﻿ ﻿Extra Drill on Difficult but Frequently Tested Topics - Develop Strategies for Success - 5 Steps to a 5: AP Physics C (2016)

## 5 Steps to a 5: AP Physics C (2016)

### Extra Drill on Difficult but Frequently Tested Topics

IN THIS CHAPTER

Summary: Drills in five types of problems that you should spend extra time reviewing, with step-by-step solutions.

Key Ideas

Tension problems

Electric and magnetic fields problems

Inclined plane problems

Motion graph problems

Simple circuits problems

Practice problems and tests cannot possibly cover every situation that you may be asked to understand in physics. However, some categories of topics come up again and again, so much so that they might be worth some extra review. And that”s exactly what this chapter is for—to give you a focused, intensive review of a few of the most essential physics topics.

We call them “drills” for a reason. They are designed to be skill-building exercises, and as such, they stress repetition and technique. Working through these exercises might remind you of playing scales if you”re a musician or of running laps around the field if you”re an athlete. Not much fun, maybe a little tedious, but very helpful in the long run.

The questions in each drill are all solved essentially the same way. Don”t just do one problem after the other … rather, do a couple, check to see that your answers are right, 1 and then, half an hour or a few days later, do a few more, just to remind yourself of the techniques involved.

Tension

How to Do It

Use the following steps to solve these kinds of problems: (1) Draw a free-body diagram for each block; (2) resolve vectors into their components; (3) write Newton”s second law for each block, being careful to stick to your choice of positive direction; and (4) solve the simultaneous equations for whatever the problem asks for.

The Drill

In the diagrams below, assume all pulleys and ropes are massless, and use the following variable definitions.

Find the tension in each rope and the acceleration of the set of masses.

(For a greater challenge, solve in terms of F, M , and μ instead of plugging in values.)

1 . Frictionless

2 . Frictionless

3 . Frictionless

4 . Coefficient of Friction μ

5 .

6 .

7 . Frictionless

8 . Frictionless

9 . Frictionless

10 . Coefficient of Friction μ

11 . Coefficient of Friction μ

12 . Frictionless

13 . Frictionless

14 . Coefficient of Friction μ

(Step-by-Step Solutions to #2 and #5 Are on the Next Page.)

1 . a = 10 m/s2

2 . a = 3.3 m/s2

T = 3.3 N

3 . a = 1.7 m/s2

T 1 = 1.7 N

T 2 = 5.1 N

4 . a = 1.3 m/s2

T = 3.3 N

5 . a = 3.3 m/s2

T = 13 N

6 . a = 7.1 m/s2

T 1 = 17 N

T 2 = 11 N

7 . a = 3.3 m/s2

T = 6.6 N

8 . a = 6.7 m/s2

T 1 = 13 N

T 2 = 10 N

9 . a = 1.7 m/s2

T 1 = 5.1 N

T 2 = 8.3 N

10 . a = 6.0 m/s2

T = 8.0 N

11 . a = 8.0 m/s2

T 1 = 10 N

T 2 = 4.0 N

12 . a = 5.0 m/s2

T = 15 N

13 . a = 3.3 m/s2

T 1 = 13 N

T 2 = 20 N

14 . a = 0.22 m/s2

T 1 = 20 N

T 2 = 29 N

Step-by-Step Solution to #2:

Step 1 : Free-body diagrams:

No components are necessary, so on to Step 3 : write Newton”s second law for each block, calling the rightward direction positive:

Step 4 : Solve algebraically. It”s easiest to add these equations together, because the tensions cancel:

F = (3m )a , so a = F /3m = (10 N)/3(1 kg) = 3.3 m/s2 .

To get the tension, just plug back into T − 0 = ma to find T = F /3 = 3.3 N.

