Physics Final

STUDY-GUIDE PROBLEMS FOR THE FINALI’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

1. The rate at which velocity changes with time is called acceleration.

2. The SI unit of acceleration is meters per second.

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.

v change in velocitya

t change in time

True

2

m s ma

s s

False

FalseAnytime velocity or direction changes, acceleration must be

changing.

FalseFor the same reason as #4:

gravity is still acting on the ball.

FalseThe distance increases

(remember the feather problem).

6. The amount of matter in an object is called the weight of the object.

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.

9. If a hockey puck slides on a perfectly frictionless surface, it will eventually slow down because of its inertia.

10. Inertia is the resistance any material object has to a change in its state of motion.

FalseThe amount of matter is the mass. WEIGHT is a force:

gF mg

FalseSee #6.

FalseForce is in NEWTONS:

2

11

kg mN

s

FalseNewton’s Laws say an object keeps going unless something

acts on the object.

True

11. The combination of all the forces that act on an object is called the net force.

12. The acceleration of an object is inversely proportional to the net force acting on the object.

13. Air resistance is caused by friction between the air and an object moving through the air.

14. The speed of an object dropped in air will continue to increase without limit until it strikes the ground.

15. When one object exerts a force on another object, the second object always exerts a force back on the first object.

True

False

If Force goes up, Acceleration goes up.If Force goes down, Acceleration goes down.

Therefore, F and a are DIRECTLY proportional.

netF ma

True

FalseWhen the upward force of air

friction equals weight, you stop accelerating.

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.

18. If a bicycle and a parked car have a head-on collision, the force of impact is greater on the bicycle.

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.

True

Almost True But FalseIf 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.

FalseNewton’s 3rd Law –

EQUAL BUT OPPOSITE FORCES.

FalseA scalar is just a number: 6.A vector is magnitude AND

direction.

FalseJust 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

FalseMass is just a number (no

direction).

TrueWeathermen report “wind is 10 mph out of the north” --- both

magnitude and direction.

25.0 kg

FIRST THING TO DO …CALCULATE FORCE VECTORS

Fg is calculated first

sinN gF F

gF mg

cosgF F

f gF F

The incline doesn't allow the full affect of

gravity.

The part of gravity pulling down the

incline is:

cosg

FF

cosg

FF

cosgF F

gF mg

Gravity is trying to pull straight down,

with a force:

gF mg

f

g

FF

There's friction on the plane, opposing the

slide with force:

f gF F

Net force on the plane is:

net fF F F

INCLINED PLANE: A Quick Overview

VECTOR ADDITIONVectors have both Magnitude and Direction (scalars are just magnitude).

ORIGINAL VECTOR

20° East of NorthVECTOR COMPONENTS

You must make a triangle by dropping the vector tail perpendicular to the x-axis.

70°

cosxv v

sinyv v

20°E of N

#1What 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 COMPONENTSYou must make a triangle by dropping the

vector tails perpendicular to the x-axis.

VECTOR COMPONENTSMake a Table (simplest way)

common mistake 1using the wrong angle

common mistake 2using the right angle, but not

including the proper sign with sin and cos.

avoid both of these mistakesdrawing the direction vectors - and

then the angles.

ADDING VECTORSHow Are Vectors Added, Geometrically? Everything is “Tip to

Tail”

Purple Vector (Tail) Added to Blue Vector (Tip)

ADDING VECTORSHow Are Vectors Added, Geometrically? Everything is “Tip to

Tail”

Red Vector (Tail) Added to Purple Vector (Tip)

THE RESULTANT VECTOR

THE RESULTANT VECTOR

ADDING VECTORSgeometrically

ADDING VECTORSWith a Table

tanopp

adj 2.4

33.4

.0719

1tan .0719 4.1

2 22 33.2 2.4m

33.4m

#2Add the following vectors:

12.3m North; 45.6m East; 78.9m West; 14.7m South.

FINDING MAGNITUDE FINDING DIRECTION

(South of West)

#3A 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.

I need

v

dv

t

8d vt v

dv

t

540

4.5

km

hr 120 /km hr

120 8 960d km

Use actual data from the trip.

540

4.5 8

km xkm

hr hr

540 8 4.5x

960x km

A Simpler Way?Using Proportions

21

2id v t at

210 2.5 7.05 2.5

2

22m

I need d. Find an equation with this, and see what I’m

missing.

#4Bill’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?

from rest

After 2.5 seconds, he’s gone 22m.

REMEMBERwhen there’s acceleration,

the distance between points increases.

#5Chuck’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?

2165.0 3.0 15 3.0

2

263m

21

2id v t at

I need d. Find an equation with this, and see what I’m

missing.

Accelerate for 3 seconds, at 15 m/s2.. The Distance Traveled during this time:

d = 263 m

Each Represents 1 Second

Constant velocity of 65 m/s for 2 seconds

#6An 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?

21

2id v t at

21

2d at

2 2dt

a

2dt

a 2 1.2

1.62 1.2s

Dropping an object from rest:

vi = 0

I need t Find an equation with this, and see what I’m

missing.

Each dot = 1/5 of a second

#7Engineers 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?

I need a. Since there are a couple equations with a,

which one includes vi and vf.

2 2 2f iv v ad

2 2fv ad

2

2fv

ad

235000

2 .02

Accelerating from rest, vi = 0

PROPER UNITS!Be careful not to just drop

in the numbers!

3.5 1000 35,000

1f

km m mv

s km s

2 1.02

1 100

cm md m

cm

230,625,000,000 /m s

#8A 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.

