Ch9

9.1

a) ��� = ���� = �0.075kg� �−6.0�� �� = −0.45kg ∙ �� � b) ��� = ���� = �0.075kg� �+4.0�� �� = +0.30kg ∙ �� � c) ∆�� = ��� − ��� = +0.75kg ∙ �� � d) Impulse

e) ��� !"#$%& '' = ∆(�)*+,--∆" = ./.0123∙45 6/./1/� = +15kg ∙ ��8 � = +15N� f) By Newton’s 3rd law ��& ''$%� !"# = −��� !"#$%& '' = −15N� g) Since the objects experience equal but opposite force they should also experience equal but opposite

changes in momentum. ∆��� !"# = −∆��& '' �� !"#∆��� !"# = −∆��& '' ∆��� !"# = −∆��& ''�� !"# ≈ −10;<1ms � For scale, consider the size of a hydrogen atom is about 10;?/m with a nucleus on the order of 10;?1m.

If we consider the earth as being initially at rest this collision would cause the earth to move about one

nucleus in 10?/s ≈ 320years. I think it is safe to assume the earth remains, for all intents and purposes,

unchanged by the collision.

9.2 The change in momentum of the ball, also known as impulse, is upwards (E� = 10.0kg ∙ �� �+F�). WATCH OUT! The initial momentum is downwards �−F� while the final momentum is upwards�+F�. E� = ∆�� = ����+F� − ����−F� E�F� = �G�� + ��HF IJ + IK = LM

The ball loses 4.00 J of energy during the collision with the ground. This tells us ΔO = −4.00J. ΔO = 12���< − 12���<

IJQ − IKQ = QRSM

From there you have two equations, two unknowns. One could solve the first bold equation for �� and plug into the

second bold equation. At first it looks like you might need the quadratic but the ��< terms drop out! �� = E� − �� T E� − ��U< − ��< = 2ΔO� E<�< − 2E� �� + ��< − ��< = 2ΔO� 2E� �� = E<�< − 2ΔO�

�� = E2� − ΔOE

�� = 10.0kg ∙ ms2�0.250kg� − �−4.00J�10.0kg ∙ ms

I think you can take it from there…

9.3

a) Rounding I’ll say ten billion (10?/).

b) To make the math easy I’ll say 100kg ≈ 220lbs. c) To relate max height to lift-off speed use ��X< = ��X< + 2GYXH∆Z.

Initial height is zero and final height is 1 m. For a leap to max height YX = −[ and ��X = 0.

We find lift-off speed is ��X = \2[ℎ ≈ 4.4�� . To make the math easier I will round this up to 5�� .

d) Doing conservation of momentum we find ��� !"#� + ��#^_ %`� = ��� !"#� + ��#^_ %`� 0 = a��� !"#� +���#^_ %`� ��� !"#� = −�a��#^_ %`�

��� !"#� = −�10?/��100kg�6 × 10<ckg �+5ms �� ≈ −10;?<ms � Even if everyone on earth weighed over 200 lbs and jumped at the same time at the same place with

incredible athletic prowess the earth’s recoil velocity is negligible.

9.4

a) Conservation of momentum gives ��?� + ��<� = ��?� + ��<� ��?�?��� = G�?�?�H� + G�<�<�H� �?�?� = �?�?� +�<�<� �?� = �?�?� −�<�<��?

�?� = �?� −�<�? �<�

b) Plugging in values gives �?� = 6.0ms − 1.00kg0.50kg 3.8ms

�?� = −1.6ms

Uhhh…wait a minute. In the problem statement I said after the collision �? moves to the right with speed �?�. Speed is a magnitude…it shouldn’t be negative. Is this messed up???!?!?!?

Turns out this is not a problem. In momentum problems like this a negative answer for speed means the

velocity is opposite the direction drawn. In this case the speed is +1.6�� moving opposite the direction

drawn. The direction drawn was “to the right”.

Therefore the velocity of �? after the collision is 1.6�� efeℎghgie = −1.6�� �. c) Initially only �? is moving so O� = ?<�?�?�< = 9.0J.

Afterwards both are moving so O� = ?<�?�?�< + ?<�<�<�< = 7. 86J ≈ 7.9J. The change in energy is ∆O ≈ −1.1J.

d) During a collision energy is lost into other forms. For instance, you can hear a sound when the blocks

collide…it must have taken some energy to make this sound. For the most part, the overwhelming majority

of the energy is transferred from the macroscopic motion of the blocks into microscopic motion of the

molecules within the blocks. When the blocks collide the molecules inside the blocks will start to wiggle

more. This takes some energy away from the translational kinetic energy. Note: when the molecules of a

solid wiggle more, we say the internal energy of the blocks goes up. If internal energy of a block increases,

we also say the temperature of the block increases. Basically, during the collision you heat up the blocks a

little bit. Eventually the blocks radiate the heat to the universe. Energy in the universe is conserved but

energy in the two block system is lost. Hope this makes sense…

Summary: During most collisions some translational kinetic energy is transferred to internal thermal

energy of the blocks molecules and lost from the system as heat.

e) Each block has an upward normal force and downwards gravitational force. These forces are both external

to the 1-2 system. If you care, the only force internal to the 1-2 system is the force of 1 on 2 and its action-

reaction pair (force of 2 on 1). For each block, however, there is no NET external force during the

collision. The net force acting on each mass during the collision comes solely from the other block.

9.5

a) Newton’s 3rd law still applies. Don’t overthink it! The force of the insect on the bus is equal in magnitude

(and opposite in direction) compared to the force of the bus on the insect.

b) Change in momentum is also the same! Remember �� = ∆(�∆" . c) By Newton’s 2nd law �� = �Y�. Ignoring direction for the moment we know Y = k_. In part a we discussed

how both objects (insect and bus) experience the same force magnitude during the collision. Clearly the

object with larger mass experiences a smaller acceleration.

Comment: to get a feel for this I like to hold a piece of paper in the air with nothing behind it. Then I let go of the

paper. At the instant I let go of the paper I punch the paper vigorously. Because the paper has such small mass

relative to my hand it will not hurt. My acceleration is tiny (my hand keeps moving with almost no discernible

change in velocity). The paper experiences a large acceleration and change in velocity. We both experience the

same magnitude of change in momentum and same force magnitude.

9.6

a) Perfectly inelastic

b) Conservation of momentum gives �?� = ��? +�<��� �� = �?�? +�< �

c) The initial kinetic energy is O� = ?<�?�<. The final kinetic energy is

O� = 12 ��? +�<���<

O� = 12 ��? +�<� T �?�? +�< �U<

O� = 12 �?<��? +�<� �<

The % change in energy is %∆O = O� − O�O�

This simplifies quickly if you notice every term has a ½ and a �<. Cancelling all these gives

%∆O = �?<��? +�<� − �?�?

%∆O = �?�? +�< − 1

%∆O = �?�? +�< −�? +�<�? +�<

%∆O = −�<�? +�< × 100%

Some people like to write " × 100%" to clarify they intend to express the answer as a percent.

d) The table is shown below. Notice energy is always lost in the perfectly inelastic collision. Mn MQ IJ %∆S � � 12 � −50%

3� � 34 � −25%

� 3� 14 � −75%

9.7

a) Elastic

b) This derivation is clever but there is no reason you can’t muscle through it the other way and still get the

correct results. If you did slog through the other mess, you can truly appreciate this style.

Conservation of momentum gives �?�?� = �?�?� +�<�<� MnGInK − InJH = MQIQJ

Conservation of energy is also valid FOR ELASTIC COLLISIONS ONLY. The ½’s will drop giving �?�?�< = �?�?�< +�<�<�< �?G�?�< − �?�<H = �<�<�< MnGInK − InJHGInK + InJH = MQIQJQ

Dividing the 2nd bold equation by the 1st bold equation gives GInK + InJH = IQJ

Plugging back into the 1st line equation (for �<�) will eliminate �<� and determine �?� in terms of �?�. �?� = �? −�<�? +�< �?� From there use this result for �?� in the 3rd bold equation to find �<� = 2�?�? +�< �?� Note: all the �’s I’m using here are technically velocities with implied �’s. Positive numbers imply to the

right, negative numbers imply to the left.

c) Trick question; no calculation necessary. Energy is conserved during an elastic collision. That implies %∆O = O� − O�O� = 0

d) The table is shown below. One way to think about the first line is to consider a cue ball hitting another

pool ball dead on. They have equal mass. Very little energy is lost in the collision. The cue ball stops and

the other ball goes flying off at the same speed the cue ball originally had. Mn MQ Ioo�nJ Ioo�QJ

� � 0 �� 3� � 12 �� 32 �� � 3� −12�� 12 ��

9.8

a) This is the first of many 3 stage problems. In this problem there are actually two separate problems.

Between the first two pictures (just before to just after collision) use conservation of momentum. ��? = �� +a��< �< = �� +a�?

Now use conservation of energy as the combined object rises to max height. O� + p� = O� + p� 12 �� +a��<< + 0 = 0 + �� +a�[ℎ �< = \2[ℎ

By setting the two result equal one finds ℎ = � �a +��< �?<2[

b) This derivation is identical to the result derived in 9.6 part c.

9.9

a) Typically we use energy if one object moving around (or a group of object’s moving around in unison).

Use energy to relate stages 1 & 2 and stages 3 & 4.

b) Typically we use momentum to compare just before to just after collision. Use momentum to relate stages

2 & 3.

c) Write down the following

Relate stages 1 & 2

using energy

Relate stage 2 & 3

using momentum

Relate stages 3 & 4

using energy

�q[ℎ� = 12�q�<<

�< = \2[ℎ� �< = \2[r sin u

�q�< = ��q +�`��v

�v = �q�q +�` �<

12�"$" '�v< +w�!�q"�$% = 0 12�"$" '�v< = −w�!�q"�$% 12�"$" '�v< = −�−x�"$" '[y� Extra minus because friction

points opposite displacement! x = �v<2[y

Now plug the pieces together x = �v<2[y

x = � �q�q +�` �<�<2[y

x = T �q�q +�`U< �<<2[y

x = T �q�q +�`U< G\2[r sin uH<

2[y

z = T M{M{ +M|UQ } ~�� ��

d) If you include the earth in the system, momentum is conserved. The block is initially at rest above the

earth distance ℎ� = r sin u. As the block comes downwards the earth must be rising up to meet it ever so

slightly. If you recall the problem with everyone on the earth jumping, the upwards velocity of earth is so

small (perhaps a billionth of a nucleon diameter per second at max speed) it is negligible. Another way to

say it, the earth exerts an upwards force on the cart-sled system during impact to cancel the vertical

component of momentum.

e) By Newton’s third law, the two experience the sled and cart exert equal but opposite forces on each other

during the collision.

f) With equal force magnitudes, the smaller mass has the larger acceleration magnitude.

g) Momentum change is given by �� = ��e. Same force maganitude implies same magnitude change in

momentum. Obviously the directions are opposite (sled momentum increases to the right while the cart

momentum decreases to the right).