Step-by-Step Solution to #5:

Step 1 : Free-body diagrams:

No components are necessary, so on to Step 3 : write Newton”s second law for each block, calling clockwise rotation of the pulley positive:

Step 4 : Solve algebraically. It”s easiest to add these equations together, because the tensions cancel:

mg = (3m )a , so a = g /3 = 3.3 m/s2 .

To get the tension, just plug back into Tmg = ma : T = m (a + g ) = (4/3)mg = 13 N.

Electric and Magnetic Fields

How to Do It

The force of an electric field is F = qE , and the direction of the force is in the direction of the field for a positive charge. The force of a magnetic field is F = qvB sinθ , and the direction of the force is given by the right-hand rule.

The Drill

The magnetic field above has magnitude 3.0 T. For each of the following particles placed in the field, find (a) the force exerted by the magnetic field on the particle, and (b) the acceleration of the particle. Be sure to give magnitude and direction in each case.

1 . an e- at rest

2 . an e- moving ↑ at 2 m/s

3 . an e- moving ← at 2 m/s

4 . a proton moving at 2 m/s

5 . an e- moving up and to the right, at an angle of 30° to the horizontal, at 2 m/s

6 . an e- moving up and to the left, at an angle of 30° to the horizontal, at 2 m/s

7 . a positron moving up and to the right, at an angle of 30° to the horizontal, at 2 m/s

8 . an e- moving → at 2 m/s

9 . a proton moving at 2 m/s

The electric field above has magnitude 3.0 N/C. For each of the following particles placed in the field, find (a) the force exerted by the electric field on the particle, and (b) the acceleration of the particle. Be sure to give magnitude and direction in each case.

10 . an e- at rest

11 . a proton at rest

12 . a positron at rest

13 . an e- moving ↑ at 2 m/s

14 . an e- moving → at 2 m/s

15 . a proton moving at 2 m/s

16 . an e- moving ← at 2 m/s

17 . a positron moving up and to the right, at an angle of 30° to the horizontal, at 2 m/s

(Step-by-Step Solutions to #2 and #10 Are on the Next Pages.)

1 . No force or acceleration, v = 0.

2 . F = 9.6 × 10−19 N, out of the page.

a = 1.1 × 1012 m/s2 , out of the page.

3 . No force or acceleration, sin θ = 0.

4 . F = 9.6 × 10−19 N, toward the top of the page.

a = 5.6 × 108 m/s2 , toward the top of the page.

5 . F = 4.8 × 10−19 N, out of the page.

a = 5.3 × 1011 m/s2 , out of the page.

6 . F = 4.8 × 10−19 N, out of the page.

a = 5.3 × 1011 m/s2 , out of the page.

7 . F = 4.8 × 10−19 N, into the page.

a = 5.3 × 1011 m/s2 , into the page.

8 . No force or acceleration, sin θ = 0.

9 . F = 9.6 × 10−19 N, toward the bottom of the page.

a = 5.6 × 108 m/s2 , toward the bottom of the page.

10 . F = 4.8 × 10−19 N, left.

a = 5.3 × 1011 m/s2 , left.

11 . F = 4.8 × 10−19 N, right.

a = 2.8 × 108 m/s2 , right.

12 . F = 4.8 × 10−19 N, right.

a = 5.3 × 1011 m/s2 , right.

13 . F = 4.8 × 10−19 N, left.

a = 5.3 × 1011 m/s2 , left.
Velocity does not affect electric force.

14 . F = 4.8 × 10−19 N, left.

a = 5.3 × 1011 m/s2 , left.

15 . F = 4.8 × 10−19 N, right.

a = 2.8 × 108 m/s2 , right.

16 . F = 4.8 × 10−19 N, left.

a = 5.3 × 1011 m/s2 , left.

17 . F = 4.8 × 10−19 N, right.

a = 5.3 × 1011 m/s2 , right.

Step-by-Step Solution to #2:

(a) The magnetic force on a charge is given by F = qvB sin θ . Since the velocity is perpendicular to the magnetic field, θ = 90°, and sin θ = 1. The charge q is the amount of charge on an electron, 1.6 × 10−19 C. v is the electron”s speed, 2 m/s. B is the magnetic field, 3 T.

