4

Kinematics

A YEAR IN REVIEW

AP PHYSICS 1

@Scott Publishing Inc.

v = 20m/s

15m

terrorist

v = 20m/s

v = terminal?

2

Kinematics

3

Chapter 1

8

Key Ideas:

Kinematics

Las Vegas rule of physics: the velocity in the direction in the x direction always stays in the x-direction. Same goes with y direction

It takes the same amount of time for a ballistic (no additional forces acting upon object after take-off) to reach the apex of flight as it does for the object to fall from the apex to the ground level it was launched off

Use the y-directions alone (not x) to calculate the time it takes for an object to fall.

To find horizontal distance traveled, multiple time by x - velocity

Vf = V0 + ½ at2

d= vot + ½ at2

Slope = Δy/ Δx Vf = at

Vx

Vy

terrorist dropped from plane

5

Chapter 1

6

Kinematics - Answer Key:

3. (A) Observe that graph given is velocity, not position. When a velocity graph is above the x-axis the position is increasing. Position is the area under the velocity graph while acceleration is the slope. Another thing to consider is speed = |velocity|. Therefore, speed is increasing when both acceleration and velocity have the same sign (either negative or positive).

4. (D) Acceleration is the slope of a velocity graph. Slope = rise/run.

5. (B) Acceleration is the slope of a velocity graph. Slope = Δy/ Δx.

6. (D) Key word in this question is average. Since the slope of velocity is linear in this graph, the average value of the function can be found by adding the highest and lowest point and diving by time.

7. (C) The package can be assumed to have an initial Vx = 50 m/s while intial Vy = 0m/s. The Las Vegas rule tells us that Vx will remain constant so we simply need to calculate the number of seconds before the package hits the ground and multiply by 50 m/s.

8. (D) Consider vertical and horizontal components. The initial vertical velocity = 10m/s(sin(30)) = 5m/s. The initial vertical velocity equals the final vertical velocity right before the ball hits the ground. We will find the time it takes for the ball to fall from its apex to the ground, then multiple by two to find total time.

d= vot + ½ at2

=

= ( ) t = √25 = 5

d = vt

250m = (50m/s)(5 s)

Kinematics

7

Vf = V0 + ½ at2 = 2 / = √1 = 1

9. (C) Since the graph is of position, the slope = velocity. When the slope = 0, velocity = 0, and the particle is at rest.

10. (D) v = at so the velocity gained in five seconds with a 3 m/s2 acceleration is 15 m/s. Since the car is going 30 m/s it must have been going 15 m/s prior to the acceleration.

11. (C) Consider the equation d = ½ at2. The acceleration (1 m/s2) and time (5 minutes = 300 seconds) are given. Plug and chug.

12. (BO d= vot + ½ at2. The velocity initial equals zero and time = 8s (8.9s – 0.9s). g equals acceleration due to gravity.

13. (D) Problems like these are often simpler to solve backwards by finding the time it takes to fall from the apex to the ground. That way v0 = 0.

Vf = at t = Vf/a t = (30m/s)/10 = 3.0 s

14. (E) Consider that when an object is accelerating linearly its average velocity = (final velocity - initial velocity)/2. Average velocity multiplied by time equals distance. On a planet with an ag = 30 m/s2 the initial velocity = 0 while the final velocity = 30m/s. Plug and chug.

15. (E) d= vot + ½ at2 150 = (0)t + ½ (3)t2 100 = t2 t = 10

16. (D) the velocity of a position time graph can be represented by the slope because velocity = ∆ / and slope = rise/run.

17. (C) Acceleration due to gravity is a constant on Earth; it takes the same amount of time to negatively accelerate to the apex on the up-climb as it does to accelerate to the initial ground level.

Chapter 1

12

Momentum

9

Momentum • Chapter 2

10

Momentum

11

Chapter 2

16

Momentum

13

Chapter 2 Answers:

14

Momentum

15

Chapter 2

20

Energy & Work

17

Chapter 2

18

Momentum

19

Chapter 2

24

EnergyEnergy and Work Review 256. (C) Change in PE= mgh Change in PE = (500kg)(10 m/s2)(10 m) 257. W= Fxd= Fcos W= Fcos d W= (10 N)(cos60)(50 m) W= 250 J 258. (C) The mass doesn’t matter. Change in gravitational potential energy is equal to a change in kinetic energy. KEi+PEi=KEf+PEf

