mid term physics review $100 $400 $300 $200 $400 $200 $100$100 $400 $200$200 $500$500 $300 $200 $500...

Mid Term Mid Term Physics ReviewPhysics Review

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Gravity and Graphs

Momentum and Acceleration

Motion

Forces

Energy and Energy and PowerPower

Gravity and Gravity and GraphsGraphs

Momentum Momentum and and

AccelerationAccelerationMotionMotion ForcesForces Energy and Energy and

PowerPower

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Gravity and GraphsGravity and Graphs - $100- $100

Gravity isA. Vector, units kgB. Vector, units NC. Scalar, units kgD. Scalar, units N


The graph represents the relationship between the displacement of an object and its time travel along a straight line. What is the average speed of the object during the first 4.0 seconds?

A. 0 m/s B. 2 m/s C. 8 m/s D. 4 m/s


If the Earth were twice as massive as it is now, then the gravitational force between it and the Sun would be

A. the same B. twice as great C. half as great D. four times as great

Gravity and GraphsGravity and Graphs - $400- $400What is the gravitational force of attraction between a planet and a 17kilogram mass that is falling freely toward the surface of the planet at 8.8 meters per second squared?

a.166.8 Nb.149.6 Nc.658.24 Nd.8.8 N

Gravity and GraphsGravity and Graphs- $500- $500Gravitational force of attraction F

exists between two point masses A and B when they are separated by a fixed distance. After mass A is tripled and mass B is halved, the gravitational attraction between the two masses is

A.FB.6FC.3/2 FD.9F

Momentum and Acceleration - $100- $100

A rocket with a mass of 1,000 kilograms is moving at a speed of 20 meters per second. The magnitude of the momentum is

A. 50 kg.m/s B. 200 kg.m/s C. 20,000 kg.m/s D. 40,000 kg.m/s

Momentum and Acceleration - $200 - $200The velocity of a car changes from 60. meters per second north to 45 meters per second north in 5.0 seconds. The magnitude of the car's acceleration is

a) 9.8 m/s2 b) 15 m/s2 c) 3.0 m/s2 d) 53 m/s2

Momentum and Acceleration - $300 - $300A 0.60-kilogram softball initially at rest is hit with a bat. The ball is in contact with the bat for 0.20 second and leaves the bat with a speed of 25 meters per second. What is the magnitude of the average force exerted by the ball on the bat?

a) 15 Nb)75 Nc)125 Nd)7500 N

Momentum and Acceleration - $400 - $400

A car moving at a speed of 8.0 meters per second enters a highway and accelerates at 3.0 meters per second2. How fast will the car be moving after it has accelerated for 23 meters?

a: 69 m/s b: 14.2 m/sc: 24 m/s d: 4 m/s

Momentum and Acceleration - $500 - $500A 20-kilogram cart traveling east with a speed of 6 meters per second collides with a 30-kilogram cart traveling west. If both carts come to rest after the collision, what was the speed of the westbound cart before the collision?

a)0 m/s b)9 m/s c)3 m/s d)4 m/s

Motion - $100 - $100

A runner completed the 100.-meter dash in 10.0 seconds. Her average speed was

a)0.100 m/s b)10.0 m/s c)100. m/s d)1,000 m/s

Motion - $200 - $200•An object that is originally moving at a speed of 20. meters per second accelerates uniformly for 5.0 seconds to a final speed of 50. meters per second. What is the acceleration of the object?

a: .3 m/s2 b: 10 m/s2 c: 6 m/s2 d: 20 m/s2

Motion - $300- $300

• The graph represents the relationship between distance and time for an object in motion. During which interval is the speed of the object changing?

a) A to B

b) B to C

c) C to D

d) D to E

Motion - $400- $400

A 4.0-kilogram rock and a 1.0-kilogram stone fall freely from rest from a height of 100. meters. After they fall for 2.0 seconds, the ratio of the rock's speed to the stone's speed is

a) 1:1 b) 2:1 c) 1:2 d) 4:1

Motion - $500- $500

The graph represents the linear motion of a car. The average velocity of the car during interval DE is

a) 0 m/s b) 10. m/s c) 20. m/s d) 40. m/s

Forces - $100Forces - $100

Which statement explains why a book resting on a table is in equilibrium?

a) There is a net force acting downward on the book.

b) The weight of the book equals the weight of the table.

c) The acceleration due to gravity is 9.8 m/s2 for both the book and the table.

d) The weight of the book and the table's upward force on the book are equal in magnitude, but opposite in direction.

