Transcript Physics Final
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
1. The rate at which velocity changes with time is called acceleration.
a
v
t
True
2. The SI unit of acceleration is meters per second.
False
a
m s
s m s
2 3. When a car rounds a corner at a constant speed, its acceleration is zero.
False
Anytime velocity or direction changes, acceleration must be changing.
4. A ball is thrown into the air. At the highest point of its path, the ball has zero velocity and zero acceleration.
False
For the same reason as #4: gravity is still acting on the ball.
5. As a ball falls freely, the distance it falls each second is the same.
False
The 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.
False
The amount of matter is the mass. WEIGHT is a force:
F
g
mg
False
See #6.
8. The SI unit of force is called the kilogram.
False
Force is in NEWTONS: 1
N
1
kg m s
2 9. If a hockey puck slides on a perfectly frictionless surface, it will eventually slow down because of its inertia.
False
Newton’s Laws say an object 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
12. The acceleration of an object is inversely proportional to the net force acting on the object.
F
False
ma
net
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.
True
17. In order to make a cart move forward, a horse must pull harder on the cart than the cart pulls on the horse.
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 greater on the bicycle.
False
Newton’s 3 rd Law – EQUAL BUT OPPOSITE FORCES.
19. A quantity that has both magnitude and direction is called a scalar.
False
A scalar is just a number: 6.
A vector is magnitude AND direction.
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
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.
True
22. Mass is a vector quantity.
23. Wind velocity can be represented as a vector quantity.
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:
A Quick Overview
F
F
g
cos Gravity is trying to pull straight down, with a force:
F
g
mg
The incline doesn't allow the full affect of gravity.
F
F
g
cos
F
g
mg F
f
F
g
FIRST THING TO DO … CALCULATE FORCE VECTORS
F
g is calculated first
F
g
mg F
F
g
cos
F
N
F
g
sin
F
f
F
g
F
The part of gravity pulling down the incline is:
F
g
cos There's friction on the plane, opposing the slide with force:
F
f
F
g
Net force on the plane is:
F
net
F
F
f
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 70°
v x
v
cos
v y
v
sin
#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
m
2
33.2
2.4
2
m
33.4
tan
FINDING DIRECTION
opp adj
2.4
33.4
.0719
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.
d
I need
v
v
d t
v
Use actual data from the trip.
v
d t
540
km
4.5
hr
120 /
d
960
km
A Simpler Way?
Using Proportions
540
km
4.5
hr xkm
8
hr
4.5
x x
960
km
#4
Bill’s motorcycle can accelerate at 7.05m/s rest, will Bill travel in the first 2.50s? 2 at a certain RPM and gear. How far, starting from
I need d. Find an equation with this, and see what I’m missing.
d from
rest v t i
1 2
at
2 2 22
m
2
REMEMBER
when there’s acceleration, the distance between points increases.
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/s while he was accelerating?
2 for 3.00s. How far did Chuck, and his car, travel
I need d. Find an equation with this, and see what I’m missing.
d
v t i
1 2
at
2
2 2 263
m
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/s 2.
. The Distance Traveled during this time: d = 263 m Each Represents 1 Second 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/s 2 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.
d
v t i
1 2
at
2 Dropping an object from rest:
v
i = 0
d
1 2
at
2
t
2 2
d a t
2
d a
1.62
1.2
s
Each dot = 1/5 of a second
#7
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?
I need a. Since there are a couple equations with a, which one includes v i and v f .
v
2
f
v
i
2
ad
Accelerating from rest,
v
i = 0
PROPER UNITS!
Be careful not to just drop in the numbers!
v f
3.5
km s
1000
m
1
km
35,000
m s d
2
cm
1 1
m
100
cm
.02
m
v
2
f
2
ad
a
v
2
f
2
d
35000
2 2
#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.
I need force of friction. There’s only one equation, so let’s see what I need.
F
f
F
g
F f F g
75
N
245
N
0.306
moving east with constant velocity
25 kg
F
friction
75
N F
gravity
mg
245
N
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/s e . If the coefficient of friction felt by the car is 0.500; what is the total force acting east on the car?