9.10

a) Here are the figures I would’ve drawn for the three stages of the problem on the previous page.

Use conservation of momentum between stages 1 & 2 (just before to just after collision). Use conservation

of energy on block only between stages 2 & 3. We will determine the speed of the block after the bullet has

passed through it using stages 1 & 2. I suppose if you really cared you could also do a separate

conservation of energy problem for just the bullet between stages 2 & 3.

b) Conservation of momentum gives ��?_ = ��<_ +a�<� �<� = ���?_ − �<_�a

�<� = �0.010kg� �1000ms − 400ms �2.00kg

�<� = 3ms

Note: the speed of sound in air is about 340�� . If you do some web research you’ll find that these numbers

are not totally ridiculous but probably imply this bullet was shot out of some high-powered rifle.

c) Ignore the bullet completely. O� + p� = O� + p� 12a�<�< + 0 = 0 + a[ℎ

ℎ = �<�<2[ ≈ 46cm

�?_�

a ℎ

�<_ �<� �v� = 0

who cares

about bullet

9.11

a) This is our first example of a 2D collision. Since the objects stick together after the collision the most

thorough way to classify it is probably “2D perfectly inelastic collision”.

b) In general we know ��?� + ��<� = ��?� + ��<�

In practice we use

Horizontal Vertical �?�� + �<�� = �?�� + �<�� �?�?�� +�<�<�� = �?�?�� +�<�<�� �?�?� + 0 = ��? +�<���� ��� = �?�? +�< �?� �?�X + �<�X = �?�X + �<�X �?�?�X +�<�<�X = �?�?�X +�<�<�X 0 + �<�<� = ��? +�<���X ��X = �<�? +�< �<�

c) The speed is given by �� = ����< + ��X<

�� = �T �?�? +�< �?�U< + T �<�? +�< �<�U<

�� = 1�? +�<\��?�?��< + ��<�<��<

� = tan;? ���X����

� = tan;? �� �<�? +�< �<��� �?�? +�< �?���

� = tan;? T�<�<��?�?�U

d) Pluggin in numbers gives �� ≈ 22. 3�� with � ≈ 50. 2° north of east.

e) In this case there is initially slightly more momentum in the Z-direction (towards north). After the collision

the same must be true. We therefore expect final velocity to be closer to north than east and it checks out.

f) As it slide to a halt across level ground we have a single object translating with external work being done to

slow it down. This looks like a classic energy problem O� + p� +w��"$!%$%;q$%�!� "��� = O� + p� 12 ��? +�<���< + 0 +w�!�q�"$% = 0 + 0

Here I assumed the initial and final heights are both zero. Since the block is sliding on a horizontal surface

we could do a quick FBD to show the normal force is � = ��? +�<�[ and the frictional force is given by i = x�� = x��? +�<�[. Since friction points opposite the direction of motion during the slide, we know

the work done by friction is negative. Furthermore, the length of the slide is y. Putting it all together gives 12 ��? +�<���< − x��? +�<�[y = 0

y = ��<2x[

g) As the coefficient increases the cars should slide less distance. Equation reflects this. Similarly, if cars

were moving faster blocks would slide more distance.

9.12 Style 1 is to do things nearly the same as the last problem. The before

and after picture now looks like

Style 2 uses the dot product. For a perfectly inelastic collision we know ��?� + ��<� = ���

Well…why not take a dot product of each side with itself!!! How would you think to do this? You

wouldn’t…unless you’ve had the wonderful experience you are having right now. ���?� + ��<�� ∙ ���?� + ��<�� = ��� ∙ ��� �?�< + �<�< + 2�?��<� cos�Y�[hg�ge�gg���?� &��<� � = ��<

Oh brother…this is basically just the law of cosines. Maybe if I had drawn the triangle shown at right I would’ve

seen this.

From dot

product ����< + �2��<��< + 2�����2��<�� cos�60.0°� = G3��0.441��H<

From law

of cosines ����< + �2��<��< − 2�����2��<�� cos�120.0°� = G3��0.441��H<

Notice all �’s drop and you get a quadratic formula for �<� �< + 4�<�< + 2��<� = 1.75�< 4�<�< − 2��<� − 0.75�< = 0

�<� = −�2�� � \4�< − 4�4��−0.75�<�2�4�

�<� = −2� � 4�8 = �+0.250�−0.750�

Only the positive root makes since, in this equation, we assumed �<� was the speed of 2 initially.

Comments: WATCH OUT!!! �?�� = �?� sin 60.0° = �?� cos 30.0°. PLEASE PLEASE PLEASE do NOT write �?�� sin 60.0°. If you do this you are essentially double counting the angle and it drives me insane.

Horizontal Vertical �?�� + �<�� = �?�� + �<�� �?�?�� +�<�<�� = �?�?�� +�<�<�� �?�?�� = ��? +�<���� �� sin 60.0° = 3��0.441� cos �� 0.8660 = 1.323 cos� � = cos;? 0.6546 ≈ 49. 11°

�?�X + �<�X = �?�X + �<�X �?�?�X +�<�<�X = �?�?�X +�<�<�X �?�?�X +�<�<� = ��? +�<���X �� cos 60.0° + 2��<� = 3��0.441� sin �� 0.500� + 2�<� = 1.323� sin � �<� = �2 G1.323 sin � − 0.500H �<� = 0.2501�

1 2

60.0°

Before After

1 & 2 �

�

0.441�

60.0°

��?� ��2� ��i

120.0°

9.13

a) A ball dropped from height ℎ has speed � = \2[ℎ at impact. Do a quick energy problem to verify.

b) I am assuming an overinflated basketball will bounce back without losing much energy. Said another way,

I think the overinflated basketball colliding with the ground can be modeled approximately as an elastic

collision.

c) Doing conservation of momentum gives �?�X + �<�X = �?�X + �<�X ��−�� + 3�� = ��?� + 3��<� QI = InJ + �IQJ

Doing conservation of energy gives ��−��< + 3��< = ��?�< + 3��<�< �IQ = InJQ + �IQJQ

Solving for �?� in the first bold equation and plugging into the 2nd bold equation gives 4�< = G2� − 3�<�H< + 3�<�< 4�< = 4�< − 12��<� + 9�<�< + 3�<�< 0 = −12��<� + 12�<�< 0 = 12�<�G�<� − �H WATCH OUT!!! People tend to snap to thinking �<� = � but in reality there are TWO possible solutions: �<� = ���0

The first solution, �<� = �, makes no sense. We know the basketball must slow down at least a little if it is

to impart some momentum to the tennis ball. Therefore the other answer must be the result!!!

How crazy is that! That implies that the basketball stops moving and all of the energy is necessarily

transferred to the tennis ball! Checking the speed we find �?� = 2�.

d) Twice the speed implies four times the height. Do a quick energy problem to verify.

9.14 DISCLAIMER: My set-up of this problem is oversimplified. The typical assumption used in the Newton’s

cradle, and the one I am assuming here, is collisions occur pairwise. For example, when one ball is raised, it first

hits only the second ball in the chain, which then hits the third, and so on. With multiple balls lifted on a single side,

the situation gets notably more complex. This assumption introduces errors, but, gives a correct feeling for the

problem.

Since the collision is elastic, we may use both conservation of momentum and energy.

Energy ���< = �����<

Momentum ��� = �����

Taking the ratio of the energy equation to the momentum equation gives � = �� .

Using this result in the momentum equation gives � = ��

Regarding parts b) and c), try it and see…

Side note: if using a standard 5-ball Newton’s cradle, write the word “LOWED” with one letter per ball. Then,

regardless of how many balls are in the down position, a valid scrabble word is formed! For example, if you pull up

two balls from the right and one from the left you get “OW”. Look up the definitions if you don’t believe me. Note:

this ignores any instances when only a single ball is in the down position. Thanks to Caleb Greene for finding the

first solution to this problem in during the Spring 2019 semester! This was a long-standing extra credit question in

may class (for almost 10 years). Caleb was the first person to find a word that worked in EVERY permutation!

9.15

a) Energy is conserved but we cannot say O� = O� . Because spring forces are conservative, we know the

energy initially stored in the springs will be transferred to the kinetic energies of the blocks. The interface

between each block and the surface contributes negligible friction according to the problem statement.

b) Momentum is conserved. The spring forces are internal to the “3 blocks plus springs” system. Normal

force and �[ balance on each block; external friction is negligible.

c) Block 3 is slightly heavier than 1. We expect its greater inertia shall keep

it in place slightly more than block 1. As a result we expect block 2 to

move slightly to the left upon release.

d) Watch out for units of kN/m on spring constant and cm for distance. Conservation of energy gives 2 T12 ��<U = 12�?�?< + 12�<�<< + 12�v�v< 2��< = �?�?< +�<�<< +�v�v< 122J = 64. 0J + �10.0kg��<< + �3.00kg��v<

 . ¡¢£Q~Q = IQQ + ¢. �¢¢I�Q

Notice you lose a sig fig during the subtraction!!!

Conservation of momentum gives 0 = −�?�? −�<�< +�v�v 0 = −8.00kg ∙ ms − �10.0kg��< + �3.00kg��v

¢. ¡¢¢£~ = −IQ + ¢. �¢¢I�

Noticing the units will work out I will now leave them off for brevity. Solve the 2nd bold equation for �<

and plug into the first bold equation gives 5. 80 = G0.300�v − 0.800H< + 0.300�v< 5. 80 = 0.0900�v< − 0.480�v + 0.640 + 0.300�v< 0 = 0.390�v< − 0.480�v − 5. 16

�v = −G−0.480H � \�−0.480�< − 4�0.390��−5.16�2G0.390H

�v = 0.480 � \0.2304 + 8. 04960.780

�v = 0.480 � 2. 8770.780 =¤¥¦¥§+4. 30ms−3. 07ms

Negative result makes no sense since we know block 3 must move to the right. From there use the second

bold equation to find �< = 0.300G4. 30H − 0.800 �< = 1. 29 − 0.800 �< = 0. 49m/s Positive result indicates block two moves in direction drawn.

