answers to physics practice problems by paradigm physics para prob ans.pdfanswers to physics...

y y v t at

y , y .

v a g .

t

t . .

= + += =

= = = −

= −

= ≈

0 012

2

0

0

2

0 3

0 10

0 3 5

6 8

Choose

We know and

F W mg ma

a g m/s .

net = ° = × =

= ≈

sin30

5

12

12

2 The correct answer is B

Answers to Physics Practice Problems by Paradigm

The Braking Car

1. Find the equation which relates position and time.

The correct answer is D, 0.77 sec.

• We could have chosen the positive direction to be down so g would be +10. This is often easier for free fallproblems; however, it is confusing to most students.

2. This could be solved using the position-time equation, but we don’t know v0. We’d have to usethe position-velocity equation (or better still, conservation of energy) to find that. The easy wayof doing the problem is to recognize that the trip up and the trip down are essentially equivalent –they must take the same time. The answer is then twice that of Problem 1. The correct answer isagain D.

3. This problem can be broken down into x and y components. If we think of the rock’sdownward trip instead of its upward trip, the y equation is exactly the same as in Problem 1. (Think of the person throwing the rock toward your direction, but a large distance away. It looksas if the rock goes straight up and straight down.) The correct answer is C.

• There are very few MCAT problems which involve two-dimensional motion.

4. We call the x-direction the direction down the plane. Then we draw a free body diagram of theblock. The only forces can be 1) gravity, W=mg, and 2) the contact (normal) force with theplane, N. We then break the gravitational force into components in the x and y directions. Thecontact force cancels the y-component of gravity, so the net force is the x-component of gravity.

• Which answers can you automatically delete?

x x v t at

x , x.

= . .

v g + .

. . t

t

= + +

= =°

= ≈

= −≈

0 012

2

0

0

2

0120

30240

0 5

0 24 25

1

Choose

We know and (Why + ?)sin

v v at

v= + =

= +×

0

0 5 1 5

2

0 064 0 64

2 64 032 4 2

02

02

0

0

a(x x ) v v

x , x= . m, v= , a= .

v ( . ) ( . )

− = −

= −

= × − × − ≈ =

Note that m/ s2

T =ma= . = .A 4 32 128×

5. We use the position-time equation again:

The correct answer is A.

6. For this part, we may use the velocity-time equation because we already know the accelerationand the time.

The correct answer is B.

• Note that if we did not know the time and acceleration, the quickest approach is to use conservation of energy.See the Block on an Inclined Plane Paradigm.

7. This problem is best solved using the position-velocity equation. (Why can’t we useconservation of energy in this problem?) We choose the +x direction to be up the plane.

The correct answer is D.

8. We should keep in mind a few things about strings and ropes: We always assume that stringsare massless. That way, they are always straight lines and the tension in the string has the samemagnitude everywhere. Let us first draw a free body diagram for theleft-hand mass. N and W cancel, so we are left with:

The correct answer is C.

T T =m a= . =

T .B A

B

− ×= + =

2 2 32 64

128 64 192

.

. .

T =ma= . = .A 7 32 224×

T mg= T mg− = ≈0 20,

T mg=ma T mg ma− = + ≈ + =, 20 4 24

T mg=ma aT

mg a− = − ≈ − = −, ,

4

210 8

9. Our approach is similar to Problem 8, the onlycomplication being that we have two tensions this time.

Thecorrect answer is C.

10. We could proceed as in Problem 9; however, let us take a little different viewpoint.Note that we can think of the tension accelerating all the masses together, so theproblem is essentially the same as Problem 8 with the mass replaced by the combinedmasses:


11. Let us draw a free body diagram and apply Newton’s Second Law:


12. The problem is identical to Problem 11. A constant velocity does not affect theforces. The correct answer is A.

13. The system now has a net acceleration which must be included in Newton’s Second Law:


14. The problem is essentially the same as Problem 13; however, the acceleration is the unknown:

• Could you choose the correct answer from qualitative considerations alone?

y y v t at

y t

v a g .

y

= + += =

= = = −

≈ × − ×

0 012

2

0

0

2

24 24

0 10

24 24 5

Choose and

Then and

2.4

, . .