0+PEi=KEf+0 mgh=(½)mv2f Cancel masses, solve for velocity, plug in 259. 266. (C) Both cannons fire balls of the same mass (m) and use the same amount powder to supply identical (F). The length of Cannon 2 is two times the length of Cannon 1. Apply work energy theorem. This theorem comes from knowing the work is change in energy. Fd1= (½)mv2 Fd2= (½)mv2

d2=2d1

F(2d1)= (½) mv2

Fd1= (¼) mv2

(¼)mv2

2= (½) mv1

2

V22 =2v1

V2 = (2v1) =1.4v1

260. (E) W= change in KE

21

Chapter 3

22

Energy & Work

23

Chapter 3

28

W= change in KE Fd= change in KE F= change in KE/d F= (500 J)/(0.125 m) F= 4,000 N 278. The cars collide in an elastic collision. Know that energy must be conserved. No potential energy because the cars start from the same height. KEA+KEB= KE’a+KE’B + HEAT (not the Sandra Bullock movie) KEA+0 = KE’a+KE’B + HEAT HEAT=KEA - KE’a- KE’B

Plug in (½)mv2=KE - All masses are the same so you can pull it out - (½)ma(vA

2-vA’2 -vB’2)

- HEAT= 0.005 J - Heat/KE= 50%

Energy & Work

25

Fd= (½) mvi 2 - (½) mvf

2 F= [(½) mvi

2 - (½) mvf 2]/d

F=[(½)m (vi 2 -vf

2 )]/d F= [(½) (1,000 kg) (02-302)]/10 261. (E) KEi=PEf

(½)vi2= 2gh (masses cancel, because everybody brought mass to the

party) vi= (2gh) vi= (2x10x0.1)= 1.44m/s2 Now use conservation of energy to calculate the bullet's initial velocity. PA+PB = Pf(A+B)

- Plug in mv=P and solve for vA by plugging in known values. 262. (B) Max kinetic energy occurs when all of the elastic potential energy is transferred to motion. US= (½)kx2 =(½)(2,500 N/m)(0.12 m)2

263. (C) Use Hooke’s Law: F=kx=(300N/m)(0.5m)=150 N 264.(A) PE=(½)kx2 = (½)(400N/m)(0.5)2 =50 J 265. (D) Use conservation of energy. KEi+PEi=KEf+PEf

0+PEi=KEf+0 (½) mvf

2= (½)kx2

vf= (kx2/m) vf= [(100 N/m)(-0.2m)2]/(1kg)= 2 m/s 266. (C) Fd1= (½)mv1

2

Fd2 = (½)mv22

d2=2d1

Chapter 3

26

Energy & Work

F(2d1)= (½)mv22

F(d1)= (1/4)mv22

(1/4)mv22= (½)mv1

2

- Cancel mass

v2= (2v12)=1.4v1

267. (C)Work= change in kinetic energy W= change in kinetic energy Fd= (½)mv2f= (½)mv2i F= ( (½)mv2f= (½)mv2i)/d Plug in given numbers and you get F= 1,375 N (approximately 1,400 N) 268. (B) According to the law of conservation energy must be conserved, so since not all of the gravitational potential energy turned into kinetic energy, then we know that 25 J - 23 J=2 J of energy that must be stored as thermal energy due to friction as the box slides down the incline. 269. (D) Use work energy theorem (basically know that work equals change in kinetic energy). Fd= (½) m(v)2

2Fd= m(v)2

m= (2Fd)/(v)2

m=( 2(2 N)(5 m))/1m/s2

m= 20 kg 270. (C) Understanding the conservation of energy, the gravitational potential energy of the stuntman gets converted to kinetic energy as he falls. KEi+PEi=KEf + PEf

0 + PEi=KEf + 0 KEf = PEi

(½) mv2f = mgh

27

W= change in kinetic energy so Fd= (½) mv2f

Fd= mgh F= (mgh)/d F= [(70 kg)(9.8m/s2)(125 m)] / (5 m) = 1.7 x 104 N 271. (E) P= W/t P= Fd/t P = [(100 N) (60 m)] / (120 s) P= 50 W 272. (E) Because points A and C are at the same height they have the same gravitational potential energy. 273. (E) The car will be move fastest at point E because it has the most KE. We know this because the law of conservation of energy tells us that PE is transferred to KE, so the lowest point would have the most KE and thus it would be moving the fastest. 274. (E) The total energy is sum of the kinetic energy and potential energy, it is the same at all points. 275. P= W/t W= Fxd Fx=Fcos W= Fcos d P= Fcos d/t P= [(20N)(cos60)(100 m)]/(300 s) 276. (D) KE= (½)mv2

(½) (2,000 kg) (343)2

277. The handgun does work on the bullet and gives it kinetic energy. All the kinetic energy is lost as heat. Calculate the the force exerted by using the work energy theorem.