Summary: block 2 moves left with speed 0. 49m/s while block 3 moves right with speed 4. 30m/s. Perhaps you are concerned you didn’t get the minus sign on the 2nd term. Check your picture and you

probably thought block 2 would move to the right by accident. No worries. Remember your result should

work out the same but in the end you will get a negative number fort the speed �<. The negative number

for speed to the right will be the same as my positive number for speed to the left…

9.16

a) When two blocks have different masses we shouldn’t expect them to split up momentum or energy equally.

This is the point I was trying to emphasize in this question.

In fact, even with equal masses they do not in general split up momentum or energy equally.

As an example, consider an elastic collision of two equal masses with one mass initially at rest.

An example of such a collision is in the game of pool.

Grind it out in a calculation rather than guessing.

b) From here the problem is identical to 9.7 where we found �?� = �? −�<�? +�< �?� = −13�

�<� = 2�?�? +�< �?� = +23�

c) The final momentum of each car is�?� = − ?v�� and �<� = + cv��.

A convenient trick for computing energies given momentum is derived below O = 12��<

O = 12��<�<

O = �<2�

Using this we find O?� = ??©��< while O<� = cª��<.

Notice the two sum to the initial energy as expected.

Recall, in an elastic collision the kinetic energy remains constant!

9.17 Doing momentum (based on directions I drew in the figure) gives

Horizontal Vertical ��� = ��� 0 = �?�?�� +�<�<�� +�v�v�� 0 = −��?� + 2��2� cos 30� + 3��� cos 30� ��?� = 7�� cos 30

�?� = 7√32 �

��X = ��X −6�� = �?�?�X +�<�<�X +�v�v�X −6�� = ��?X + 2��2� sin 30� − 3��� sin 30� −2��2� sin 30� + 3��� sin 30� − 6�� = ��?X ��?X = −�� sin 30 − 6�� �?X = −132 �

Notice the final speed is

�? = ��7√32 ��< + T132 �U< = √79�

The direction is

� = tan;?¬ 132 �7√32 �­ = tan;? T 137√3U ≈ 47.0° Notice the sketch helps us ignore all the possible ways to screw up thinking about the signs.

Summary: the speed of the third fragment (mass �) is about 8.89� directed 47.0° below the negative x-axis.

In part b) it asks about the energy added. Remember, in an explosion some kind of potential energy is converted to

kinetic. There should be more kinetic energy in the fragments than in the original bomb. Initially we have O� = 12 �6���< = 3��<

Afterwards the kinetic energy in all three fragments is O� = 12 ���G√79�H< + 12 �2���2��< + 12 �3�����< = 45��< ∆O = 42��<

Notice I asked for the change (not percent change). As a percent the change is >1300%!!! Notice the overwhelming

majority of the energy is in the smallest fragment of the bomb. Those tiny masses can get going quickly and, since

energy depends on speed squared, carry a lot of energy.

7√32 � ≈ 6.06�

132 � = 6.5� √79� ≈ 8.89� 47.0°

9.17½

a) The smaller mass has larger acceleration when the applied forces are equal in magnitude.

Since both particles travel distance y, the mass with larger acceleration should finish the race in less time.

In equation form:

Smaller mass M Larger mass ® Y? = ��

Δ�? = 12Y?e?<

e? = �2Δ�?Y? = �2�y�

Y< = �a

Δ�< = 12Y<e<<

e< = �2Δ�<Y< = �2ay�

If a = 2� one finds e< = √2e? ≈ 1.41e? (e< is about 40% longer than e?).

b) For each mass, the same force is applied over the same displacement.

This implies the work done on each mass is the same.

Since the work done on each mass is the same, the change in kinetic energy of each mass must be the same.

Since both started with zero initial kinetic energy, the two masses have the same final kinetic energy!

In equation form:

Smaller mass M Larger mass ® w? = ΔO? w? = O�? − 0 �y = O�?

w< = ΔO< w< = O�< − 0 �y = O�<

When the same force is applied over the same displacement, �&aexperience the same ΔO!

c) The two masses have the same final kinetic energy.

Kinetic energy is directly proportional to mass and speed (squared).

Therefore, the smaller mass must have larger speed (when they have the same kinetic energy).

In equation form:

Smaller mass M Larger mass ® O = 12��?<

�? = �2O�

O = 12a�<<

�< = �2Oa

If a = 2� one finds �< = ?√< �? ≈ 0.71�< (�< is about 30% slower than �?).

d) We know from the previous part the masses have the same final kinetic energy. We also know � < a.

This implies we should try to write a formula that directly relates momentum, mass, and energy.

In this instance, it is much easier to understand the concepts by looking at the equations.

Smaller mass M Larger mass ®

O = 12��?< = 12� ��?��< = �?<2� �? = √2O�

O = 12a�<< = 12a ��<a�< = �<<2a �< = √2Oa

If a = 2� one finds �< = √2�? ≈ 1.41�? (�< is about 40% larger than �?).

The larger mass gains more momentum EVEN THOUGH THE SAME WORK WAS DONE!

9.18 This one is sneaky because the given velocity is relative to thrower instead of relative to a stationary observer.

a) In all of our problems thus far we have always talked about the momentum of objects relative to a

stationary observer. If the alien woman moves left with speed �? relative to earth and the ball moves to

the right with speed �< relative to the earth we know these speeds are related since ��°± = ��°² + ��²± ��°± = ��°² − ��±² �� = �<� − �−�?�� � = �< + �? IQ = I − In

WHEW! Now that that is settled we can do conservation of momentum just as before.

Comparing Stage 1 to Stage 2 (using only velocities relative to earth/stationary observer) 0 = −�±�? +�°�< 0 = −�±�? +�°�� − �?� �? = �°�± +�° �

Plugging back into the bold equation gives �< = � − �°�± +�° � = �±�± +�° �

As a check, THINK! If the alien is much bigger than the ball the alien should have almost no recoil

velocity AND the speed of the ball relative to the earth should be approximately �. Our equations match

this reasoning.

Now we can compare Stage 2 to Stage 3. Here I will assume �v is the speed of the 2nd alien and ball after

the catch. I will ignore the first alien as she is not involved in the collision. �°�< = ��± +�°��v �v = �°�± +�° �<

�v = �±�°��± +�°�< �

This is the answer to part a. Next part is on the next page…

b) Keep grinding away to get to the interesting part. I will assume �c is the speed of the 2nd alien after the 2nd

throw (relative to stationary) and �1 is speed of ball after 2nd throw (relative to stationary). Again we must

deal with the fact the ball is thrown with speed � relative to the thrower. We again find ��°± = ��°² + ��²± ��°± = ��°² − ��±² −�� = −�1� − ��c�� −� = −�1 − �c I  = I − I�

Momentum between stages 3 & 4 using velocities relative to stationary gives ��± +�°��v = −�°�1 +�±�c ��± +�°� �±�°��± +�°�< � = −�°�� − �c� + �±�c �±�°�± +�° � = −�°� + ��± +�°��c

T �±�°�± +�° +�°U � = ��± +�°��c

� �±�°�± +�° +�°��± +�°��± +�° �� = ��± +�°��c

��°�2�± +�°��± +�° � � = ��± +�°��c

�c = ��°�2�± +�°���± +�°�<

Plugging this result into the bold equation from relative velocities gives �1 = � − �c �1 = � − ��°�2�± +�°���± +�°�<

�1 = � ³1 − �°�2�± +�°���± +�°�< ´ �1 = � �±<��± +�°�<

Compare to �< (speed of ball after first throw). Notice the factor is squared after the second toss (for �1).

c) Is it fast enough to get there? Only if �1 > �?. This is true if � �±<��± +�°�< > �°�± +�° � �±< > �°��± +�°� Doing a quadratic one finds the ball will reach the 1st alien a 2nd time if �± > �° ?.√1< ≈ 1.618�°

FAR OUT! Not what I intended but totally bad ass. That funky number is the golden ratio �¶ = ?.√1< �!

d) I suspect (but have not proven) that each time the ball is thrown it slows a little bit as it is multiplied by

another factor of _·_·._¸. Therefore, after the �"# toss the I suspect the speed of the ball should be

�1 = � T �±�± +�°U%

If you repeat this process for the next two tosses you could hopefully verify this as well as coming up with

speeds of the left and right alien after each toss. After each toss the aliens speed up a little bit while the ball

slows down. At some point the game will end and the ball will be moving towards one of the aliens (but

not moving fast enough to catch it).

9.19 Momentum is conserved whenever there is no net external force OR whenever the collision time is small. If

the collision time is small, the external force acts for such a small amount of time the objects will not accelerate that

much during the collision.

9.20 Energy is conserved whenever there is no external work. Of course, from a practical standpoint, we usually

include the external work by writing O� + p� +w��" = O� + p�

9.21 While this may seem odd, consider the figure showing the trajectories of the objects in space whipping past

each other.

This is actually modeled well by an elastic collision since they never touch or lose energy by smashing into each

other. After they pass by each other they are attracted to each other by gravity which causes them to reverse

direction. Since gravity is a conservative force, the collision is elastic!

If you run the numbers in the collision you will see that momentum is conserved. Check it; did you remember the ±

on each velocity? When you check if energy is conserved you will see that the final energy is actually greater than

the initial energy. This is not possible for an elastic collision.

9.22 There are no external forces in the �-direction as rain falls into the box. This implies momentum is conserved

in the �-direction. Since mass is increased the speed must decrease to maintain constant momentum.

Going further: notice that momentum is not conserved in the vertical direction as the rain initially has downward

momentum but after the collision the box and rain combined have no vertical momentum component. This implies

that the normal force and �[ are not exactly equal! Strictly speaking the normal force upwards is slightly greater

then �[ since it not only must support the weight of the box and all drops in it but also slow down and stop each

drop that is impacting at any given instant.