.

y y v t at

y h y t

v a g .

h

h

= + += = =

= = = −≈ − ×≈ × × ≈ × ≈

0 012

2

0

0

2

0 24

0 10

0 5

5 25 25 5 625 30

Choose and

Then and

2.4

, , . .

. . .

T W ma T W Ma

T g a T g a

T g a g a

a g

a g

l l r r− = − =− = + − = −= + = −=

= ≈

,

,

.

2 2 4 4

2 2 4 4

6 2

3313

v v at

v t

v v

= += =

= − × ≈

0

0 0

0 24

0 10 24 24

At the top, and

. .

. ,

15. We sketch free body diagrams for the two blocks. Note that thetensions will be the same and the accelerations will of the samemagnitude, but opposite in direction. Newton’s Second Law gives:


• MCAT problems usually do not require you to solve two equations simultaneously. Also note how we handled thesigns of the accelerations.

16. Sprays of water in the air, as all liquid flows which maintain a constant pressure, obey theusual laws of motion. If it helps, think of a ball instead of a fluid flow. (Note at any point ofcontact with the atmosphere, liquids are at atmospheric pressure. More will be said on that topicin the fluid mechanics section.) We therefore need to use the velocity-time equation:


17. For this part, we can use either the position-time equation or the velocity-position equation.We choose the former.

Note that the computation is messy and requires a correct solution of the previous problem aswell. As in Problem 2, it is easier to calculate the distance from the time it takes the water to fallthan to rise:


18. After the cannonball leaves the cannon, the only force on the cannonball is that of gravity.(Ignoring air friction.) Because of that, the acceleration is always 9.8 m / s2 in the downwarddirection. The correct answer is D.

Top: =

Bottom:

E mgh

E mv

mv mgh

v gh

0

0

2 2 10 3 75

12

2

12

2

+

= +

=

= ≈ × × ≈ .

Top: =

Half way:

E mgh

E mv

mv mgh

v gh

0

0

2 2 10 15 55

12

2

12

2

+= +

=

= ≈ × × ≈. .

Top: =

Bottom of hole:

E mgh

E mv

mv mgh

v gh

0

0

2 2 10 6 11

12

2

12

2

+

= +

=

= ≈ × × ≈

Hand: =

Top:

E mv

E mgh

mv mgh

hv

g

12

2

12

2

2

0

0

2

36

2018

+= +

=

= ≈ = .

Block on an Inclined Plane

1. We apply conservation of energy to the block at the top of its trajectory and as it reaches theground.


2. The problem is done exactly the same way as Problem 1. It is convenient, however, to choosethe origin of the coordinate system at the point half way down. That eliminates the need tocalculate two separate potential energies.


3. This is again the same problem. Note that there is nothing special about the ground level.Choose the bottom of the hole to be y=0.


4. This is again a typical conservation of energy problem.


Hose: =

Top:

E mv

E mgh

mv mgh

hv

g

12

2

12

2

2

0

0

2

16

2008

+= +

=

= ≈ = .

Initial:

Final: =

E kx

E mv

mv kx

vk

mx

= +

+

=

= = × = × ≈ × =

0

0

64

5002 128 002 35 002 007

12

2

12

2

12

2 12

2

.

.. . . . . .

Top: =

Bottom:

E mgh

E mv

mv mgh

hv

g

0

0

2

625

2031

12

2

12

2

2

+

= +

=

= ≈ ≈.

.

U kx= = × × =12

2 12

264 02 000128. (. ) .

5. A spray of water into the air behaves the same way as a ball being tossed into the air (ignoringthe effects of air friction). Therefore:


6. We simply have to know the equation for the potential energy of a spring.


• Spring problems are not commonly found on MCAT exams. Be careful with decimal points.

7. This problem is very similar to gravitational energy problems – we just have a differentequation for the potential energy.


8. This is very similar to dropping a block in free fall.


E U U mg h hf i f i f= − = − ≈ × × =( ) . .7 10 002 14

E mc= = × × × = ×

× = ×

− −

− −

2 28 8 2 12

12 11

25 10 3 10 225 10

1125 10 112 10

. ( ) .

. .

The energy of each photon is half that, or

W Fd Fd mv

vFd

m

= = =

= =× ×

= ≈

cos

. .

θ 12

2

2 2 4 12

1096 32

9. From the result of Problem 8, we can see that the speed of the block at the bottom of the planedepends only on the height of the plane. The correct answer is C.

