Physics Final

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



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



mg

False

See #6.

8. The SI unit of force is called the kilogram.

False

Force is in NEWTONS: 1

 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,

and

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

Net force on the plane is:

F

net



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



cos

v y



sin

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

ADDING VECTORS

geometrically

ADDING VECTORS

With a Table

Add the following vectors: 12.3m North; 45.6m East; 78.9m West; 14.7m South.

FINDING MAGNITUDE

m



33.2

  2.4



m

 33.4

tan

FINDING DIRECTION



opp adj

  2.4

 33.4

 .0719

  tan  1



.0719



 4.1



(South of West)

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



d t

v

Use actual data from the trip.



d t

 540

4.5

 120 /

d

  960

km

A Simpler Way?

Using Proportions

540

 4.5

hr xkm

 4.5

x x

 960

km

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

2   2  22

m

 

REMEMBER

when there’s acceleration, the distance between points increases.

After 2.5 seconds, he’s gone 22m.

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.



v t i

 1 2

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

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.



v t i

 1 2

2 Dropping an object from rest:

i = 0

 1 2

2  2

d a t

 2

d a

 1.62

 1.2

s

Each dot = 1/5 of a second

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

Accelerating from rest,

i = 0

PROPER UNITS!

Be careful not to just drop in the numbers!

v f

 3.5

km s

 1000

 35,000

m s d

 2

1  1

100

 .02

v

 2



 35000

 

)

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

F

forward

 18

N

2 kg



F



ma 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.



v t i

 1 2

2   2

I need

What is acceleration?

  



t v f



v i t

 4.3

  1.28 / 2



   

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?

 195.7

N

 36.75

N

I need v

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

I need

v

2  

a ad

 40

a v



F

net



ma

110.65



a





 177.04

F

net



  





 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

. There’s only one equation with it:

d y



from rest

 1 2

d y

 1 2

v x



d x t

 45

I need



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

. There’s only one equation with it:

v x



d x t t total

 2

v y g

 12.5

t



d



v t

I need

9.8

d



  

 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!

 4.0

km hr

 1000

 1

3600



#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

1 100

a t

 300

s

 3

t a

 





v f t



v i

 

 3

#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

  1

1000

   .003

kg v b

 2

 1000

 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

 

v



#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?

m

1   1250

kg m s

 125

kg p

p

 2574



60.9

 if no collision:

1 =1250

m

2   2250 225

kg

kg m s

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.

i = 2 m/s

15 kg

Additional Force:

F = ?

15 kg

2 m

v

5 2  2 2  2

a

25 4 4

a F



ma

I need

 15

a

 15 21 4

 21 4  78.75

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.



W t

W



Fd



Fd t

F



mg

 

   

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?

i = 3.25 m/s

24.5 kg

I push the ball with force 10N

24.5 kg

2 m

I need v

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

v

 3.25

2 2

 2

a

v

 12.1945

 2

f = ?

24.5 kg