9.23

a) This is another one of those strange problems where momentum and energy are conserved even though it is

some kind of explosion (two fragments separate). Think: during the separation the only external forces

acting on the block-wedge system is gravity on each object and normal force of floor on wedge. Note the

normal force between block and wedge is internal and problem says friction is negligible. Because gravity

is a conservative force, we may choose to account for it in the energy equation and say energy is conserved.

Last note: normal force on wedge acts perpendicular to motion of wedge and does no external work.

In the table below I choose to let �? be the speed of the wedge after they separate (relative to earth).

I also choose �< be the speed of the ice cube after they separate (relative to earth).

Momentum Energy ��� = ��� 0 = −5��? +��<  In = IQ

�[ℎ = 12 �5���?< + 12��<< 2[ℎ = 5�?< + �<<

Use the bold equation to eliminate �< in the energy equation and solve for �?. One finds 2[ℎ = 5�?< + �5�?�< 2[ℎ = 30�?<

�? = � 115[ℎ

b) Plug this result back into the bold equation to find

�< = �53[ℎ

Notice this speed is less than if the ramp was fixed in place. If the ramp was made to remain motionless,

by bolting it to the hockey rink, expect the speed of the ice cube to be \2[ℎ.

c) The speed of the cube relative to the wedge is ��°¹ = ��°² + ��²¹ ��°¹ = ��°² − ��¹² ��°¹ = �<� − �−�?�� ��°¹ = ��< + �?�� ��°¹ = �� 115 [ℎ + �53[ℎ� � ��°¹ = �� 115 [ℎ + �2515[ℎ� �

��°¹ = �[ℎ15 �1 + 5�� ��°¹ = �3615[ℎ� ��°¹ = �125 [ℎ�

Notice this last result is actually faster than if the block slid down a ramp fixed in place. If the ramp was

fixed in place we would expect the speed for the cube to be \2[ℎ. This seems very strange to me…

9.24

a) Think: we know momentum is conserved in an explosion as long as external forces are negligible. In this

instance we begin with no vertical momentum. After the explosion the ball is moving upwards and has

positive momentum in the Z-direction. Nothing seems to be moving downwards to provide a negative

momentum in the Z-direction which allows for no net vertical momentum after the collision.

REMEMBER: the entire earth is moving downwards after the collision (albeit at an infinitesimal rate). It is

impractical to measure the change in the earth’s momentum so it is pointless to use momentum in the

vertical direction to calculate the speed of the ball.

b) The problem says a stationary observer sees the ball leave with speed �. That is good news! We need not

think of relative velocity this time. Remember that when do momentum problems we have been using the

momentum of the objects relative to the stationary observer. ��� = ��� 0 = �?�?�� +�<�<�� 0 = −a�q +�� cos u �q = �a� cos u

c) If you really wanted to, we could assume the initial energy given to both objects came from the spring in

the gun. I suppose we could in theory calculate the spring constant if we knew the amount of compression

and used 12 ��< = 12�?�< + 12�<�q<

9.25

a) See the four stage pictures at right.

Conservation of energy applies between stages a and b. p� + O� +w ��"%$%;q$% = p� + O�

�?[º + 0 + 0 = 0 + 12�?�?&< �?& = \2[º

Conservation of momentum applies between stages b and c. �?�� + �<�� = �?�� + �<��

For perfectly inelastic the final mass stick together and move off

together as a single unit. Both have the same final speed. �?�� + �<�� = �� �?�?& + 0 = ��? +�<��q �q = �?�? +�< �?& = �?�? +�<\2[º

Conservation of energy applies between stages c and d. p� + O� +w ��"%$%;q$% = p� + O�

0 + 12 ��? +�<��q< + 0 = ��? +�<�[ℎ + 0 12 �q< = [ℎ 12 �q< = [º�1 − cos u� cos u = 1 − �q<2[º

u = cos;? �1 − �q<2[º�

� = »¼~;n �n − T MnMn +MQUQ�

Think about this. If �< = 0 the block reaches 90° which implies it reaches the same height as the initial

relase if it hits no mass…seems reasonable.

If �< = ½p¾¿ the block reaches ≈ 0° which implies it doesn’t go up at all hits no mass…seems

reasonable.

WATCH OUT! If �? = �< the angle is u = cos;? �vc� = 41.4° for a final height of ℎ = Àc. Notice the

final energy is not equal to the initial energy. In fact, 50% energy was lost!!!

Part b on next page…

º

�?

�<

ÁÂþ¿Y

�?& �? �<

ÁÂþ¿�

�q �Ä$"

ÁÂþ¿Å

u �Ä$" º ÁÂþ¿y

�Æ = 0 ℎ = º�1 − cosu�

b) If the collision was instead elastic then stage c, just after the collision

changes dramatically. I will assume �? reverses direction upon

impacting �<. If we get a negative value for �?q that simply means the

magnitude of �?q is correct but the direction it moves is opposite the

direction drawn.

The math derving �?& is unchanged from part a. �?& = \2[º

Conservation of momentum applies between stages b and c. This time I

will keep all the vectors in place so to show how the minus sign shows up. ��?� + ��<� = ��?� + ��<� �?�?&� + 0 = −�?�?q� + �<�<q� Here I want to stress that �?q is a speed…we expect it is positive. We

explicitly included a minus sign from the picture out front in the

equation…

Again, if we get a negative �?q that means the magnitude of �?q is correct

for the speed but the direction of velocity is opposite the direction drawn.

Cancelling the unit vectors and dividing by �? gives InÇ = −In{ +MQMn IQ{ UH-OH! We HAVE TWO UNKNOWNS (�?q & �<q)!!!

Fortunately, for an elastic collision, energy is conserved as well O?� + O<� = O?� + O<� 12�?�?&< + 0 = 12�?�?q< + 12�<�<q<

InÇQ = In{Q +MQMn IQ{Q

In this problem we know �< = 2�? or the mass ratio is _8_È = 2. The

bold equations above can be rewritten InÇ = −In{ + QIQ{ InÇQ = In{Q + QIQ{Q

Solving for In{ in the first bold equation and plugging it into the second

gives �?q = 2�<q − �?& �?&< = �2�<q − �?&�< + 2�<q< �?&< = 4�<q< − 4�?&�<q + �?&< + 2�<q< 0 = 6�<q< − 4�?&�<q

Watch out! When doing momentum, I like to factor out �<q instead of moving the negative term to the

other side and dividing by �<q. Wondering why? Re-visit the basketball problem (9.11) after finishing… 0 = 2�<q�3�<q − 2�?&� Notice there are two possible solutions: �<q = <v �?& = <v\2[º or �<q = 0. Just after the collision we

expect object 2 to change speed. Only the first option makes sense. In 9.11 the zero answer corresponded

to the change in velocity!

Note: Using this result in one of our above equations �?q = 2�<q − �?& gives �?q = ?v �?& = ?v\2[º

Problem continues on next page…

u< �<

º ÁÂþ¿y �<Æ = 0 ℎ< �?q

�?q �?

ÁÂþ¿Å

�< �<q

�?& �? �<

ÁÂþ¿�

º

�?

�<

ÁÂþ¿Y

u? �<

ÁÂþ¿g �?� = 0 ℎ? �<q

Now the problem will finally come together. Notice in stage d �< has reached max angle (max height) on

the right side while in stage e �? has reached max angle (max height) on the left side.

We can now consider conservation of energy for each block separately since no collision is taking place! ÉK = ÉJ for MQ between stages c and d ÉK = ÉJ for Mn between stages c and e p� + O� +w ��"%$%;q$% = p� + O�

0 + 12�<�<q< + 0 = �<[ℎ< + 0 12 �<q< = [ℎ< 12 �<q< = [º�1 − cos u<� cos u< = 1 − �<q<2[º

u< = cos;? �1 − �<q<2[º�

u< = cos;?¬1 − �23\2[º�<2[º ­

�Q = »¼~;n T ÊU ≈  Ë. �°

p� + O� +w ��"%$%;q$% = p� + O�

0 + 12�?�?q< + 0 = �?[ℎ? + 0 12 �?q< = [ℎ? 12 �?q< = [º�1 − cos u?� cos u? = 1 − �?q<2[º

u? = cos;? �1 − �?q<2[º�

u? = cos;?¬1 − �13\2[º�<2[º ­

�n = »¼~;n T¡ÊU ≈ QÌ. �°

Part c on next page…

c) In part c the equation the situation has changed. In part b we had a moving light mass hitting a stationary

heavy mass with an elastic collision and found the light mass got bounced backwards. Now we have the

opposite situation, a moving heavy mass impacts a stationary light mass. Our equations are still valid but

we find the two bold equations now read as follows with the mass ratio flipped InÇ = −In{ + nQIQ{ InÇQ = In{Q + nQIQ{Q

Solving for In{ in the first bold equation and plugging it into the second gives �?q = 12�<q − �?&

�?&< = T12 �<q − �?&U< + 12�<q<

�?&< = 14�<q< − �?&�<q + �?&< + 12�<q<

0 = 34 �<q< − �?&�<q

0 = �<q T34 �<q − �?&U

Notice there are two possible solutions: �<q = cv �?& = cv\2[º or �<q = 0. Just after the collision we expect object 2 to change speed.

Only the first option makes sense. Note: Using this result in one of our

above equations �?q = ?<�<q − �?& gives �?q = − ?v�?& = − ?v\2[º.

Recall this derivation assumed �?q was the speed of �? to the left just

after the collision (see stage c at right).

Negative motion to the left implies positive motion to the right.

We know �? will move negative to the left which is the same thing

as saying it moves positive to the right. Not sure where it will be

when �< flies off the track. It is probably already moving back down

from it’s max angle on the right quarter circle but I don’t know that

for sure. After doing chapter 10 you could come back and figure this

out…

ÉK = ÉJ for MQ between stages c and d p� + O� +w ��"%$%;q$% = p� + O�

0 + 12�<�<q< + 0 = �<[ℎ< + 12�<�<Æ< 12 �<q< = [ℎ< + �<Æ< 12 T43\2[ºU< = [º + �<Æ< 169 º[ = º[ + �<Æ<

�<Æ = �79[º

Finally, speed of 1 just after collision is same as part b; 1 reaches same max angle except on the right side!