10. At a given height, the blocks must be traveling the same speed; however, the block travelingdown plane A has a considerably shorter distance to travel. The block therefore reaches thebottom of A first. Think of extreme limits in this problem. If the angle of the planes were 1� and89�, it would be intuitively clear that the block sliding down the steeper plane would arrive first.The correct answer is A.

11. Rolling objects can present a little problem in energy conservation problems because there isrotational kinetic energy as well as translational kinetic energy. We will look at that in the Ball onan Inclined Plane Paradigm. Here, however, the two points of interest involve zero kineticenergy. Energy conservation then requires that the final potential energy is smaller than the initialpotential energy by the amount of energy lost to friction:


12. If you try to use the formula K = ½mv2, you get zero for an answer. That’s because ofrelativistically, photons can have effective mass. What you should know, however, is that themass of the �

0 is all converted to energy through Einstein’s equation. Conservation of energygives us:


• Note that all you need to do is roughly get the right order of magnitude to choose the correct answer.

13. This is an example of the work-energy theorem, which essentially says that work is energy.We also need to remember the formula W = F d cos

�. Here the angle

� is zero, as is most often

the case in MCAT problems, hence:


W Fd Fd Fd mv

vFd

m

= = ° = =

= =×

= ≈

cos cos

. .

θ 60

4 12

1048 22

12

12

2

F d F d

F d F d

F F mg

p p b b

p b b b

p b

=

=

= = ≈ × × =

2

10 10 5012

12

12

14. Now we have to use the more general version of the formula, which essentially says that onlythe component of the force parallel to the direction of motion can do work on the box.


15. The work done by the person equals the work done in moving the block because the kineticenergy of the system remains a constant. Note that the person must pull the rope twice thedistance the block moves, however. Thus:


• There are a few assumptions you make in the problem, such as the direction the person is pulling the rope is notimportant. There are a few MCAT problems where you have to worry about this kind of detail. But if you’ re indoubt, just ask yourself what you’d have to do to work out the problem without the assumptions. If its at all hard,they probably expect you to make the simplifying assumption. – Note that this problem can also be worked usingNewton’s Laws. Just draw a free body diagram of the lower pulley and remember the tension in the string will bethe same everywhere.

PE

t

E E

t

mv

tave

f i= =−

= − = − ××

= =∆∆ ∆ ∆

12 0

2 1

2

1600 400

32

640000

64100000

. .

P IVE

tE IV t

= =

= = × × =

∆∆

∆ ∆ 2 100 3600 720000

PE

tE P t

=

= = × =

∆∆

∆ ∆ 100 60 6000

PE

t

E E

t

mgh

tave

f i= =−

≈−

=× ×

=× ×

=∆∆ ∆ ∆

0 300 10 28

12

100 10 7

17000

.

.

PE

t

E E

t

mgh

tave

f i= =−

≈−

=× ×

=×

=∆∆ ∆ ∆

0 10 10 12

36

10 12

3633

..

Porsche

1. As with most power problems, we need to find the change in energy in the time specified.


2. This is an important type of problem. To determine voltage and current in terms of resistance,see the Series-Parallel Circuits Paradigm.


3. This is another simple problem, but typical of MCAT problems on power.


4. Again, we resort to the definition of power:


5. This is a slight variation on Problem 4. Note that the distance along the plane doesn’t matter.


FW

d

E

d

mgh

d= = =

−≈

× ×= ≈

∆ 0 10 10 12

24

100

250

.

.

v v ad

v

202

2

2

2 16 2 64

− == × × =. .

PE

t

E E

t

mgh mv

tave

f i= =−

≈+ −

≈ +× ×

=∆∆ ∆ ∆

12

2 120

3310 64

3643.

..

6. This isn’ t a power problem; however, many readings which involve power also have questionson work, force, etc. Here, the easiest method is to relate force to work:


7. The twist here is that we need to know the final kinetic energy as well as the final potentialenergy. We use the velocity-position equation:

Then we apply the definition of power:


• This is more complicated than most MCAT problems.