ForcesForces - $200- $200

Which is the most likely mass of a high school student?

a) 1 kg b) 5 kg c) 60 kg d) 250 kg


Four forces are acting on an object as shown in the diagram. If the object is moving with a constant velocity, the magnitude of force F must be

a) 0 N b) 10 N c) 30 N d) 40 N

ForcesForces - $400- $400In the diagram, surface B of the

wooden block has the same texture as surface A, but twice the area of surface A. If force F is required to slide the block at constant speed across the table on surface A, approximately what force is required to slide the block at constant speed across the table on surface B?

a)4 Fb)2 Fc) Fd)½ F


A 2,400-kilogram car is traveling at a speed of 20. meters per second. Compared to the magnitude of the force required to stop the car in 12 seconds, the magnitude of the force required to stop the car in 6.0 seconds is

a) half as great b) twice as great c) the same d) four times as great

Energy and PowerEnergy and Power - $100 - $100

A 10-kilogram cart moving with a energy of 5 J is brought to a stop in 2 seconds. What was the magnitude of the power used to stop the cart?

a) 0 wattsb) 2.5 wattsc) 5 wattsd) 25 watts

Energy and PowerEnergy and Power - $200- $200

A 60 kg student running at 3.0 m/s has a kinetic energy of:

a) 180 J

b) 270 J

c) 540 J

d) 8100 J


If a 5 Kg Object is raised 2 meters from the surface in 10 s, its gain in potential energy is:

a)0 Jb)10 Jc)20 Jd)100 J


The diagram below which represents a 3.0-kilogram mass being moved at a

constant speed by a force of 6.0 newtons.The magnitude of the work done on the mass is:

a: 54 J b: 24 Jc: 18 J d: 180 J


In the diagram, 55 joules of work is needed to raise a 10 N weight

5.0 meters in 5 s. How much work is need to overcome friction?

a: 1 Jb: 5 Jc: 50 Jd: 55 J

5 m

F

10 N

Gravity and Graphs- $100Gravity and Graphs- $100

b Vector, Newtonsb Vector, Newtons

Gravity is a Force (Vector) Gravity is a Force (Vector) and uses Newtons for a and uses Newtons for a

UnitUnit


B. 2 m/sB. 2 m/s

v = d / tv = d / tv = 8 m / 4sv = 8 m / 4sv = 2m/sv = 2m/s

Gravity and GraphsGravity and Graphs - $300- $300B: Twice as B: Twice as greatgreat

Using the Using the Gravitational Gravitational formula Mass is formula Mass is directly directly proportional to the proportional to the Gravitational force Gravitational force between the between the ObjectsObjects


B: 149.6 NB: 149.6 N

F = ma = 17 kg 8.8 m/s2 = 149.6 N

Gravity and GraphsGravity and Graphs - $500- $500 If the mass of A increases by a factor of 3, the force F also increases by a factor of 3. If the mass of B is halved, the force F is also halved. Increases by a factor of 3/2.

Mass A Mass B

Mass A

3X

Mass B

1/2

3m

3m

Momentum and Acceleration - $100- $100

C: C: 20,000 kg-m/s

P = mvP = mv = 1000Kg ( 20 m/s)= 1000Kg ( 20 m/s) = 20,000 kg-m/s= 20,000 kg-m/s


c: 3.0 m/sc: 3.0 m/s22

a = a = v / v / tta = 60 m/s – 45 m/s / 5.0 a = 60 m/s – 45 m/s / 5.0 ssa = 15 m/s / a = 15 m/s / 5.0 s5.0 sa = 3.0 m/sa = 3.0 m/s22