Chapter 3

32

Newton’s Laws

29

Chapter 3

Energy and Work Review

Terms:

Energy - The ability to do work.

Work - A force applied over a distance.

Joule - The units of work, equivalent to a Newton-meter. Also units of energy.

Kinetic Energy - The energy of motion.

Power - Work done per unit time.

Watt - Unit of power; equal to joule/second.

Law of Conservation of Energy - energy cannot be created or destroyed, but only changed from one form into another or

transferred from one object to another.

30

Energy & Work

Equations:

W= Fdcos(θ) - theta is the angle between the force and the displacement which it causes

P= W/t so P=(Fdcos(θ))/t so P=Fvcos(θ)

PEgrav= mgh

KE= 1/2mv2

KEi + PEi + Wnc = KEf + PEf when Wnc= work which is done to it by non-conservative forces

31

Chapter 4 Newton’s Laws

36

55. (D) a= Fnet/m. Acceleration is inversely proportional to mass, so mass must be ½ as much (for the same net force). 56. (C)26 degrees is found from the arctan9.5)=26. 57. (C) Take 9.8 (gravity) and multiply by mass (1,000) to get force of gravity. Divide this by 6, because the tension is continuous throughout the pulleys. 58. (D) The weight of the object (mg) acts in the downward direction. If you have trouble with this idea visualize the mass as any number, and gravity as -9.8, and understand the force of gravity always pulls back towards the surface of Earth. Since the object isn’t moving know that the force of the spring acts in the upward direction to maintain equilibrium. 59. (E) Gravity is always acting on an object unless otherwise mentioned in a question. The perpendicular force, is another way to say the normal force. We know normal force is acting because if we summed the y-components they would equal zero. Fun note:

h k l f l f f

Newton’s Laws

60. (B)

33

Chapter 4

34

Newton’s Laws

35

Chapter 4

Newton’s Laws: Dynamics 51. (A) The first, which mathematically states the the net force equals zero.When the net force equals zero, then the acceleration must also equal zero, as shown when you rewrite F=ma to 0(F)/m= 0(a). Equilibrium is the state of motion when something isn’t accelerating. 52. (D) There is a gravitational field on Earth than is about of Earth’s so both objects will accelerate. You don’t really have to know this though, just knowing that they are both on the same planet we can assume that gravity is the same. Using F=ma, which becomes mg/m=a, we can understand that the masses cancel, and the objects hit the ground at the same time. Basically, the more massive sphere has a greater gravitational force, but it also is more massive so that ratio simplifies to the same free fall acceleration 53. (C) Sliding friction (kinetic friction) acts in the direction opposite that of motion (you can always remember this by going minus slowing forces). Friction is directly proportional to Normal Force. You know this became Force of Friction= Force Normal x Mu. As Fn goes up the so must Ff. 54.( Essentially you can look at this a two separate ropes.First step, label your angle as 45 degrees. Next knowing trig, you know that your force of tension = 1000cos(45)=approx 52. Double that since you have “two” ropes and you have 100 N.

40

Rotation

37

To solve for the force of tension, we break into components and solve for the force of gravity in the x direction, know this is equal to the total tension (2000sin(43)). Then since there are two ropes we divide the tension between the ropes giving us 682 N. 61. (B). First eliminate all the answers that say to the left, because we know the box will go right because James box has 2N more of force. After the collision the box will have a net force of 2 N, which we can divide by the mass of the box (5kg) to get 0.4 acceleration. 62. (D) First step is to realize the force of gravity must be resolved into components and to know that force normal equals mu times force of friction. Since the object doesn’t move up or down, the Fnety= 0. We can set FN=mgcos . Fnetx= ma, Fnetx= mgsin .- Ff= a. Replace Ff with μFN, then replace FN with mgcos . Everyone brought mass to the party so cancel the masses. a= g(sin - μcos ). 63. (E) The force that you exert is equal to the rope’s tension (T). The acceleration due to gravity is 10 m/s2 and the acceleration of the box is given at 10 m/s upward. Fnet= ma so T-mg= ma so T=m(a+g) T= (10kg)(10m/s2+ 10 m/s2) T=200 N 64. (A) Resolve into components. Fx= Fcos Fx= 20,000cos = 10,000 N a= Fnetx/m a= (Fx- Fwind)/(90,000kg)

Chapter 4

a= 0.1 m/s2

65. (D) An object only moves when the forces acting on it are unbalanced, as is the case at t=2. (1.2N-1N=0.2N). We know F=ma so a=F/m. So a=0.2/0.1= 2 m/s2

.