Colliding Blocks

1. Conservation of energy and momentum equations can always be used together in elasticcollision problems; however, the algebra becomes quickly difficult. In MCAT problems, the bestapproach is to logically think how energy and momentum can be conserved in a collision. In thiscase, the initial momentum is zero: +mv – mv = 0 so the final momentum must also be zero. Theballs must have the same speed after the collision. – This fact is probably obvious by the symmetryof the problem. – Also, if no kinetic energy is lost, each ball must have the same kinetic energyafter the collision as before the collision. Hence, the direction of the velocity of each ball mustchange, but the speed of each ball remains the same after the collision.


2. We may apply a similar analysis here; however, we cannot use conservation of energy. Bydefinition, totally inelastic collisions are where the masses are stuck together after the collision.Since the initial momentum is zero, the final momentum must also be zero – which means nothingis moving.


K K

mv mv

v v

i f

f i

f i

=

=

= ≈ × =

12

12

2 14

2

12 7 24 17. . .

p mv mvl l r r= + = × + × − =2 4 1 8 0( )

p mv m vl l r r= + = × + + × − =2 4 1 8 0( ) ( )

p mv m vl l r r= − = × + − × + =2 4 1 8 0( ) ( )

3. Again, we know that the initial and final momenta must both be zero, since conservation ofmomentum applies in any collision process. This means that the left ball has the same speed andthe same energy as the right ball after the collision; however, the speed is less than what it wasinitially. We see then that each ball must have half its initial kinetic energy, so:


• It is useful to remember that 2 14 07≈ ≈. . and 12

4.We remember to add the momenta as vectors:


• Note that there are two ways of handling signs in this problem: a. We take vr to be a negative number and put a + sign in the equation. (The method we used.)

b. We can put the + and – signs in the equation and take each v to be a positive number,

There are advantages to each, but you should choose one method and stick with it.

5. If the total momentum is zero, the problem is fairly to analyze. In this case, we need a solutionwhere the final momentum is zero and the final energy is the same as the initial energy. Theobvious solution is that both balls bounce back with the same speed (but opposite direction) theyhad initially.


6. The final momentum must be zero. Since the balls stick together, each must have zero velocity.


p p

mv mv mv

v v v

v v v

i f

l r f

l r f

f l r

=+ =

+ =

= + = − = −

2

2

0 30 1512

12( ) ( . ) .

mv mv mv mv mv mv mv mv

v v v v v v

v v v v v v v v v

v v v v

l l r r l l r r l l r r l l r r

r l r r l r

l r r r r r r r r

r r r r

+ = ′ + ′ + = ′ + ′= ′ + ′ = ′ + ′

′ = − ′ = − ′ + ′ + ′− ′ + ′

,

,

( ), ( )

12

2 12

2 12

2 12

2

2 2 2

2 2 2 2

2 2

2 2 2 2

2 2 4 2 2

2 8 6 =

′ + ′ + =′ = −

0

6 24 18 0

1

2v v

vr r

r

v v

v vl r

l r

= + = −′ = − ′ = +

2 1

2 1

,

,

v v

v vl r

l r

= = −′ = − ′ = −

0 3

4 1

,

,

7. Again, a full solution requires conservation of energy and momentum. Think about what couldhappen to conserve energy and momentum. In this case, the entire energy and momentum of theright ball could be transferred to the left ball, leaving the right ball at rest.


• Since this particular problem often appears, it is good to remember the solution.

8. By the arguments above, the correct answer is B.

9. We apply conservation of momentum directly:


10. You probably won’t see this on the MCAT, but you should understand the physics behind it.

If this comes easy to you, try the following method. It’s mathematically easier, but conceptuallymore challenging. View the collision from the center of mass. Note that if the left ball is at x=0and the right ball at x=a, the center of mass is at x=2a/3. Hence the velocity of the center ofmass is 2/3 the velocity of the right particle, or –2m/s. With respect to the center of mass, thetotal momentum is zero. Hence (compare Problem 5):

Moving this back to the original reference frame, we add the center of mass velocity to eachvelocity:

( )

0

0 3

36

6

12

2 12

2

12

2 32

2

13

12

2 32

19

2 12

16

2 23

2

2

= + = += + = +

= − = + × = + =

==

mv MV U mv MV

v V v V

V v v v v v

v

v

,

,

,

24

24

mv MV m M v

vmv MV

M m

+ = + ′

′ =+

+=

× − + × −+

= − = −

( )

. ( . ) . ( . )

. .