�< ÁÂþ¿y �<Æ

ℎ< = º

? �feÎÏÐg�ℎgÐg�?ÎℎfÏhy�g, yf�ÒeÅYÐgÐ�[ℎe�f�

�?q �?

ÁÂþ¿Å

�< �<q

d) Elastic collision with mass ratio Friction does negative work on �? as it slides to the right across the flat

section. The energy equation comparing stage a to just before impact is p� + O� +w ��"%$%;q$% = p� + O�

�?[º + 0 + �−x�?[º� = 0 + 12�?�?&< �?& = \2[º�1 − x� Again the collision is elastic with �? = ?c�<. We would plug in our new mass ratio (

_8_È = 4) to find InÇ = −In{ + �IQ{ InÇQ = In{Q + �IQ{Q

Solving for In{ in the first bold equation and plugging it into the second gives �?q = 4�<q − �?& �?&< = �4�<q − �?&�< + 4�<q< �?&< = 16�<q< − 8�?&�<q + �?&< + 4�<q< 0 = 20�<q< − 8�?&�<q 0 = 4�<q�5�<q − 2�?&� Notice there are two possible solutions: �<q = <1 �?& = <1\2[º�1 − x� or �<q = 0. Just after the

collision we expect object 2 to change speed. Only the first option makes sense. Note: Using this result in

one of our above equations �?q = 4�<q − �?& gives �?q = v1 �?& = v1\2[º�1 − x�.

According to the problem statement, there is no friction on the arcs, only the flat section.

We are told �< reaches a max angle of 15°. Therefore, from an energy problem on �< comparing just after collision to its max angle 12�<�<q< = �<[ℎ< �<q< = 2[º�1 − cos 15°�

T25\2[º�1 − x�U< = 2[º�0.03704� 825[º�1 − x� = 2[º�0.03704� �1 − x� = 254 �0.03704� x ≈ 0.787

Now we know the coefficient of friction!!!

Does it have enough speed to cross the flat section before stopping? Do an energy problem for �? as it

travels to the left across the flat portion on its way back to the left quarter circle.

I will assume it makes it across and try to compute �?� as it exits.

• If I get a positive �?�, I know it makes it across and I can find the max angle on the left side.

• If I get a negative �?�, it will not make it across. I then assume it travels distance � < º across the

flat section and comes to rest. I can solve for � to see where it stops.

Work on the next page…

Assuming it makes it across we find

12�?�?q< − x�?[y = 12�?�?�< �?q< − 2x[º = �?�<

T35\2[º�1 − x�U< − 2x[º = �?�< 1825[º�1 − x� − 2x[º = �?�< 1825[º�1 − x� − 2x[º = �?�< 1825[º − 6825 x[º = �?�< 1825[º − 6825 �0.787�[º = �?�<

This gives a negative result for a squared number…makes no sense. Block didn’t make it across.

Reworking assuming it travels distance � then comes to rest. 12�?�?q< − x�?[� = 0

� = �?q<2x[

� = �35\2[º�1 − x��<2x[

� = 1825[º�1 − x�2x[

� = 1825 �1 − x�2x º

� = 1825 G1 − �0.787�H2�0.787� º Ó ≈ ¢. ¢ÊÌ�Ô

Remember this distance is from the bottom of the right quarter-circle.

e) QUICKIE: for an equal mass, 1D, elastic collision the objects switch speeds! Think about a perfect shot in

pool where the cue ball stops and the ball impacted by the cue ball goes flying off…this is a great

approximation to an 1D elastic collision of two objects of equal mass with one initially stationary.

We expect m1 will stop, transferring all momentum to the m2. Since m2 has the same momentum as m1

did originally, it will just barely reach the top of the right arc. It will come back down and smack into m1

again. This time m2 will stop and m1 will go flying off again. The process repeats as long as the work

done by friction is negligible. In real life this would start to slow down after a few oscillations.

Notice this is very similar to a Newton’s cradle.

Think: would this work with balls rolling on a curved track? Would the rotational motion somehow mess

with our solutions?

9.26 Do a web search and you can easily find some online. Try making one and compare to what is already out

there…

9.27

a) My guess is below and to the left of the origin. Why? More mass below origin than above. More mass left

of origin than to the right. While this reasoning is not always true, it seems plausible in this case. I

suppose my gut tells me the center of mass is approximately at the coordinate �−20cm,−20cm�. b) For the �-direction we use �ÕÖ = �?�? +�<�< +�v�v�? +�< +�v

�ÕÖ = �1kg��−20cm� + �2kg��40cm� + �3kg��−60cm�6kg �ÕÖ = −20cm

ZÕÖ = �?Z? +�<Z< +�vZv�? +�< +�v

ZÕÖ = �1kg��60cm� + �2kg��−20cm� + �3kg��−40cm�6kg ZÕÖ ≈ −17cm

As Tom Petty says, “Even the losers get lucky sometimes.” My guess is pretty close.

Notice: the units on top and bottom for mass should always drop out. We can use grams, kilograms, etc

without worrying about it.

9.28 Note: we can use weights instead of mass in the center of mass equation. Why? We can multiply all numbers

in the top and bottom of �CM = �1�1+�2�2�1+�2 by [. Note: this is only valid as long as all objects experience the same

value of [. This might not be true for a system of very large clouds or groups of spaceships spread out in the upper

atmosphere. More on this in Chapter 13.

a) For a mobile to be balanced we want the center of mass of each rod to be located at the point where a string

connects to the rod. It is often convenient to call the center of mass � = 0. For the bottom rod we find �ÕÖ = �v�v + �<�<�v + �<

0 = �v�v + �<�<�v + �< 0 = �v�v + �<�< �v = −�< �<�v

In center of mass problems it is always important to notice the distinction between distance and position.

Remember we said origin of the coordinate system for the lower rod was at the center of mass (little star).

For the lower rod we know �< is distance y< = 42cm to the left of the origin. Therefore �< = −42cm.

Plugging in we find �v = −�8lbs� �−42cm��+56cm� = 6lbs b) The total weight hanging from the left end of the upper rod is �v + �< = 14lbs. Based on the sketch the

distance from the left end to the upper rod’s center of mass (upper rod’s star) is yv = 56cm. This means

the distance from the star to the mass on the right end of the upper rod is r − yv. I will once again set the

origin at the center of mass of the upper rod. �ÕÖ = ��v +�<��Ù + �?�À�v + �< + �?

0 = ��v +�<��Ù + �?�À�v + �< + �? 0 = ��v +�<��Ù + �?�À 0 = ��v + �<��−yv� + �?�r − yv� r = Ú�v + �<�? + 1Û yv

r = Ú14lbs16lbs + 1Û 56cm = 105cm

9.29

a) The slice has thickness yZ. Thus we will eventually find ZÕÖ.

b) For a uniform 3D object the density is given by Ü = efeYh�YÎÎefeYh�fhÏ�g

Ü = a�12 r½�w

Ü = 2ar½w

c) The height varies linearly. The equation of a line in the ZÝ-plane is Ý = Îhf�g ∙ Z + ��egÐÅg�e Îhf�g = Ð�ÎgÐÏ� = −½r ��egÐÅg�e = ½ Ý = −½r Z + ½

Ý = ½r �r − Z� Note: it is easy to flip something or forget a minus sign. Always check by plugging in a point or two.

When Z = 0 we get Ý = ½ as expected. Simlarly, when Z = r we get Ý = 0 as expected.

In case you are wondering why I bothered to rewrite in that last form…see the next result.

d) A 3D object makes a 3D slice. The mass of a 3D slice is given by y� = ÜyÞ y� = T 2ar½wUÝwyZ

y� = T 2ar½wU Ú½r �r − Z�ÛwyZ

y� = T2ar< U �r − Z�yZ

Now perhaps you understand why I bothered to do that extra step to rearrange Ý in the previous part. Not

that important, but it does keep things neatly organized.

e) The center of mass integral is written as ZÕÖ = 1aßZy�

ZÕÖ = 1aßZ T2ar< U �r − Z�yZ

ZÕÖ = 2r<ß�rZ − Z<�yZ

Though it may seem silly, why not check the units before you do the integral. If the units are wrong for

some reason you might be able to catch a mistake before wasting your time doing a worthless integral.

Don’t forget: yZ has units of meters!!! The units check out.

Think: what are the limits we should use for this problem? Smallest possible value of Z to biggest: 0 to r. ZÕÖ = 2r<ß �rZ − Z<�yZÙ/

ZÕÖ = 2r< ³rZ<2 − Zv3 ´/Ù

ZÕÖ = 2r< �rv2 − rv3 � = 2r< �rv6 � = 13 r

w r

½

Z � F

�à

Part e) continued from previous page:

I remember that, for right triangles, the center of mass is “1/3 from the fat end”.

Notice thickness w dropped out! Therefore this is true for 2D or 3D right triangles.

Notice height ½ dropped out! That means this is valid for tall or short right triangles.

f) Your gut instinct is most likely to think there is equal mass on either side of the center of mass. I know I

thought that way. I mean, c’mon, the term even says CENTER. In general this is NOT true. Even in this

simple case you can show there is not equal mass on either side of the center of mass.

The volume to the right of the center of mass is 29½re while the volume to the left of the center of mass is 518½re. You can easily check this sums to the total volume of the piece. Since the piece has uniform

density we know volume is directly proportional to mass. Notice there is more mass on the left side than on

the right!

What’s up with this? Center of mass really relates directly to torque or moments. We will learn more in

chapter 10 on this. It suffices to say, for now, a light mass far from the center of mass can balance a heavy

mass close to the center of mass.

Random side note: in statistics similar integrals exist where you take moments about the mean. The

process is similar. Check out kurtosis today!

g) Rather than redo all this math, notice that determining ÝÕÖ is trivial. We already know that the center of

mass is located ?v from the fat end of a right triangle. Immediately write down ÝÕÖ = áv .

Since the object is uniform, we expect the �ÕÖ is in the middle of the object. Write down �ÕÖ = −¹<

The center of mass position is thus �−¹< , Ùv , áv �.