B: 75 N

F = ma = ( 0.60 kg)(125 m/s2) = 75 N


b: 14.2 m/s

Vf2 = Vi

2 + 2ad = ( 8m/s )2 + 2 ( 3m/s2 ) ( 23m) = 64 m2/s2 + 138 m2/s2

Vf2 = 202 m2/s2

Vf = 14.2 m/s


d: 4 m/s d: 4 m/s

Pbefore = Pafter

m1 v1+ m2v2 = m1 v1/+ m2v2 /

20kg (6 m/s)- 30kg (v2) = 0 + 0-30kg (v2) = -120 kg-m/s(v2) = 4 m/s

Motion - $100- $100

b: 10 m/sb: 10 m/s

v = d/tv = d/tv = 100 m / 10.0 sv = 100 m / 10.0 sv = 10 m/sv = 10 m/s

Motion - $200- $200

c: 6.0 m/sc: 6.0 m/s22

a = v / t

a = (50 m/s – 20 m/s) / 5.0 s

a = 30 m/s / 5.0 s

a = 6.0 m/sm/s22

Motion - $300- $300

d: D to Ed: D to E

On a distance – time On a distance – time graph, a curve line is graph, a curve line is

changing speedchanging speed ( acceleration)( acceleration)

Motion - $400- $400

a: 1:1a: 1:1

All Objects fall at the All Objects fall at the same rate!same rate!

Motion - $500- $500

c: 20 m/sc: 20 m/s

n a Velocity time graph, the n a Velocity time graph, the average velocity is mathematical average velocity is mathematical

average of the two points D and E. average of the two points D and E. v = 20 m/s + 20 m/s /2v = 20 m/s + 20 m/s /2

Forces - $100- $100

c: 60 kgc: 60 kg

1kg = 2.2 pounds1kg = 2.2 pounds60 kg = 132 pounds60 kg = 132 pounds

Forces - $200- $200d: d: The weight of the book and the table's upward force on

the book are equal in magnitude, but opposite in

direction.

For every action there is For every action there is an equal and opposite an equal and opposite reaction, Newton’s 3reaction, Newton’s 3rdrd

LawLaw

Forces - $300- $300

C: 40 NC: 40 N

If an object is moving at a If an object is moving at a constant velocity, then constant velocity, then

forces must be equal and forces must be equal and opposite.opposite.

Forces - $400- $400

c: Fc: F

If the surface area If the surface area doubles (B), but the force doubles (B), but the force

is ½ because of the is ½ because of the weight distribution, then weight distribution, then

the friction will not the friction will not change!change!

Forces $500$500

b : twice as great

J = F (t)

2400 N-s = F (12 m/s)

2400 N-s = 2F (6 m/s)


B: 2.5 watts

P = W / tP = W / t= 5 J / 2s= 5 J / 2s=2.5 watts=2.5 watts


b: 270 Jb: 270 J

Ke = ½ mvKe = ½ mv22

= ½ 60 kg (3.0 m/s)= ½ 60 kg (3.0 m/s)22

= 30 kg (9 m= 30 kg (9 m22/s/s22)) = 270 J= 270 J


d: 100 Jd: 100 J

Pe = mghPe = mgh= (5 kg) (10 m/s= (5 kg) (10 m/s22) (2 m)) (2 m)= 100 J= 100 J


b: 24 Jb: 24 J

W = F x DW = F x D = 6 N ( 4 m)= 6 N ( 4 m) = 24 J= 24 J


b: 5 Jb: 5 J

W = F x dW = F x dW = 10 N (5.0 m) = 50 J,W = 10 N (5.0 m) = 50 J,If the total work = 55 J, If the total work = 55 J, therefore 5 J to overcome therefore 5 J to overcome frictionfriction

FINAL FINAL CATEGORY:CATEGORY:MomentumMomentum

Place you Bet nowPlace you Bet now

FINAL CATEGORYFINAL CATEGORY

A Cart has a mass M on a frictionless A Cart has a mass M on a frictionless track starts from rest at the top of the track starts from rest at the top of the hill having a height hhill having a height h1, 1, Express the Express the kinetic energy when it reaches the top kinetic energy when it reaches the top of the next hill, having hof the next hill, having h22? ? (in terms of M, g, h(in terms of M, g, h1 1 , h, h2 2 , or h, or h33))

Answer: Ke (hAnswer: Ke (h22) = mg (h) = mg (h11- - hh22))

The difference in height The difference in height will divide the total will divide the total energyenergy

KeKemaxmax = ½ mv = ½ mv2 2 = mgh= mgh11

FINAL CATEGORYFINAL CATEGORY

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