..

10 48 20 24

10 20

96

332

11. This is an “explosion” type of problem. It is just a collision in reverse. In this case, energy isconserved.


• Note than when quite of bit of algebra appears, it can be best to put in numbers fairly early.

12. At first glance, this doesn’t seem like a conservation of momentum problem. However, youshould remember that momentum conservation shows that internal forces cannot change themotion of the center of mass of an object. From the center of mass’s viewpoint, the totalmomentum is zero before the explosion, so it must remain zero after the explosion. The center ofmass goes on as if no explosion had occured.


13. The question with this problem is probably how the gravitational acceleration complicates theconservation of momentum. The answer is that if restrict our attention to the moment beforecollision to the moment after the collision, gravity doesn’t have time to do anything, so it canignored completely. Thus:


U mgh mv I v r

mgh mv Iv

r

mgh mI

mrv

vgh

I mr

= = + =

= +

= +��

��

=+

12

2 12

2

12

2 12

2

2

22

2

2 1

2

1

ω ω,

/

U mgh I

mgh

I

= =

= ≈× × ×

= = ≈

12

2

2 2 12 10 15

36

30

310 32

ω

ω. .

..

τ = = = ≈ × × × =⊥r Fl

F lmg2

3 12 10 1812

12 .

Ball on an Inclined Plane

1. The inertia of an object is its “resistance” to change in linear motion, or in other words, itsmass. The “rotational inertia” of an object is its resistance to rotational motion. (For a mass on astring, it just I = mr2. ) A larger mass on a longer string has more rotational inertia, so the correctanswer is A.

2. This is a qualitative energy conservation problem. Both spheres start with the same potentialenergy. This energy is converted into 1) trnaslational kinetic energy (the energy of the center ofmass motion) and 2) rotational kinetic energy (the energy of rotational motion about the center ofmass). The sphere with the smaller rotational kinetic energy will have the larger translationalkinetic energy and will arrive at the bottom first. This is the sphere with the smallest rotationalinertia (the sphere which has the mass distributed more toward the center), the solid sphere.


• Note the math:

3. We apply conservation of energy. Let the origin of the y-axis correspond to the center of masswhen the rod is in a vertical position.


• Note that you can easily eliminate two answers on the basis of units.

4. The torque is the force multiplied by the moment arm (torque arm). The contact force of thepivot has zero moment arm, so the torque is just:


τ α

ατ

=

= = =

I

I

18

365

.

5. From the definition of angular acceleration:


6. The torque is the force times the moment arm. The contactforce at the pivot produces no torque about the pivot. Thegravitational force remains constant. The moment arm, as seen inthe figure, gets smaller.


V IR IV

R= = = =,

10

52

IQ

tI t= = × =

∆∆

∆, Q = 2 1 2

R R R= + = + =1 2 4 2 6

V IR IV

R= = = =,

12

62

P IV= = × =2 12 24

Series-Parallel Circuit

1. This is the simplest possible circuit. We need Ohm’s Law:


2. Thus uses the definition of current:


3. There is no voltage drop across a wire. If the voltage across the battery is 10V, the voltageacross this resistor s also 10V.

The correct answer is C. 4. Resistors in series add:


5. We can use the result of Problem 5 with Ohm’s Law:

Current travels from + to –, so the current is going to the left.


6. The current is the same through every component as long are there are no junctions in between.(In series current is the same, voltages add.) The correct answer is B.

7. We use the power equation:


P IV I R= = = × =2 4 2 8

PE

tE P t= = = × =

∆∆

∆ ∆, 8 60 480

1 1 1 1

2

1

4

3

4

4

31 2R R RR= + = + = =,

V IR IV

R= = = =,

12

26

P IVV

R= = = =

2 144

272

8. First, we find the power dissipated by the resistor:

The power-energy relation is:


• Note that the power supplied by the battery is total power dissipated by the resistors.

9. The reciprocals of the resistances add in series:


10. There is no voltage drop across the wires, so the voltage across each resistor is the same asthe voltage across the battery, 12 V. (In parallel voltages are the same, currents add.)


11. Using Ohm’s Law:


12. We use the usual relationship again:


• Note that the power dissipated by the 4� resistor is 36 W.