25 kg

25 9.8

245N

moving east with constant velocity

75frictionF N

gravityF mg

f gF F

f

g

F

F 75

245

N

N 0.306

I need force of friction. There’s only one equation,

so let’s see what I need.

REMEMBERThe amount of matter in an object is the mass; WEIGHT is the gravity applied to

mass (i.e., F = ma)

#9A 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?

f gF F

0.5 4292

2146NgravityF mg

438 9.8

4292.4N

F ma

engine frictionF F 438 2.55 1117N

2146 1117engineF

3263engineF N

3263engineF N

Accelerating east

THINK ABOUT THE PROBLEMThe 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!

#10How 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?

2 kg 10fF N

2 4.0 8forward frictionF F N

10 8forwardF N N

18forwardF N

F ma

18forwardF N

#11A man with a mass of 1.0 x 102kg 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?

21

2id v t at

215.5 4.3 4.3

2a

va

t

f iv v

t

0 5.5

4.3

12m

I need d. Find an equation with this, and see what I’m

missing.

I need

a

What is acceleration?

21.28 /m s

215.5 4.3 1.28 4.3

2d

#12A 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?

25.0 kg

37°

37°

25 9.8 245gF mg N

245gF N

245cos 37 147.4F N

245sin 37 195.7NF N

147.4

F

N

.15 245 36.75f gF F N

36.75f

F

N

net fF F F

147.4 36.75

110.65N

2 2 2f iv v ad

0 2 20a

40a

netF ma

110.65 25 a

24.426 /a m s

2 40 4.426 177.04fv

13.3 /fv m s

CALCULATE FORCE VECTORSFg is calculated first

The force of friction literally says, "Not so fast", slowing down the block with force

36.75N.

The block is pulled down the plane with the part of gravity

parallel to the plane: 147.4N

I need vf. Find an equation with this, and see what I’m missing.

I need

a

What is the NET FORCE?

#13With 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?

75.5 kg

gF mg 75.5 9.8 740N

22.8°

22.8

°

sin 22.8gF F 740 sin 22.8 287N

740gF N287

F

N

Heading up the frictionless ramp

#14Tom 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?

2 ydt

a 2 20

9.8 2.02s

21

2y iyd v t at

21

2yd at

xx

dv

t

45

t

45

2.02xv 22.48 /m s

I need vx. There’s only one equation with it:

Each dot = .25 seconds

I need

t

from rest

#15Nicole throws a ball at 25m/s at an angle of 60˚ above the horizontal. What was the range of the

ball?

2 ytotal

vt

g

2 21.7

9.8 4.43s

xx

dv

t

x xd v t 12.5t

12.5 4.43 55.375xd m

I need dx. There’s only one equation with it:

I need

t

25cos 60 12.5

25sin 60 21.7

x

y

v

v

Each dot = .50 seconds

#16Calvin is walking down the street at 4.0km/hr. If he has a mass of 70.kg, what is his momentum?

4.0 1000 11.11 /

1 3600

km m hrv m s

hr km s

p mv

70 1.11

78kg m

s

PROPER UNITS!Be careful not to just drop

in the numbers!

#17A 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.

va

t

f iv v

t

0 3

t

3

t

F ma

100 10,000a

1 100a

31 100

t

300t s

I need

a

#18A 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?

b bi w wi b bf w wfm v m v m v m v

.003 2000 8 0 .003 500 8 wfv

6 1.5 8 wfv

.5625 /wfv m s

13 .003

1000b

kgm g kg

g

2 1000 2000

1 1b

km m mv

s km s

Law of Conservation of MomentumPROPER UNITS!

Be careful not to just drop in the numbers!

8 kg

.003

2000 /b

bi

m kg

v m s

8

0 /w

wi

m kg

v m s

BEFORE COLLISION

8 kg

.003

500 /b

bf

m kg

v m s

8

?w

wi

m kg

v

AFTER COLLISION

#19A 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?

2 225m kg

1 125m kg1 1250

kg mp

s

2 2250kg m

ps

if no collision: p1=1250

if n

o co

llis

ion:

p2=

2250

2 2 21 2tp p p

2 21 2tp p p

2 21250 2250

2574tp 1&2 1&2tp m v

1&22574 125 225 v

1&2 7.35 /v m s

tanopp

adj

2250

1250 1.8

1tan 1.8 60.9 (North of East)

Angle

2574kg m

s

After-Collision Velocity

p 2=22

50

tp

After-Collision Momentum

60.9

#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.

F ma 15a

2 2 2f iv v ad

2 25 2 2 2a

25 4 4a

21

4a

2115 78.75

4F N

I need

a

15 kg

vi = 2 m/s

15 kg 15 kg

vf = 5 m/s

2 m

Additional Force:

F = ?

#21A 13.0kg box is lifted to a ledge that is 3.50m high in 8.00s. Calculate the power generated

when moving the box.

WP

t W Fd

F mgFd

Pt

mg dP

t

13 9.8 3.5

8

55.7J

#22A 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?

2 2 2f iv v ad

23.25 2 2a

10.5625 4a

F ma

Fa

m

2 10.5625 4 .408 12.1945fv

12.1945fv 3.49 /m s

I need vf. Find an equation with this, and see what I’m

missing.

I need

a

210.408 /

24.5m s

24.5 kg

vi = 3.25 m/s

24.5 kg 24.5 kg

vf = ?

2 m

I push the ball with force

10N