9.30

a) The slice has thickness yZ. Thus we will eventually find ZÕÖ.

b) A thin plate is typically considered a 2D object. Use â. Note: we are told the plate is uniform so we may

assume â is constant.

c) Consider the figure at right. We see �< + Z< = º< � = �\º< − Z<

Note: here use the positive root as we want the width of the slice (a + quantity). Also,

do a quick check…when Z = 0 the width of the slice is 2º. When Z = º the width is

zero. Both limits produce results we expect. yà = 2�yZ yà = 2\º< − Z<yZ

d) y� = âyà = 2â\º< − Z<yZ

e) The center of mass integral is written as ZÕÖ = 1aßZy�

If I am working with density I usually re-write in this form ZÕÖ = ãZy�ãy�

ZÕÖ = ãZG2â\º< − Z<yZHã2â\º< − Z<yZ

If the density is uniform, only then can it factor out.

ZÕÖ = 2â ãZ\º< − Z<yZ2â ã\º< − Z<yZ

ZÕÖ = ãZ\º< − Z<yZã\º< − Z<yZ

On an exam I would suspect you would use a table…but why not have some fun!

For the upper integral a substitution of Ï = º< − Z< seems likely to work.

For the lower integral, I recall we used a triangle (Pythagorean theorem) to get that equation. That

makes me think I should try a trig sub. Redrawing the triangle at right I see Z = º sin u pop out.

Work continues on the next page.

º Z

�

2�

º Z

u

Top of fraction Bottom of fraction

ß Z\º< − Z<yZÀ/

Let Ï = º< − Z< yÏ = −2ZyZ ZyZ = −12yÏ

When Z = 0 get Ï = º<, when Z = º get Ï = 0 −12ß Ï?/<yÏ/À8

12ß Ï?/<yÏÀ8/

12 Ïv/<32 ä/À8 = 13ºv

ß \º< − Z<yZÀ/ Z = º sin u yZ = º cos u yu

Z< = º< sin< u \º< − Z< = \º< − º< sin< u = º\1 − sin< u = º\cos< u = º cos u

Notice u goes from å/4 to 0 in radians

If Z = 0 then u = 0

If Z = º then u = æ<

ß �º cos u��º cos u yu�æ/</

º<ß cos< u yuæ/</

º<ß 1 + cos 2u2 yuæ/</

º< Ú12 u + sin 2u4 Û/æ/<

The second term is zero with either limit giving å4 º<

From all this we get

ZÕÖ = ãZ\º< − Z<yZã\º< − Z<yZ

ZÕÖ = 13 ºvå4 º<

ZÕÖ = 43å º ≈ 0.4244º

f) After all that this will seem like a joke. Since the density is uniform â = efeYh�YÎÎefeYhYÐgY

â = a12åº<

a = 12åº<â ≈ 1.571º<â

g) The angle corresponding to Z = cvæº is given by Z = º sin u giving u ≈ 25.11° = 0.4383rad. The mass

portion of the plate below the center of mass line is thus

ß 2â\º< − Z<yZcvæÀ/ = 2âº< Ú12 u + sin 2u4 Û//.cv©v = 0.8226âº< = 0.524a

9.31

a) Split the thing into three pieces as shown at right.

Each piece is labeled with a number (1, 2, or 3) for

ease of communication. Notice the triangle’s center

of mass is located “1/3 from the fat end” both

horizontally and vertically. The mass of each piece

should be area times 2D density (â). �? = YÅâ �< = �Åâ �v = 12Yyâ

�? = − Å2

�< = − Å2

�v = +y3

�ÕÖ = �?�? +�<�< +�v�v�? +�< +�v

�ÕÖ = �YÅâ� �− Å2� + ��Åâ� �− Å2� + �12Yyâ� �+y3�YÅâ + �Åâ + 12 Yyâ

First thing I would do is multiply every term (in both numerator and denominator) by 2 to reduce the

number of fractions. Notice every term (in both numerator and denominator) has â…that will drop out.

�ÕÖ = �YÅ��−Å� + ��Å��−Å� + �Yy� �y3�2YÅ + 2�Å + Yy

�ÕÖ = Yy<3 − �Y + ��Å<2�Y + ��Å + Yy

b) Based on the figure there appears to be much more mass on the left. I would expect �ÕÖ < 0 if the figure

is drawn to scale.

c) If the center of mass is to be above the origin we know �ÕÖ = 0. Remember, the only way a fraction can

equal zero is if the numerator is zero. This occurs when Yy<3 = �Y + ��Å<

y = �3�Y + ��Y

If � = 0 we find y = Å√3. Therefore we need Å < y < 2Å if � = 0.

If � = Y we find y = Å√6 ≈ 2.45Å. Therefore we need y > 2Å if � = Y.

Note: if Y = vc � then y = 2Å.

y/3

2Y3

Å/2 Y2 �2 n

Q

�

9.32 I would think of this object as shown in the figure at right. Notice

that the diagonal of the square is º. Since the diagonal of a square is √2

times the side we know º = √2Î or Î = À√<. �? = åº<â �< = Î<â = º<2 â �? = 0 �< = + Î2 = º2√2

To use the center of mass equation we must remember we have subtracted out the mass of the square. One way to

handle this is to explicitly change all the masses for the square to include negative signs. �ÕÖ = �?�? −�<�<�? −�<

Note: I notice in advance the 2D density â will appear in all terms (both numerator and denominator). It will drop.

�ÕÖ = −º<2 T º2√2Uåº< − º<2

�ÕÖ = − 14√2å − 12 º

�ÕÖ ≈ −0.0669º

Notice this is essentially a 7% shift to the left (and down). The shift is always away from the hole.

9.33

a) I would expect it to be closer to the dense end (right end). We expect �ÕÖ > Ù<. b) Doing unit math: èéê = èëêè�ê èëê = èéêè�ê = kgm<

c) The integral is based on the figure shown at right a = ßy�

a = ß ë�y�Ù/ = 12ër<

d) Doing the center of mass integral we find �ÕÖ = 1aß�y�

�ÕÖ = 112 ër< ß ��ë�y��Ù/

�ÕÖ = 13ërv12 ër< =23 r

e) A right triangle with sharp tip at origin has mass that increases proportional to distance from origin (just

like a linear mass density). Maybe you can think of a better way to say it than that…I’m in a hurry to print.

− = n Q

�

y� y� = éy�

9.33n�

a) The density function changes. The limits of integration also change!

�ÕÖ = ã ��ë�<y��.Ù/<;Ù/<ã �ë�<y��.Ù/<;Ù/<

Perhaps you notice, by symmetry, �ÕÖ = 0. No need to do any calculation at all!!!

b) If one plugs in a value for � to the left of the origin (� < 0) one gets a negative value for density! This

makes no sense. When we create mathematical models of matter, we must check that our models and

coordinate systems do not produce unphysical nonsense.

9.33Q�

�ÕÖ = ã ��Y�%y��Ù/ ã �Y�%y��Ù/

�ÕÖ = Y ã �%.?y�Ù/ Y ã �%y�Ù/

The Y’s cancel (do this AFTER pulling them out of the integral).

�ÕÖ = �%.<� + 2ì/Ù�%.?� + 1ì/Ù

In this case, plugging in the zero limit drops out each of those terms.

�ÕÖ = r%.<� + 2 − 0� + 2r%.?� + 1 − 0� + 1

�ÕÖ = r%.<� + 2 ∙ � + 1r%.?

�ÕÖ = � + 1� + 2r

�ÕÖ�� + 2� = �� + 1�r

From there, plug in #’s. A common error is in misunderstanding the 14.29 cm comment in the problem statement.

The center of mass is �ÕÖ = 1.00m − 0.1429m = 0.8571m.

I found � = 5.00.

9.34

a) We want to approximate this density change as being linear. We therefore expect the density to be of the

form é��� = Îhf�g ∙ � + ��egÐÅg�e At the left end (� = 0) the density is éÙ while at the right end (� = r)

density is éÀ. Making a plot can help make sense of this. See figure at right.

Notice the slope is given by Îhf�g = Ð�ÎgÐÏ� = éÀ − éÙr

while the intercept is éÙ. é��� = éÀ − éÙr � + éÙ

Before moving on, check your work by plugging in both � = 0 and� = r to

verify the equation gives appropriate results. In this case it does.

b) We could use the center of mass integral like �ÕÖ = 1aß�y�

But I prefer to set up this one like this �ÕÖ = ã�y�ãy�

Before moving on, make sure you see this is essentially the same equation.

As before we expect y� = éy� giving

�ÕÖ = ã �éÀ − éÙr �< + éÙ�� y�Ù/ã �éÀ − éÙr � + éÙ� y�Ù/

Inspect the numerator closely. Notice I already multiplied in the additional factor of �! To eliminate

fractions, I would multiply all terms in the numerator and denominator by r.

�ÕÖ = ã G�éÀ − éÙ��< + éÙr�Hy�Ù/ã G�éÀ − éÙ�� + éÙrHy�Ù/

�ÕÖ = �éÀ − éÙ� �v3 ì/Ù + éÙr �<2 ì/Ù�éÀ − éÙ� �<2 ì/Ù + éÙr�|/Ù

�ÕÖ = �éÀ − éÙ� rv3 + éÙ rv2�éÀ − éÙ� r<2 + éÙr<

�ÕÖ = r 13 �éÀ − éÙ� + 12 éÙ12 �éÀ − éÙ� + éÙ

If desired, multiply each term by 6 to reduce the number of fractions. �ÕÖ = r 2�éÀ − éÙ� + 3éÙ3�éÀ − éÙ� + 6éÙ

c) Plugging in numbers gives �ÕÖ : 0.595r = 29.7cm. This is closer to the high density end as expected.

��

� r

éÙ éÀ

9.35 The point of this is to help you practice your set-ups. You would want to slice it vertically and find y� = âyà y� = â�Zy�� y� = â TÐ�ÎgÐÏ� ∙ � + ��egÐÅg�eU y�

y� = â T−½r ∙ � + ½U y�

y� = �ë − î�<� T−½r ∙ � + ½U y�

y� = 1r �ë − î�<��−½� + ½r�y�

From there use the center of mass integral in the form

�ÕÖ = ã�y�ãy�

�ÕÖ = ã� �1r �ë − î�<��−½� + ½r�y��ã �1r �ë − î�<��−½� + ½r�y��

�ÕÖ = ã�ë − î�<��−½�< + ½r��y�ã�ë − î�<��−½� + ½r�y�

Lastly, notice you can factor out the ½ from the second term in both the numerator and denominator giving

�CM = ã �ë − î�2��r� − �2�y�r0ã �ë − î�2��r − ��y�r0

9.36 Split this into three pieces as shown in the figure. Note: alternate coordinate system

solution shown below. Recall the problem statement indicated that �< = �v!!! �ÕÖ = �?�? +�<�< +�<�v�? + 2�<

The problem also stated we want to balance on the edge of the disk. Based on this

information we know �ÕÖ = º. Finally, notice that �? = 0 while �< = �v = Ù< cos u.