Accelerating east
F
engine
3263
N F
gravity
mg
4292.4
N
F
ma F
engine
F
friction
F
engine
F
engine
3263
N
1117
N
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!
F
f
F
g
2146
N
#10
How much total force is needed to accelerate a 2.0kg block of wood at 4.0m/s 2 along a rough table, against a force of friction of 10.N?
F
forward
18
N
2 kg
F
ma F
f
10
N F
forward
F
friction
8
N F
forward
10
N
8
N F
forward
18
N
#11
A man with a mass of 1.0 x 10 2 kg 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.
d
v t i
1 2
at
2 2
a
2
I need
a
What is acceleration?
a
v
t v f
v i t
4.3
1.28 / 2
d
2
2 12
m
#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
The force of friction literally says, "Not so fast", slowing down the block with force 36.75N.
F
f
36.7
5
N
37°
F
147 .4
N F
g
245
N F
g
CALCULATE FORCE VECTORS
F
g is calculated first
mg
245
N F
147.4
N F
N
F
f
F
g
195.7
N
36.75
N
I need v
f
. Find an equation with this, and see what I’m missing.
I need
a
v
2
f
v
i
2
a ad
40
a v
2
f
F
net
ma
110.65
a
177.04
F
net
F
F
f
110.65
N a
2
v
f
#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?
F
22.8°
287
N F
g
740
N
Heading up the frictionless ramp
F
g
mg
F
F
g
740
N
287
N
#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?
I need v
x
. There’s only one equation with it:
d y
from rest
1 2
at
2
d y
1 2
at
2
v x
d x t
45
t
I need
t
t
v x
45 2.02
2
d y a
9.8
2.02
s
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?
v x v y
12.5
21.7
I need d
x
. There’s only one equation with it:
v x
d x t t total
2
v y g
12.5
t
d
x
v t
x
I need
t
9.8
d
x
55.375
m
4.43
s
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?
p
mv
78
kg m s
PROPER UNITS!
Be careful not to just drop in the numbers!
v
4.0
km hr
1000
m
1
km
1
hr
3600
s
#17
A 1.0x10
4 kg freight car is rolling along a track at 3.0m/s. Calculate the time needed for a force of 1.0x10
2 N to stop the car.
F
ma
a
I need
a
1 100
a t
300
s
3
t a
v
t
v f t
v i
t
3
t
#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 5.0x10
2 m/s. How fast is the block moving after the collision?
m b
v bi
.003
kg m w
8
kg v wi
0 /
8 kg
m w
8
kg v wi
?
m b
.003
kg v bf
500 /
8 kg BEFORE COLLISION PROPER UNITS!
Be careful not to just drop in the numbers!
m b
3
g
1
kg
1000
g
.003
kg v b
2
km
1
s
1000
m
1
km
2000
m s
AFTER COLLISION Law of Conservation of Momentum
m v
b bi
m v
w wi
m v
b bf
m v
w wf
v
wf
v
wf
v
wf
#19
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?
p
1
m
1 1250
kg m s
125
kg p
t
p
t
2574
60.9
if no collision:
p
1 =1250
p
2
m
2 2250 225
kg
kg m s
After-Collision Momentum
p
t
2
p
1 2
p
2 2
p
t
p
1 2
p
2 2 1250 2 2250 2 2574
kg m s
After-Collision Velocity
p
t
m v
1&2 1&2 2574
v
1&2
v
1&2 tan
Angle
opp
adj
2250 1250 1.8
tan 1 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.
v
i = 2 m/s
15 kg
Additional Force:
F = ?
15 kg
2 m
v
2
f
v
i
2
ad
5 2 2 2 2
a
25 4 4
a F
ma
I need
15
a
a
F
15 21 4
a
21 4 78.75
N
v
f = 5 m/s
15 kg
#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
t
8 55.7
J
#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?
v
i = 3.25 m/s
24.5 kg
I push the ball with force 10N
24.5 kg
2 m
I need v
f
. Find an equation with this, and see what I’m missing.
v
2
f
v
i
3.25
2 2
ad
2
a
10.5625 4
a
I need
a
v
2
f
F
ma
a
F m
10 24.5
12.1945
v
f
12.1945
2
v
f = ?
24.5 kg