IV

R= = =

12

4 39

/

P IV= = × =9 12 108

1 1 1 1

2

1

21 1

12 1 2R R RR= + = + = =,

R R R= + = + =12 3 1 4 5

1 1 1 1 1

6

1

6

1

6

1

22

1 2 3R R R RR= + + = + + = =,

R R R12 1 2 1 2 3= + = + =

1 1 1 1

3

1

3

2

315

12 12R R RR

||.= + = + = =,

R R R R= + + = + + =|| . .1 2 15 1 05 3

13. First, we calculate the current passing through the battery:

Then we determine the power:

The correct answer is C

14. The general rule is to add everything in series first, then everything in parallel, and so forth. Inthis problem, there are no series combinations, so we begin by adding the two resistors in parallel:

.

Then we add resistances in series:


• Note the general result that adding two similar resistances in parallel gives half the value of each resistanceseparately.

15. These are three 6� resistors in parallel (somewhat disguised, however).


16. Here we go through a few steps of series-parallel reduction. First, the two branches:

Then the parallel branches:

And finally, everything in series again:


+ �

1�

12V3�

4�

3�

1�+ �

1�

12V3�

4�

3�

1�+ �

12V4� 4�

4�

+ �

12V

4�4V1A

4�4V1A

17. Let’s go through a series of simplifications:

Now we find voltages and currents, and add them to the diagrams in reverse order:


18. The total resistance is 4+3+2+1=10 � . The current is then 0.5A, and the voltages across theresistors are 0.5 V, 1.0 V, 1.5 V, and 2.0 V.


F kq q

r= 1 2

2

EV

d= = = = =

6

0024

60000

24

10000

42500

.

Other Concepts in Electr icity and Magnetism

1. The direction of the electric field from a single point charge is directly away from a positivecharge and directly toward a negative charge. If there are multiple charges, the electric field at apoint is the vector sum of the fields from the individual charges. Another way of putting it is thatdirection of the electric field is the direction of the force on a small positive test charge.


2. To find the direction of the magnetic field from a wire: Put your right thumb in the direction ofthe current and the magnetic field lines form circles around the wire in the direction of yourfingers.


3. The direction of the force on a positive test charge is given by another right-hand rule. See theparadigm sheet for a description of this rule. Right thumb up, right fingers to right, palm out.Since the electron is negatively charged, the force is in the opposite direction.


4. Coulomb’s Law states:

So the distance decreasing by a factor of two is offset by the charge dropping by a factor of four.Since the third sphere may be on the same side or the opposite side of the first charge, we cannotdetermine its sign from the information given.


5. The necessary relationship is:


6. For the electric field of a capacitor, the voltage and electric field are related by:


• Note that this relationship is remembered easily by the units of electric field, V / m. (An alternate name for thesame units is N / C).

IV

RP IV

= = =

= = × =

12

34

4 12 48

C C C= + = +1 2 4 2

1 1 1 14

12

34

1331 2C C C

C= + = + = =, .

7. Steady state means that the current, voltage, etc., are not varying in time. When you first hookup a capacitor and the capacitor is charging, these quantities will be changing. The capacitorcharges until its voltage is equal to that of the applied voltage.


8. When the capacitor becomes fully charges, no more current will flow in the circuit. Anotherway of looking at it is that, with the same voltage as the battery, the capacitor tends to drivecurrent around the circuit with an equal and opposite “force” to the battery.


9. Since no current flows through a capacitor in steady state, the right branch is as if it were noteven there. The circuit is then just a battery and resistor in series:


10. Since no current flows in the right branch, there will be no voltage drop across the resistors inthis branch. The voltage across the capacitor is the same as the voltage across the battery.


11. Capacitors in parallel add like resistors in series:

We need to be sure we remember the units are µF.


12. Capacitors in series add like resistors in parallel:


13. Root mean squared voltage is related to the maximum voltage as .V Vrmsmax = 2


P PF

AP

mg

A

P P Pmg

Ag

= + = +

= − = ≈×

=

0 0

0

1 10

001010000

.

Surface:

Point A:

H P

H P gd

P P gd

P P gd

= + += + −

= −

= + ≈ × + × × = ×

0

0

05 5

0 0

0

10 10 1000 10 6 16 10

ρρρ . .

