  
There's friction on the
plane, opposing the
slide with force:
Ff   Fg
Net force on the plane
is:
Fnet  F  Ff
cos


FIRST THING TO DO …
CALCULATE FORCE VECTORS
Fg is calculated first
Fg  mg
F  Fg cos  
FN  Fg sin  
Ff   Fg
VECTOR ADDITION
Vectors have both Magnitude and Direction (scalars are just magnitude).
ORIGINAL VECTOR
20° East of North
VECTOR COMPONENTS
You must make a triangle by dropping the
vector tail perpendicular to the x-axis.
20°
E of N
vy  v sin  
70°
vx  v cos  
#1
What is the magnitude of the sum of the
horizontal components of the following vectors?
12m @ 34˚E of N + 56m @ 78˚N of W +
91m @ 23˚ S of E
VECTOR COMPONENTS
You must make a triangle by dropping the
vector tails perpendicular to the x-axis.
VECTOR COMPONENTS
Make a Table (simplest way)
common mistake 1
using the wrong angle
common mistake 2
using the right angle, but not
including the proper sign with
sin and cos.
avoid both of these mistakes
drawing the direction vectors - and
then the angles.
ADDING VECTORS
How Are Vectors Added, Geometrically? Everything is “Tip to Tail”
Purple Vector (Tail) Added to Blue Vector (Tip)
ADDING VECTORS
How Are Vectors Added, Geometrically? Everything is “Tip to Tail”
Red Vector (Tail) Added to Purple Vector (Tip)
THE RESULTANT VECTOR
THE RESULTANT VECTOR
ADDING VECTORS
geometrically
ADDING VECTORS
With a Table
#2
Add the following vectors:
12.3m North; 45.6m East; 78.9m West; 14.7m
South.
FINDING MAGNITUDE
m2   33.2    2.4 
2
m  33.4
2
FINDING DIRECTION
tan   
opp
2.4
 .0719

adj
33.4
  tan 1 .0719   4.1
(South of West)
#3
A car travels 540km in 4.5hr. How far will it go
in 8 hrs at the same average speed?
I need d. Find an
equation with this, and see
what I’m missing.
v
d
t
d  vt  v 8
I need
v
Use actual data
from the trip.
d
t
540km
 120km / hr

4.5hr
v
d  120 8  960km
A Simpler Way?
Using Proportions
540km xkm

4.5hr 8hr
5408  4.5x
x  960km
#4
Bill’s motorcycle can accelerate at 7.05m/s2 at a
certain RPM and gear. How far, starting from
rest, will Bill travel in the first 2.50s?
REMEMBER
when there’s acceleration,
the distance between points increases.
I need d. Find an equation
with this, and see what I’m
missing.
from
rest
1
d  vit  at 2
2
  0  2.5 
 22m
1
2
 7.05 2.5
2
After 2.5 seconds, he’s
gone 22m.
#5
Chuck’s car is traveling at 65.0m/s when he
suddenly accelerates his car at 15.0m/s2 for
3.00s. How far did Chuck, and his car, travel
while he was accelerating?
I need d. Find an equation
with this, and see what I’m
missing.
1
d  vit  at 2
2
  65.0  3.0  
1
2
15 3.0
2
 263m
Each
Represents 1 Second
5
0
0
65
Distance Traveled
Constant velocity of
65 m/s for 2 seconds
130
195
260
325
Accelerate for 3 seconds, at 15 m/s2..
The Distance Traveled during this time:
d = 263 m
390
455
#6
An astronaut drops a feather from 1.2m above
the surface of the moon. If the acceleration due
to gravity is 1.62m/s2 on the moon; how long
does it take the feather to reach the ground?
I need t Find an equation
with this, and see what I’m
missing.
1
d  vit  at 2
2
Dropping an
object from rest:
vi = 0
1
d  at 2
2
t2 
2d
a
t
2d
2 1.2 
 1.2s

a
1.62
Each dot
= 1/5 of a second
#7
PROPER UNITS!
Be careful not to just drop
in the numbers!
Engineers are developing new types of guns that
might someday be used to launch satellites as if
they were bullets. One such gun can give a small
object a velocity of 3.5km/s moving it through a
distance of only 2.0cm. What is the acceleration
of the object?
vf 
3.5km 1000m 35,000m


s
1km
s
d
2cm
1m

 .02m
1 100cm
I need a. Since there are a
couple equations with a,
which one includes vi and vf.
Accelerating
from rest, vi = 0
v  v  2ad
2
f
2
i
v 2f  2ad
v 2f
350002
 30,625,000,000m / s 2
a