º = 2�< �r2 cos u��? + 2�< ºr = �<�? + 2�< cos u

cos u = ºr ∙ �? +�<2�<

u = cos;? �ºr T2 + �?�<U�

Alternate solution with origin of coordinate system at center of mass.

Using the red coordinate system (shown at right) we know �ÕÖ = 0.

Furthermore, notice �? = −º while �< = �v = Ù< cos u − º.

0 = �?�−º� + 2�< �r2 cos u − º��? + 2�<

From this point, algebra gives the same result.

r2 cosu

n

Q

�

b

n

Q

�

b

9.37

a) First I assumed Z��� = Å/ + Å?� + Å<�<. When � = 0 (left boundary) we find Z�0� = Å/. But looking at

the figure you can see Z�0� = Y. This tells us Å/ = Y. Now we know Z��� = Y + Å?� + Å<�<.

Because the minima occurs at the left end we know ÆXÆ� = 0 when � = 0. yZy� = Å? + 2Å<�

Since ÆXÆ� = 0 when � = 0 we see Å? = 0. Now we know Z��� = Y + Å<�<.

Using the right boundary (� = r) we see Z�r� = Y + Å<r< = �. We find Å< = &; Ù8 . Putting it all together

we find Z��� = Y + &; Ù8 �<.

Note: you don’t have to do things in the same order I did. By using all three conditions you get three

equations and three unknowns. From that point you can solve in any order you like.

b) To get mass use

� = ß y���% '�%�"� '

� = ß âZy�Ù/

� = ß â TY + � − Yr< �<U y�Ù/

� = âY� ì/Ù + â T� − Yr< U �v3 ï/

Ù

� = ârY3 T2 + �YU

Think: if � = Y we have a rectangle and mass is � = ârY as expected. With tricky problems it is always

good to think of simple cases to give you ways to check your work!

c) Center of mass integral gives

�ÕÖ = ã ��âZ�y�Ù/ã �âZ�y�Ù/

Since density is uniform it can pull out of both the top and bottom integral and cancel out.

�ÕÖ = ã �Zy�Ù/ã Zy�Ù/

�ÕÖ = ã � �Y + � − Yr< �<� y�Ù/ã �Y + � − Yr< �<� y�Ù/

�ÕÖ = Y �<2 + � − Yr< �c4Y� + � − Yr< �v3 ð/

Ù

�ÕÖ = 3r4 �Y + ���2Y + �� �ÕÖ = 3r4 �1 + �Y��2 + �Y�

Think! Is � = Y we have a rectangle and center of mass should be at �ÕÖ = Ù<. Answer checks.

d) You’re on your own.

9.38

a) We know é = Y�?/v which implies èéê = èYêè�ê?/v or èYê = èñêè�êÈ/ò = óô4 �È/ò = 23�õ/ò b) I would do this: �!$Æ = ßy�

�!$Æ = ßéy�

�!$Æ = ß Y�?/vy�Ùö<À/

�!$Æ = ÷Y�c/v43 ø/<À

�!$Æ = 34Y�2º�c/v

Sure…you could try to clean this up…but why? The point here is to practice the procedure…

c) I would do this:

�ÕÖ = ã�y�ãy�

�ÕÖ = ã �GY�?/vHy�Ùö<À/ 34 Y�2º�c/v

�ÕÖ = ã Y�c/vy�Ùö<À/ 34 Y�2º�c/v

�ÕÖ =÷Y�0/v73 ø

/<À

34 Y�2º�c/v

�ÕÖ = 37Y�2º�0/v34 Y�2º�c/v

�ÕÖ = 47 �2º�0/v�2º�c/v

�ÕÖ = 47 �2º� = 87º : 1.143º

Think: this makes some sense because we know the density function is increasing as � increases. This

implies the right end of the rod is more dense than the left end of the rod. Therefore we expect the center

of mass of the rod should be to the right of the middle of the rod. The middle of the rod is º…our answer

is 8/7º…1/7º to the right of the center. Seems reasonable.

d) Without hole the mass of the rod is a = âåº<.

e) The chunk of metal you cut out of the disk (the hole) has mass � = âÎ< = â �Àc�< = ùÀ8?ú . If you want to,

you could think of the hole as having negative mass or negative density…whatever floats your boat.

CONTINUES NEXT PAGE…

f) The center of mass equation becomes �ÕÖ = �`$'�ÆÆ�`��`$'�ÆÆ�`� − �#$'��#$'� + �!$Æ�!$Æ�`$'�ÆÆ�`� −�#$'� +�!$Æ

Óûü = �−Ô��ýþÔQ� − �−Ô� TýÔQnË U + �¡ÌÔ� �����QÔ��/��ýþÔQ − ýÔQnË + ����QÔ��/�

NOTE: this could be cleaned up…but why? The point is to practice the set-ups here…

Note: if you said the mass of the hole is negative you would write �ÕÖ = �`$'�ÆÆ�`��`$'�ÆÆ�`� + �#$'��#$'� + �!$Æ�!$Æ�`$'�ÆÆ�`� +�#$'� +�!$Æ

Upon plugging in your numbers you should still get the exact same bold equation shown above.

g) THINK. A bigger hole on the left implies less mass on the left which, in turn, implies the center of mass

shifts to the right!

h) Challenge is no joke. To be honest I’m not sure right away. My intuition says increasing the exponent

should tend to shift mass away from the origin and shift the center of mass to the right. That said, I’m

pretty good with thinking about how integer powers affect the density function, I’m not so hot with

thinking about how fractional powers affect the density function. What I would do is redo part c:

�ÕÖ = ã�y�ãy�

�ÕÖ = ã �G��?/<Hy�Ùö<À/ã ���?/<�y�Ùö<À/

�ÕÖ = � ã �v/<y�Ùö<À/� ã �?/<y�Ùö<À/

�ÕÖ =÷�1/<52 ø

/<À

÷�v/<32 ø/<À

�ÕÖ = 65º = 1.2º

Looks good. Increasing the exponent does tend to shift the mass to the far end of the rod.

9.39 Search for it on the internet.

9.40 Answer given in problem. Here is how I would

try to set it up.

Mass of slice is y� = ÜåZ<

y� = Üå ��º< − º?�½ ∙ � + º?�<

�ÕÖ = ã � �Üå T�º< − º?�½ ∙ � + º?U<� y�á/ã �Üå T�º< − º?�½ ∙ � + º?U<� y�á/

�ÕÖ = Üå ã � �T�º< − º?�½ ∙ � + º?U<� y�á/Üå ã �T�º< − º?�½ ∙ � + º?U<� y�á/

�ÕÖ = 1½< ã � �G�º< − º?� ∙ � + º?½H<� y�á/1½< ã �G�º< − º?�� + º?½H<� y�á/

�ÕÖ = ã � �G�º< − º?� ∙ � + º?½H<� y�á/ã �G�º< − º?�� + º?½H<� y�á/

From there you are off to the races. Since this is so gnarly, it might be worth checking some simple cases as

mentioned in the problem.

Think: a case you should check with the internet is º? = 0 and º< = º.

Think: a case you should check with the internet is º? = º and º< = 0.

Think: what should happen if º? = º< = º?

Think: I’m showing a side view here. Slice is a disk.

Radius of slice is Z. Notice the radius changes linearly

from one end to the other. � = |���� ∙ Ó + K���{��� � = K|�� ∙ Ó + K���{��� � = �ÔQ − Ôn�� ∙ Ó + Ôn

9.41 Procedural hints and answer given in problem.

ZÕÖ = ã Zy�.æc;æcã y�.æc;æc

Watch out! Because the angle is to the vertical we should use Z = ºÅfÎu!!!

ZÕÖ = ã º cos u éºyu.æ/c;æ/cã éºyu.æ/c;æ/c

Density and radius º are constant and factor out. After factor out we can cancel all but a single R in the numerator.

ZÕÖ = º ã cos u yu.æ/c;æ/cã yu.æ/c;æ/c

ZÕÖ = º sin u|;æ/cæ/cu|;æ/cæ/c

ZÕÖ = º sin å4 − sin �−å4�å4 − �−å4�

You might have realized we could have used an even function trick. If you saw that…awesome. If you want to

know what I mean, come to office hours. �ûü = ¢. Ê¢¢Ô

9.42

a) I found the result �CM = 13 �r + ��. Hint: consider a right triangle with a right triangular hole cut from it.

b) When � = 0 you are back to a normal right triangle…the result should be 1/3 from the fat end. Check!

c) When � : r this is very similar to a non-uniform rod with linear mass density. The result should be 1/3

from the high density end (�CM : 23 r). Check.

9.43 Pretty much done for you in the problem statement.

9.44

a) No NET external force on person-board system. True, the floor exerts an external normal force. True, the

earth exerts an external gravitational force. These two forces balance. Remember, they balance even

though they are NOT action-reaction pair!!! Friction was claimed to be negligible. No NET external force.

b) The center of mass OF THE SYSTEM remains motionless if momentum is conserved.

c) You could use conservation of momentum to relate the speeds of the board and the person (relative to

earth). For more on this, consider the next problem!!!

d) The figure at right gives the distances between various points.

Notice the position of the board’s center of mass is �? = −Tr4 − �U = � − r4

Giving �ÕÖ = �?�? +�<�<�? +�<

�ÕÖ = � �� − r4� + a�� +a

Don’t forget we initially set the system center of mass at zero. This

means, in the after picture, the system center of mass is STILL at zero.

Rearrange to find � = r _c�_.�� = 0.125m.

�

r4 r4 − �

9.45

a) One finds ���² = ���° + ��°².

b) One finds a���² +���°² = 0. Note: in this equation we expect to find ��°² < 0 since it moves to the left.