Tank with Narrowing Pipe

1. You may have memorized the equation . If you did, you may just use it;P P gh= +0 ρhowever, this is a special case of Bernoulli’s Equation. Letting H be “total head,” we have:


2. The only question about this one is whether the depth should be 9 m because B is 9 m belowthe surface or 3 m because B is 3 m below the ceiling. The correct answer is 9 m. The reason isthat the pressure inside the cave at the ceiling is not atmospheric pressure, it is the same pressureas at point A.


3. The pressure only depends on the depth, not on the shape. The liquid exerts a greater totalforce in B because it has a greater area; however, the pressure is the same.


4. This is just to test your understanding of pressure. F = PA. Note that the horizontalcomponent of the force on surface A is the same as for surface B.


5. Although this problem can be solved using , it is more easily done byP P gh= +0 ρrecognizing the pressure at the bottom of the cylinder is atmospheric pressure plus the pressurecause by the weight of the water. (If the sides were did not go straight up, we couldn’ t do that, asthey would contribute some vertical force to the water! We’d have to use .) ThisP P gh= +0 ρalso uses the definition of gauge pressure, the total pressure – the atmospheric pressure :


� Be careful of units. Remember 1 cm2 = 0.0001 m2.

W W

m g m g

V V

h A h A

h h

w c

w c

w w c c

w w b c

w cc

w

===

=

= = × =

ρ ρρ ρ

ρρ

1 075 cm750

1000 cm.

Surface:

Top of straw:

H P

H gh

P gh

hP

g

= + += + +

=

= ≈×

×=

0

0

05

0 0

0 0

10 10

1000 1010

ρρ

ρ.

Surface:

Out of hose:

H P gh

H P v

P gh P v

P P v gh

= + +

= + ++ = +

= + − = × + × × − × × = × + × − × = ×

0

0

10 10 1000 100 1000 10 2 1 10 5 10 2 10 13 10

012

2

012

2

012

2 5 12

5 5 5 5

ρρ

ρ ρρ ρ . . . .

Av A v

vA

Av

d

dv

1 1 2 2

21

21

1

2

2

1 4 12 48

=

= =�

��

�

�� = × =

7. By Archimedes’ Principle, we know that the weight of the displaced water is the buoyant force.This must equal the weight of the entire block:

This is height of the region of displaced water or the amount of the cube in the water. Thus, thecorrect answer is B.

8. We apply Bernoulli’s Equation with the pressure at the upper surface of the water equal tozero. We take the origin of the coordinate system at the water surface. We also assume the wateris barely moving as it comes out of the straw.


9. For this, we must apply the continuity equation:


10. Another application of Bernoulli’s Equations: We choose the origin of the coordinate systemat the hose. Note that the pressure at the hose end is atmospheric pressure.


Surface:

Out of hose:

H P gh

H P v

gh v

v gh

= + +

= + +=

=

0

012

2

12

2

0

0

2

ρρ

ρ ρ

Flow Av r v= = ≈ × × × =π 2 12

23 01 8 0006( . ) .

n n1 260

10 60 15

3

2

3

21

3

sin sin

. sin . sin

sin

sin

° = ′° = ′

= ′

′ =

ββ

β

β

11. The problem is similar to the preceding one:

Note that the velocity in no way depends on the diameter of the hose. The continuity equation isapplicable only to flow through a single hose of changing diameter.


12. The flow rate is:


Ball and Wall/Cart and Sandpaper

1. First note that angles in optics are always measured withrespect to the normal, rather than with respect to thesurface. For reflection, the angle of reflection equals theangle of incidence, so � = 30�.


2. We could use Snell’s Law, but since we don’t have calculators, taking an arcsin is rough. Weshould qualitatively recognize that The refracted ray bends toward the normal when the beamtravels from a less optically dense material (small index of refraction) into a more dense material.We are still left with two possible choices, however. Looking at Snell’s Law:

Since you should know sin and cos for 30�, 45�, and 60�, youshould recognize that the answer cannot be 45�.


n nc

c

c

1 2 90

18 10 10

1

18

10

18

5

9

sin sin

. sin . .

sin.

θθ

θ

= °= ×

= = =

vc

n= =

×= ×

3 10

152 10

88

.