F = (1.6 × 10−19 C)(2 m/s)(3 T)(1) = 9.6 × 10−19 N

The direction is given by the right-hand rule. Point your fingers in the direction of the electron”s velocity, toward the top of the page; curl your fingers in the direction of the magnetic field, to the right; your thumb points into the page. Since the electron has a negative charge, the force points opposite your thumb, or out of the page.

(b) Even though we”re dealing with a magnetic force, we can still use Newton”s second law. Since the magnetic force is the only force acting, just set this force equal to ma and solve. The direction of the acceleration must be in the same direction as the net force.

9.6 × 10−19 N = (9.1 × 10−31 kg)a
a
= 1.1 × 1012 m/s2 , out of the page

Step-by-Step Solution to #10:

(a) The electric force on a charge is given by F = qE . The charge q is the amount of charge on an electron, 1.6 × 10−19 C. E is the electric field, 3 N/C.

F = (1.6 × 10−19 C)(3 N/C) = 4.8 × 10−19 N

Because the electron has a negative charge, the force is opposite the electric field, or right.

(b) Even though we”re dealing with an electric force, we can still use Newton”s second law. Since the electric force is the only force acting, just set this force equal to ma and solve. The direction of the acceleration must be in the same direction as the net force.

4.8 × 10−19 N = (9.1 × 10−31 kg)a
a
= 2.8 × 108 m/s2 , left

Inclined Planes

How to Do It

Use the following steps to solve these kinds of problems: 1) Draw a free-body diagram for the object (the normal force is perpendicular to the plane; the friction force acts along the plane, opposite the velocity); 2) break vectors into components, where the parallel component of weight is mg (sin θ ); 3) write Newton”s second law for parallel and perpendicular components; and 4) solve the equations for whatever the problem asks for.

Don”t forget, the normal force is NOT equal to mg when a block is on an incline!

The Drill

Directions: For each of the following situations, determine:

(a) the acceleration of the block down the plane

(b) the time for the block to slide to the bottom of the plane

In each case, assume a frictionless plane unless otherwise stated; assume the block is released from rest unless otherwise stated.

1 .

2 .

3 .

4 .

5 .

6 .

7 .

8 .

Careful—this one”s tricky.

(A Step-by-Step Solution to #1 Is on the Next Page.)

1 . a = 6.3 m/s2 , down the plane.
t = 2.5 s

2 . a = 4.9 m/s2 , down the plane.
t = 2.9 s

3 . a = 5.2 m/s2 , down the plane.
t = 2.8 s

4 . a = 4.4 m/s2 , down the plane.
t = 3.0 s

5 . Here the angle of the plane is 27° by trigonometry, and the distance along the plane is 22 m.a = 4.4 m/s2 , down the plane.
t = 3.2 s

6 . a = 6.3 m/s2 , down the plane.
t = 1.8 s

7 . a = 6.3 m/s2 , down the plane.
t = 3.5 s

8 . This one is complicated. Since the direction of the friction force changes depending on whether the block is sliding up or down the plane, the block”s acceleration is NOT constant throughout the whole problem. So, unlike problem #7, this one can”t be solved in a single step. Instead, in order to use kinematics equations, you must break this problem up into two parts: up the plane and down the plane. During each of these individual parts, the acceleration is constant, so the kinematics equations are valid.

• up the plane:

a = 6.8 m/s2 , down the plane.

t = 0.4 s before the block turns around to come down the plane.

• down the plane:

a = 1.5 m/s2 , down the plane.

t = 5.2 s to reach bottom.

So, a total of t = 5.6 s for the block to go up and back down.