66. (D) a= Fnet/m a= (F1-F2)/m

● We know that F2 is the force of friction, because the second force is always the slowing force (in this case, friction)

Reorder to F2= F1-ma Plug in: F2=(10N)-(10kg)(0.1m/s2)

38

Newton’s Laws: Dynamics

First Law:

An object at rest will remain at rest unless acted on by an unbalanced force. An object in motion continues in motion with the same speed and in the same direction unless acted upon by an unbalanced force.

This law is often called "the law of inertia".

What does this mean? This means that there is a natural tendency of objects to keep on doing what they're doing. All objects resist changes in their state of motion. In the absence of an unbalanced force, an object in motion will maintain this state of motion.

Second Law:

Acceleration is produced when a force acts on a mass. The greater the mass (of the object being accelerated) the greater the amount of force needed (to accelerate the object). (F=ma)

Newton’s Laws

39

What does this mean? Everyone unconsciously knows the Second Law. Everyone knows that heavier objects require more force to move the same distance as lighter objects.

Third Law:

For every action there is an equal and opposite re-action.

What does this mean? This means that for every force there is a reaction force that is equal in size, but opposite in direction. That is to say that whenever an object pushes another object it gets pushed back in the opposite direction equally hard.

Chapter 4

44

Rotation

41

Chapter 5

42

Rotation

43

Chapter 5

48

DC Circuts

45

Chapter 5

46

Rotation

47

Chapter 5

52

DC Circuts1. (B) R =pL/A, where p is the resistivity of the wire material.

Consider that Area = pi(radius)2 . 2. (D) V = IR. 12V/0.6amps = 20ohms. Since know that the

number of volts in total is 12, we must consider figure out the true resistance so the that the equation balances. 12v = 0.6A * resistance so R = 12/0.6 = 20 ohms. Subtract 6 ohms for the reactor in the circuit.

3. (C) Things to consider: when in parallel voltage is constant and current changes, when in series voltage changes and current remains constant.

4. (C) P = I∆ . However, since it tells us the resistance is constant, we know that the voltage also doubles. Thus power 4x.

5. (A) I = ∆ . Therefore current = or 0.55 A. 6. (D) Parallel: voltage differential is constant while current is

different. Consider that different resistance (think Maple and Cinnamon) allow for different amounts of current (or customers) to go through. However delta-V is constant (all the customers end up in the same place).

7. (C) Δ = . Know this equation people. 8. (A) The river takes the path of least resistance. If the

resistance of one route in a parallel circuit decreases, the current in the route increases.

9. (C) Slope equals rise over run. V/R equals what? Modify the original V = IR.

10. (D) If you got this wrong, I am honestly embraced for you. This is a DC Circuit review. But in all seriousness, Kirchhoff’s junction law simple states that the current going into a junction must come out on the other side.

49

Chapter 6

50

DC Circuts

51

Chapter 6

56

Electrostatics

53

11. (A) Going back to one of the fundamental principles, current is constant when circuit is in series. Remember this.

12. (D) Consider a water fall. The voltage or potential difference is the height fallen by the water. In order to have a waterfall, it is necessary to have a point of high potential so the water can fall to a low potential.

13. (D) Common sense cannot be explained. 14. (C) I personally find this an ambiguous question. The

potential difference between C and D would be 6V. However, they are considering the PD between A and a point between C and D which equals 3 *6V = 18V.

15. (D) = . Therefore Amps*seconds just equates to charge.

16. (B) Process of elimination. Current CAN flow through resistance (light bulbs) but it CAN also flow through a perfect conductor (like copper wire) which is the same as absence of resistance.

17. (B) If the wire remained the same diameter but 4x in length, the resistance would be 4x. In order for the resistance to maintain constant, the cross-sectional area of the wire must 4x. The area of a circle is pi*r^2 so the radius must double.

18. (D) The total resistance increases while the voltage remains constant. I = V/R so the total current decreases. Since the resistors are in series, the current is constant through out. Therefore, the current going through the initial resistor decreases.

19. (A) V=IR. I = . = 15 amps. 20. (A) Resistance is ohms. This is a hard question, if you have

questions about the algebra ask Gris.

Chapter 6

54

21. (D) I = V/R. More R means a smaller I. Math people. 22. (A) I = V/R. To make I bigger, V must increase or R must

decrease. If both those things happen, current increases even more.

23. (C) Total current decreases because total resistance increases. Since current is constant, current decreases. Since the resistance of the resistor remains constant and current decreases, so does voltage.