2d
2 .02 
#8
A box with a mass of 25.0kg is moving at a
constant velocity across a horizontal surface
because of a 75.0N force. Calculate the
coefficient of friction acting on the box.
moving east with
constant velocity
25 kg
Ffriction  75N
Fgravity  mg
I need force of friction.
There’s only one equation,
so let’s see what I need.
Ff   Fg

Ff
Fg

75 N
 0.306
245 N
  25 9.8
 245N
REMEMBER
The amount of matter in an object is the
mass; WEIGHT is the gravity applied to
mass (i.e., F = ma)
#9
A 438kg car is accelerating east at 2.55m/se. If
the coefficient of friction felt by the car is 0.500;
what is the total force acting east on the car?
Accelerating east
Fengine  3263N
THINK ABOUT THE PROBLEM
The car is accelerating east, but
encountering huge friction forces trying to
slow it down. The force exerted by the
engine to overcome this must be huge!
Ff   Fg
  0.5 4292
Fgravity  mg
  438 9.8
 4292.4N
 F  ma
Fengine  Ffriction   438 2.55  1117N
Fengine  2146  1117
Fengine  3263N
 2146N
#10
How much total force is needed to accelerate a
2.0kg block of wood at 4.0m/s2 along a rough
table, against a force of friction of 10.N?
Fforward  18N
2 kg
Ff  10 N
 F  ma
Fforward  Ffriction   2 4.0  8N
Fforward  10 N  8N
Fforward  18N
#11
What is
acceleration?
102kg
A man with a mass of 1.0 x
slides across a
frozen lake with an initial speed of 5.5m/s.
Friction slows him, and after 4.3s he comes to a
stop. How far did he slide across the lake?
I need d. Find an equation
with this, and see what I’m
missing.
a

v
t
v f  vi
t
0  5.5

4.3
1
d  vit  at 2
2
  5.5 4.3 
1
2
 a  4.3
2
 1.28m / s 2
I need
a
d   5.5 4.3 
 12m
1
2
 1.28 4.3
2
#12
A 25.0kg box is on a 20.0m long incline that is at
an angle of 37.0˚, with a coefficient of friction of
.15. What is the boxes velocity at the bottom of
the incline?
What is the NET FORCE?
The block is pulled
down the plane with
the part of gravity
parallel to the plane:
147.4N
F
f
3
6.7
5N
Fnet  F  Ff
37°
F
1
47.
4N
Fg  245N
 147.4  36.75
I need vf. Find an
equation with this, and
see what I’m missing.
F  245cos  37  147.4 N
FN  245sin  37  195.7 N
Ff   Fg  .15 245  36.75N
Fnet  ma
v2f  vi2  2ad
CALCULATE FORCE VECTORS
Fg is calculated first
Fg  mg   25 9.8  245N
The force of friction
literally says, "Not so
fast", slowing down
the block with force
36.75N.
 0  2a  20 
I need  40a
 110.65N
110.65   25 a
a  4.426m / s 2
a
v2f   40 4.426  177.04
v f  13.3m / s
#13
With what force does a 75.5kg person need to be
pushed in order to go up a 22.8˚ frictionless
incline at a constant velocity?
22.8°
F
Heading up the
frictionless ramp
Fg  740 N
2
87 N
Fg  mg   75.5 9.8  740N
F  Fg sin  22.8   740 sin  22.8  287N
#14
Tom kicks a rock horizontally off of a 20.0m
high cliff. How fast did he kick the rock if it hits
the ground 45.0m from the base of the cliff?
from
rest
1
d y  viyt  at 2
2
I need vx.
There’s only one
equation with it:
vx 
dx
45

t
t
1
d y  at 2
2
t
2d y
a

 2  20 
9.8
 2.02s
I need
t
vx 
45
 22.48m / s
2.02
Each dot = .25 seconds
#15
Nicole throws a ball at 25m/s at an angle of 60˚
above the horizontal. What was the range of the
ball?
vx  25cos  60   12.5
v y  25sin  60   21.7
I need dx.
There’s only one
equation with it:
d
vx  x
t
d x  vxt  12.5t
ttotal 