Step-by-Step Solution to #1:

Step 1: Free-body diagram:

Step 2: Break vectors into components. Because we have an incline, we use inclined axes, one parallel and one perpendicular to the incline:

Step 3: Write Newton”s second law for each axis. The acceleration is entirely directed parallel to the plane, so perpendicular acceleration can be written as zero:

mg sin θ − 0 = ma .
F Nmg cos θ = 0.

Step 4: Solve algebraically for a . This can be done without reference to the second equation. (In problems with friction, use F f = μF N to relate the two equations.)

a = g sin θ = 6.3 m/s2

To find the time, plug into a kinematics chart:

Solve for t using the second star equation for kinematics (**): Δx = vo t + ½at 2 , where vo is zero;

Motion Graphs

How to Do It

For a position–time graph, the slope is the velocity. For a velocity–time graph, the slope is the acceleration, and the area under the graph is the displacement.

The Drill

Use the graph to determine something about the object”s speed. Then play “Physics Taboo ”: suggest what object might reasonably perform this motion and explain in words how the object moves. Use everyday language. In your explanation, you may not use any words from the list below:

velocity

acceleration

positive

negative

increase

decrease

it

object

constant

Note that our descriptions of the moving objects reflect our own imaginations. You might have come up with some very different descriptions, and that”s fine … provided that your answers are conceptually the same as ours.

1 . The average speed over the first 5 s is 10 m/s, or about 22 mph. So:

Someone rolls a bowling ball along a smooth road. When the graph starts, the bowling ball is moving along pretty fast, but the ball encounters a long hill. So, the ball slows down, coming to rest after 5 s. Then, the ball comes back down the hill, speeding up the whole way.

2 . This motion only lasts 1 s, and the maximum speed involved is about 5 mph. So:

A biker has been cruising up a hill. When the graph starts, the biker is barely moving at jogging speed. Within half a second, and after traveling only a meter up the hill, the bike turns around, speeding up as it goes back down the hill.

3 . The maximum speed of this thing is 30 cm/s, or about a foot per second. So:

A toy racecar is moving slowly along its track. The track goes up a short hill that”s about a foot long. After 2 s, the car has just barely reached the top of the hill, and is perched there momentarily; then, the car crests the hill and speeds up as it goes down the other side.

4 . The steady speed over 200 s (a bit over 3 minutes) is 0.25 m/s, or 25 cm/s, or about a foot per second.

A cockroach crawls steadily along the school”s running track, searching for food. The cockroach starts near the 50 yard line of the football field; around three minutes later, the cockroach reaches the goal line and, having found nothing of interest, turns around and crawls at the same speed back toward his starting point.

5 . The maximum speed here is 50 m/s, or over a hundred mph, changing speed dramatically in only 5 or 10 s. So:

A small airplane is coming in for a landing. Upon touching the ground, the pilot puts the engines in reverse, slowing the plane. But wait! The engine throttle is stuck! So, although the plane comes to rest in 5 s, the engines are still on. The plane starts speeding up backwards! Oops …

6 . This thing covers 5 meters in 3 seconds, speeding up the whole time.

An 8-year-old gets on his dad”s bike. The boy is not really strong enough to work the pedals easily, so he starts off with difficulty. But, after a few seconds he”s managed to speed the bike up to a reasonable clip.

7 . Though this thing moves quickly—while moving, the speed is 1 m/s—the total distance covered is 1 mm forward, and 1 mm back; the whole process takes 5 ms, which is less than the minimum time interval indicated by a typical stopwatch. So we”ll have to be a bit creative:

In the Discworld novels by Terry Pratchett, wizards have developed a computer in which living ants in tubes, rather than electrons in wires and transistors, carry information. (Electricity has not been harnessed on the Discworld.) In performing a calculation, one of these ants moves forward a distance of 1 mm; stays in place for 3 ms; and returns to the original position. If this ant”s motion represents two typical “operations” performed by the computer, then this computer has an approximate processing speed of 400 Hz times the total number of ants inside.

8 . Though this graph looks like #7, this one is a velocity–time graph, and so indicates completely different motion.