3. When light passes from a material which is more optically dense into a material which is lessoptically dense, the refracted beam bends away from the normal. As the angle of incidenceincreases, the refracted beam eventually exits at 90�. This angle of incidence is called the criticalangle. If the angle of incidence increases beyond the critical angle, all the light reflects back intothe first medium. All we need is Snell’s Law:


4. Nothing travels faster than light in a vacuum. The index of refraction is a measurement of howmuch light slows down.


f = ±10 cm, o = 20 cm

1. The focal length is half the radius of curvature for mirrors. (For lenses there is a morecomplicated equation, the lensmaker’s equation, which you needn’t memorize.) Concave lensesare converging, so the sign of the focal length is positive.


2. The best way to do this is to make a quick ray sketch. Since f is positive, the lens is converging.

The image is inverted, real (light rays actuallypass through it), and of the same size as theobject.


� Note that for single lenses or mirrors, real imagesare always inverted and virtual images are alwayserect.

1 1 1 1 1 1 1

25

1

50

1

5050

f o i i f oi= + = − = − = =, ,

Mi

o M

o

io

o f

o

f

M

= − = − = −�

��

�

�� = − = −

−=

= +

, 1 1 1

1 124

242

12

3. From the ray sketch, we see the answer should be 50 cm; however, it is good to verify theresult with the equation:

The sign of f isgiven, o is positive as the distance to real objects is always positive.


4. A convex mirror is diverging (whereas a convex lens is converging). A diverging lens or mirroralways forms a virtual image. A ray sketch is helpful:


5. You might be able to trust your sketch alone. To do the algebra, we need the relation formagnification:

The 1/M step is just a trick to make computation easier. f is negative because the mirror isdiverging.


� Recall that the sign of the magnification tells us whether the image is erect (+) or inverted (–).

Pf o i

= = + =∞

+−

= −1 1 1 1 1

0502

.

Pf o i

= = + = +−

= +1 1 1 1

025

1

0502

. .

Mf

fo

i

= = =200

1020

6. First, you must recall that power in diopters is 1 / f if distances are in meters. Then you needto realize that to correct vision for a nearsighted person, you must take an object at infinity andmake an image of it at a point where the person can focus, namely the far point. (The far point isthe most distant point the viewer can see clearly.) All we need to do is use the lens equation. Keepin mind that the image is virtual. (The virtual image is what the person with glasses sees. It iserect, light does not actually pass through the image, you can not put a screen where the image islocated and see the image on the screen, it is on the same side of the lens as the object.) Thereforethe image distance is negative.


� Note that you do not have to consider the optics of the eye itself in doing this problem.

7. To correct vision for a farsighted person, you must take an object at a comfortable readingdistance (usually 25 cm) and make an image of it at the near point. (The near point is the closestpoint the viewer can see clearly.)


8. The overall magnification is the product of the two powers.


9. The magnification is the ration of focal lengths:


f = = =1 1

254

τ .

v f= = × =λ 12 4 48. .

fc

= =××

= ×−λ3 10

4 1075 10

8

715.

Simple Waveform

1. The wavelength is the distance between crests. The correct answer is B.

2. The period is the number of seconds per cycle, 0.25 sec. The frequency is the number of cyclesper second:


3. The velocity is the product of frequency and wavelength.


4. We use the same relationship as in C.


5. An increase in sound intensity of a factor of two increases the number of dB by about 3. Anincrease of sound intensity by a factor of 10 increases the number of dB by 10.


Carbon-14 Decay

1. 94

23524

92231Pu→ +α X


2. 93

24010

94240Np→ +−

−β X


3. 47

10710

46107Ag→ ++

+β X


4. 47

10700

47107Ag→ +γ X


5. In one half-life, one half of the original material remains. In two half-lives one-fourth of theoriginal material remains.


6. Since one fourth of the original Fe-55 remains after two half-lives, the decay rate also drops toone fourth of its original value.


7. After the time has elapsed only 1/9.2 of the original material must remain. Therefore the timemust be between 3 half-lives (1/8 remaining) and 4 half-lives (1/16 remaining). This correspondsto somewhere between 7.8 and 10.4 years.


8. Since � and � radiation have little ability to penetrate matter, � and

� radiation are often

stopped in the surface layer of the skin, posing little health threat. � radiation, on the other hand,is highly penetrating, and could pose significant health risks.


• Note the situation is significantly different if the material is ingested.

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