24. (A) The voltage doubles because another voltage step-up (battery) is added in series. Current is the same because there is only one path (series).

25. (D) V = I*R. R = (2.22*10-8)(0.) Look more at t 26. (D) Material is the same; Resistivity is the same. 27. (C) Since the 3 ohm resistor burned out, the current does

not have two paths (3 and 6 that it can flow through) it only has one the 6. However, the total current decreases by 67% because total resistance increases from 6 to 10 ohms. Overall the current going through the 6 ohm reactor increases by 300% while the total current only decreases by 67%.

28. (A) 1 amp = 1 C/s. # of electrons = (340 * 10-3 C/1s)(1 electron/1.6 * 10-19 C)

29. (B) R=V/I. Slope = rise over run. 30. (D) Going back to the good ‘ol V=IR. Remember that if

nothing else

DC Circuts

55

Key Ideas:

Chapter 6

Δ = =

Kirchhoff’s junction law simple states that the current going into a junction must come out on the other side

When resistors are in series, the current remains constant while the voltage changes.

When resistors are in parallel, the current changes while voltage remains constant.

The resistance of a wire is directly related to the cross sectional area and the length.

60

Key Ideas:

Electrostatics

Charge = # electrons * electron charge (Q = n(e-)) =

=

W = f*d Remember that radius has a squared inverse effect to

Force, there will be several questions related to this. Newton’s third law still applies, any charge apply a

force will have the same amount of force acting on it.

57

Chapter 7

58

Electrostatics

1. (C) Charge = # electrons * electron charge. Plug and chug.

Q = n(e-) n = Q/(e-)

2. (C) Consider the equation for F due to electricity. Since r is both in the denominator the radius has a squared inverse relationship with charge. In English, if you plug in four for the radius, the force will be 1/16 of what it was when the radius was one.

3. (D) To calcite this force you must know both the electro static and gravitational force.

4. (B) Newton’s third law combined with common sense should give you this answer. Remember, Gris can’t touch anybody without being touched back.

5. (D) n = Q/e- Remember that charge = # electrons * electron charge. Next step is algebra.

6. (C) Since the radius has an inverse squared relationship with force, if radius halves its radius the force is 4x.

7. (D) Since both the charges are doubled, the numerator will increase by 4 times. However, since the radius is squared the denominator will also increase by 4 times. 4/4 is what people?

8. (A) Particles with an opposite charge attract. However, the F due to electricity decreases as the radius (or distance) increases.

=

= 1.49 ∗32,400

=

= 0.18 ∗(32,400 )

59

Chapter 7

9. (B) Newton’s third law remains true. Gris can’t poke without getting poked back.

10 (D). =

11. (C) Q = ne- n = Q/e- n = (8*10-6)/(1.6*10-19)

12. (A) An object with a charge can attract another object of opposite charge or a neutral charge. Like charges cannot attract, but the type of charge does not matter otherwise.

13. (D) = = ( . ∗ )( . ∗ ). =

W = f*d

14. (D) In order for q2 to be in static equilibrium, the repellent force of Q must equal to the attractive charge of q1. Since q1 has twice the radius from q2 as Q, then its charge must be the opposite sign and 4x as large as Q’s.

15. (C) If one charge was doubled, the force of electricity would double. Doubling both is the same as saying 22 so the force 4x.

16. (C) Since the NA has a positive charge, it is necessary for the bonding element to be negative. That narrows it down to two choices. Now you must now a little chemistry – since Oxygen has more valence electrons then Hydrogen so it can more easily gain electrons

17. (C) charge = #electrons*electron charge. So charge = (30)(-1.6*10-19) = -4.8*10-18

18. (B) As the test charge is repelled, the radius of the electrical field increases and therefore the force of electricity and the acceleration due to electricity decrease.

19. (C) Coulomb is a unit of charge.

20. (A) The best way to approached this question is by process of

61

+ -

Magno-helment: saving lives one magnetically attracted bullet at the time

terrorist

Good luck on the AP Test!!! - your friendly physics review makers

@Scott Publishing Inc.

Some Tips:on physics and other things...1. Don’t stress if you don’t understand something the fi rst time; it will likely make sense eventually. 2. Watching cat-physics videos helps retention.3. Forces in a closed non-accelerating system always add to zero.4. Th e AP test is not any harder, and probably easier, than Gris’s tests.5. Cheating is always a risk; there’s a chance the person you’re cheating off didn’t study either.6. If you are a Junior, DO NOT do you’re fi nal project with Seniors. Th ey WILL graduate and shaft you.7. One thing you learn looking around a physics classroom — there’s a fortune to be made in dermatology.