A small child pretends he is a bulldozer. Making a “brm-brm-brm” noise with his lips, he speeds up from rest to a slow walk. He walks for three more seconds, then slows back down to rest. He moved forward the entire time, traveling a total distance (found from the area under the graph) of 4 m.

9 . This stuff moves 300 million meters in 1 s at a constant speed. There”s only one possibility here: electromagnetic waves in a vacuum.

Light (or electromagnetic radiation of any frequency) is emitted from the surface of the moon. In 1 s, the light has covered about half the distance to Earth.

10 . Be careful about axis labels: this is an acceleration –time graph. Something is accelerating at 1000 cm/s2 for a few seconds. 1000 cm/s2 = 10 m/s2 , about Earth”s gravitational acceleration. Using kinematics, we calculate that if we drop something from rest near Earth, after 4 s the thing has dropped 80 m.

One way to simulate the effects of zero gravity is to drop an experiment from the top of a high tower. Then, because everything that was dropped is speeding up at the same rate, the effect is just as if the experiment were done in the Space Shuttle—at least until everything hits the ground. In this case, an experiment is dropped from a 250-ft tower, hitting the ground with a speed close to 90 mph.

11 . 1 cm/s is ridiculously slow. Let”s use the world of slimy animals:

A snail wakes up from his nap and decides to find some food. He speeds himself up from rest to his top speed in 10 s. During this time, he”s covered 5 cm, or about the length of your pinkie finger. He continues to slide along at a steady 1 cm/s, which means that a minute later he”s gone no farther than a couple of feet. Let”s hope that food is close.

12 . This one looks a bit like those up-and-down-a-hill graphs, but with an important difference—this time the thing stops not just for an instant, but for five whole seconds, before continuing back toward the starting point.

A bicyclist coasts to the top of a shallow hill, slowing down from cruising speed (∼15 mph) to rest in 15 s. At the top, she pauses briefly to turn her bike around; then, she releases the brake and speeds up as she goes back down the hill.

Simple Circuits

How to Do It

Think “series” and “parallel.” The current through series resistors is the same, and the voltage across series resistors adds to the total voltage. The current through parallel resistors adds to the total current, and the voltage across parallel resistors is the same.

The Drill

For each circuit drawn below, find the current through and voltage across each resistor.

Note: Assume each resistance and voltage value is precise to two significant figures.

1 .

2 .

3 .

4 .

5 .

6 .

7 .

(A Step-by-Step Solution to #2 Is on the Next Page.)

1 .

2 .

(Remember, a kΩ is 1000 Ω, and a mA is 10–3 A.)

3 .

4 .

5 .

6 .

7 .

Step-by-Step Solution to #2:

Start by simplifying the combinations of resistors. The 8 kΩ and 10 kΩ resistors are in parallel. Their equivalent resistance is given by

which gives R eq = 4.4 kΩ.

Next, simplify these series resistors to their equivalent resistance of 6.4 kΩ.

6.4 kΩ (i.e., 6400 Ω) is the total resistance of the entire circuit. Because we know the total voltage of the entire circuit to be 10 V, we can use Ohm”s law to get the total current

(more commonly written as 1.6 mA).

Now look at the previous diagram. The same current of 1.6 mA must go out of the battery, into the 2 kΩ resistor, and into the 4.4 kΩ resistor. The voltage across each resistor can thus be determined by V = (1.6 mA)R for each resistor, giving 3.2 V across the 2 kΩ resistor and 6.8 V across the 4.4 kΩ resistor.

The 2 kΩ resistor is on the chart. However, the 4.4 kΩ resistor is the equivalent of two parallel resistors. Because voltage is the same for resistors in parallel, there are 6.8 V across each of the two parallel resistors in the original diagram. Fill that in the chart, and use Ohm”s law to find the current through each:

1 For numerical answers, it”s okay if you”re off by a significant figure or so.

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