2v y
g
 2  21.7 
 4.43s
9.8
I need
t
d x  12.5 4.43  55.375m
Each dot = .50 seconds
#16
Calvin is walking down the street at 4.0km/hr. If
he has a mass of 70.kg, what is his momentum?
PROPER UNITS!
Be careful not to just drop
in the numbers!
v
p  mv
  70 1.11
 78
kg m
s
4.0km 1000m 1hr


 1.11m / s
hr
1km 3600s
#17
A 1.0x104kg
freight car is rolling along a track at
3.0m/s. Calculate the time needed for a force of
1.0x102N to stop the car.
a
F  ma
100  10,000a
1  100a
v v f  vi

t
t
03

t
3

t
I need
a
 3 
1  100  
 t 
t  300s
#18
A 3.0g bullet moving at 2.0km/s strikes an 8.0kg
wooden block that is at rest on a frictionless
table. The bullet passes through and emerges on
the other side with a speed of 5.0x102m/s. How
fast is the block moving after the collision?
mb  .003kg
mw  8kg
vbi  2000m / s
vwi  0m / s
8 kg
BEFORE COLLISION
PROPER UNITS!
Be careful not to just drop
in the numbers!
 1kg 
mb  3g 
  .003kg
1000
g


2km 1000m 2000m
vb 


1s
1km
s
mw  8kg mb  .003kg
vbf  500m / s
vwi  ?
8 kg
AFTER COLLISION
Law of
Conservation of Momentum
mbvbi  mwvwi  mbvbf  mwvwf
.003 2000  8 0  .003500  8 vwf
6  1.5  8vwf
vwf  .5625m / s
#19
After-Collision Momentum
A 125kg cart with a momentum of 1250kgm/s
east collides with a 225kg cart whose
momentum is 2250kgm/s north. The two carts
lock together. What is the velocity of the carts
after the collision?
pt  p12  p22
pt2  p12  p22
kg  m
p1  1250
s
m1  125kg
ptp
2574
t
p2=2250
if no collision:
p2=2250
 12502  22502
kg m
 2574
s
60.9
if no collision:
p1=1250
kg  m
p2  2250
m2  225kg s
After-Collision Velocity
pt  m1&2v1&2
2574  125  225 v1&2
v1&2  7.35m / s
Angle
tan   
opp 2250

 1.8
adj 1250
  tan 1 1.8  60.9
(North of East)
#20
. A 15kg ball is rolling across the floor at 2.0m/s.
A force is applied for 2.0m which increases its
velocity to 5.0m/s. Calculate the magnitude of
the force.
vf = 5 m/s
vi = 2 m/s
Additional
Force:
15 kg
F=?
15 kg
15 kg
2m
v2f  vi2  2ad
52  22  2a  2 
25  4  4a
F  ma  15a
a
21
4
I need
a
 21 
F  15    78.75 N
 4
#21
A 13.0kg box is lifted to a ledge that is 3.50m
high in 8.00s. Calculate the power generated
when moving the box.
P
W
t
W  Fd
P
Fd
t
F  mg
P
 mg  d
t
13 9.8 3.5

8
 55.7J
#22
A 24.5kg ball is rolling at 3.25m/s across a
frictionless plane. If a 10.0N force is exerted for
2.00m on the ball, what will the new velocity of
the ball be?
vf = ?
vi = 3.25 m/s
24.5 kg
I push the ball
with force
10N
24.5 kg
2m
I need vf. Find an equation
with this, and see what I’m
missing.
v2f  vi2  2ad
 3.252  2a  2 
 10.5625  4a
I need
F  ma
F
m
10

 .408m / s 2
24.5
a
a
v2f  10.5625  4 .408  12.1945
v f  12.1945  3.49m / s
24.5 kg