From the perspective of an observer on the earth there are no external forces on the system. Therefore

momentum is conserved (Newton’s 3rd law applies) for velocities relative to the earth. If one considered

the person as a system, there would be an external force from the board and thus momentum wouldn’t be

conserved. Newton’s 2nd law would then apply.

c) From step b) we find ���² = −_� ��°². Plugging this result into step a) and solving for ��°² gives

��°² = ;??. � ���° = ;??.È�óô��óô2.0�� � = −1.75�� �. Likewise one finds ���² = ??.� ���° = +0.25�� �. d) We find ��ÕÖ = ��o���._�o�¸��._ = ��; ��o�¸��._�o�¸��._ = 0.

9.46 I just updated problem statement to read (“tension in the upper string is

not �"$"[”). The original problem statement was total screw-up on my part.

Students with editions before April 2020 have my apologies.

Consider everything within the purple box (dotted line) as a system.

We know

��$%`X`"�_ = �`X`"�_Y�`X`"�_

Assuming everything has F implied, we can ignore the �′Î. Â ((�! −�`X`"�_[ = �`X`"�_Y`X`"�_

Here, Y`X`"�_ is the same thing as Y��!

Tio be clear, Y�� also has an implied F on it in this case. ���� = M|�|��M�� + ��®� By taking two derivatives (with respect to time) of Z�� equation one finds Y�� = Y?�? + Y<�<�? +�<

Because of the implied F, Y?&Y< are acceleration vectors (not magnitudes).

We must include � signs. In this case Y? � Y<.

Furthermore, from the basic Atwood’s machine problem in Ch 5 we know

acceleration magnitude is Y = [�< −�?�< +�? = [�< −�?�`X`"�_

Since both blocks have equal acceleration magnitude but opposite directions I will use Y? = Y & Y< = −Y. Y�� = Y?�? + Y<�<�? +�<

Y�� = Y�? − Y�<�`X`"�_

Y�� = Y�? −�<�`X`"�_

Here is makes sense to factor out a minus sign on top (because �< µ �?). Y�� = −Y�< −�?�`X`"�_

Now plug in Y = [ _8;_È_�����

Y�� = −�[�< −�?�`X`"�_ ��< −�?�`X`"�_

Y�� = −[��< −�?�`X`"�_ �<

Plug this into the bold equation above: ���� = M|�|��M�� + ��®�  ((�! = �`X`"�_ [ − [ ��< −�?�`X`"�_ �<!

 ((�! = �`X`"�_[ "1 − ��< −�?�`X`"�_ �<# Finally, using �? = �&�< = 2� gives

���� = �M� "n − � MM|�|��M�Q#

Mn MQ

����

�

9.47 On my to do list…

9.48

a) The initial height is 2.0 m as read from the graph. It is debatable how many sig figs one could get reading

the plot so I will err on the side of caution and assume only 2 sig figs but keep my extra digits just in case.

The initial slope of the Ze-plot is zero. From this we can tell the initial velocity is zero. Impact speed is

estimated using an energy problem: �[ℎ� + 12���< +w��" = �[ℎ� + 12���<

�[ℎ� + 0 + 0 = 0 + 12���<

�� = \2[ℎ� = �2�9.8 ms<� �2.0m� ≈ 6. 26ms

Note: I ignored air resistance. The max speed occurring just before impact is about ≈ 6�� ≈ 12mph.

Drag starts to play a role but ignoring drag seems appropriate for a rough estimate valid only to 2 sig figs…

WATCH OUT! After careful inspection of the plot appears the ball hits the ground somewhere around

0.25m, not zero! How can one tell? We see the ball start to accelerate upwards at about this position.

More on this later. Using this we get the slightly improved value of ℎ� on finds �[ℎ� + 12���< +w��" = �[ℎ� + 12���<

�[ℎ� + 0 + 0 = �[ℎ� + 12���<

�� = �2[Gℎ� − ℎ�H = \−2[Δℎ = �2 �9.8 ms<� �1.75m� ≈ 5. 86ms

b) Just after impact to the final height we see a similar situation. The final slope on the Ze-plot is again zero

implying the final speed is again zero. I will again ignore drag. 0 + 12���< + 0 = �[ℎ� + 0

�� = �2[ℎ� = �2 �9.8 ms<� �0.75m� ≈ 3. 83ms

Again, the more refined estimate using the non-zero initial height looks like �[ℎ� + 12���< + 0 = �[ℎ� + 0

�� = �2[Gℎ� − ℎ�H = \+2[Δℎ = �2 �9.8 ms<� �0.50m� ≈ 3. 13ms

c) Here I had a major advantage over you because I could see the raw data table. My hope was you would

look on the Ze-plot for a region positive concavity. Concave up on a Ze-plot indicate positive acceleration.

This can only occur when the ground is in contact with the ball. I estimated the time interval as 140msec using the two times where the ball is at position Z = 0.25m. See plots next page after part f…

d) Impulse is E� = Δ�� = ����&^`" �"�!�_( q" − ��&^`"&��$!��_( q" � = 0.500kg �3. 13ms �+F� − 5. 86ms �−F�� ≈ 4. 50 kg ∙ ms F e) Average force is given by

�� �' = E�Δeq$''�`�$% = 4. 50 kg ∙ ms F140msec ≈ 32. 1NF f) The average acceleration is thus

Y� �' = �� �'� = 32. 1N0.500kg ≈ 64 ms< F ≈ 6.4[Òsdirectedupwards

g) Plots of position, velocity, and acceleration versus time are shown below. The red dot indicates the start of

impact and the purple dot indicates the end of impact on each plot. Notice the average acceleration of

about 64 ��8 F corresponds to a positive slope. Notice the peak acceleration is much larger than the average.

0.00

0.25

0.50

0.75

1.00

1.25

1.50

1.75

2.00

0.00 0.25 0.50 0.75 1.00

y (t)

t (s)

y vs t

-6.0

-5.0

-4.0

-3.0

-2.0

-1.0

0.0

1.0

2.0

3.0

4.0

0.00 0.25 0.50 0.75 1.00

v (m/s)

t (s)

v vs t

-10

0

10

20

30

40

50

60

70

80

90

0.00 0.25 0.50 0.75 1.00

a (m/s2)

t (s)

a vs t

9.49

a) Imagine you are an observer at rest watching the ship go by. The fuel does not move backwards with speed �g. As an example, if the rocket moves to the right at � = 100�� and is ejecting fuel backwards �g =500 ms (relative to the rocket) you observe the fuel to move backwards at speed �g − � = 400 ms relative to

you!

b) a� = �a − y���� + y�� − y���� − �� c) Now eliminate y� using ya = −y�. a� = �a + ya��� + y�� + ya��� − ��

Remember, the mass of the rocket is decreasing because ya < 0 while the speed of the rocket is increasing y� > 0.

d) Factoring it all out gives a� = a� +ay� + �ya + yay� + ��ya − �ya

Cancelling stuff gives 0 = ay� + yay� + ��ya

The term yay� is negligible compared to the other terms. Why? This term has two differentials (two tiny

quantities) while the other terms each have only one differential (one tiny thing). If for some reason it was

non-negligible, one could simply cut up the fuel into smaller and smaller chunks and eventually it would be

negligible. Dropping this term and rearranging gives ay� = −��ya

e) Starting from the previous step we find ay� = −��ya y��� = −yaa

ß y����*�* = −ß yaa�*

�*

�� − �� = −�� lna�a� �� = �� + �� ln a�a�

Notice this is not ��e�. Of course, if one knew the burn rate one could, in theory, rewrite a� in terms of

time and obtain ��e�. Keep in mind that, for rockets, a� < a�. This implies ln �*�* > 0 and thus the rocket

should speed up (�� > ��). Note: we assumed the fuel is ejected at a constant rate �� relative to the ship. It seems plausible that, as

fuel runs out, there will be less pressure inside the tank. I suspect fuel will be ejected at slower and slower

rates as the fuel is consumed. To be honest, I have no clue. Still, this is a good start for seeing how you

might handle a problem with changing mass. Seems like an ideal problem for numerical simulation, no?

f) By dividing each side of the equation in part c by ye we find ay�ye = −�� yaye

aY!$q��" = −�� yaye

ÂℎÐÏÎe = �� T−yaye U

The term − Æ�Æ" is the rate of fuel consumption (burn rate) in 23� .

9.50 This is a common problem and you can probably find the solution on the internet. I love it but don’t have the

time to write it up yet. The main thing to think about here? The scale reads the force to support the chain AND to

stop the motion of the chain. As the last link hits the scale it will actually read three times the weight of the chain.

Do a search for “chain falling on scale”.

9.51 Considering the horizontal direction only, momentum must be conserved. This is true because there is

negligible force in the horizontal direction. As mass is added to the box velocity must decrease if momentum is to

remain constant! Conservation of momentum gives a� = �a + y���� + y�� Here I am assuming y� is positive (mass is added to box) and y� is negative (box slows down). I chose a random

point in time for the left side of the equation (not the initial point in time). Why? Now this equation is valid for any

point in time as the box slides across the frozen lake (not just the initial time). Notice that M is not the mass of the

box. It is the mass of the box plus some rain…I’ll return to this point later. a� = a� + y�� + ay� + y�y�

Last term on the right side is negligible (product of two differentials is always much smaller than terms with only a

single differential). First two terms on each side cancel. Cleaning up gives y�� = −ay�

Notice the explicit negative sign on the right side will offset the implicit minus sign hidden inside y� < 0. To get Æ_Æ" involved, divide both sides by ye to find

� y�ye = −ay�ye

While this might seem strange, remember that º = ÐYeg = Æ_Æ" = Åf�ÎeY�e. The problem statement said mass is

added to the box at a constant rate! This means the mass as a function of time is a = � + ºe. Plugging this in

gives �º = −�� + ºe� y�ye

From here do separation of variables. y�� = − ºye�� + ºe� ß y���*�� = ß − ºye�� + ºe�"

/

I think you can handle it from there. If not, re-read the examples on separation of variables back in Chapter 2. A

similar example was given there.

9.52 Momentum is conserved during the earth rock collision. Due the earth’s enormous mass, changes in its speed

after the collision are imperceptible.