Final Exam - Siena Science - Please Discover the School …mmccolgan/GP130F12/Exams/final_all.pdfequation of projectile motion to ﬁnd the time it takes the sand to fall to the top

Physics 130 General Physics Fall 2012

Final Exam

December 14, 2012

Name:

Instructions

1. This examination is closed book and closed notes. All your belongingsexcept a pen or pencil and a calculator should be put away and yourbookbag should be placed on the floor.

2. You will find one page of useful formulae on the last page of the exam.

3. Please show all your work in the space provided on each page. If youneed more space, feel free to use the back side of each page.

4. Academic dishonesty (i.e., copying or cheating in any way) willresult in a zero for the exam, and may cause you to fail theclass.

In order to receive maximum credit,

each solution should have:

1. A labeled picture or diagram, if appropriate.2. A list of given variables.3. A list of the unknown quantities (i.e., what you are being asked to find).4. One or more free-body or force-interaction diagrams, as appropriate,

with labeled 1D or 2D coordinate axes.5. Algebraic expression for the net force along each dimension, as appro-

priate.6. Algebraic expression for the conservation of energy or momentum equa-

tions, as appropriate.7. An algebraic solution of the unknown variables in terms of the

known variables.8. A final numerical solution, including units, with a box around it.9. An answer to additional questions posed in the problem, if any.

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1. A tennis player hits a ball 2.0 m above the ground. The ball leaves his racquet with aspeed of 20.0 m{s at an angle of 5˝ above the horizontal. The horizontal distance to thenet is 7 m and the net is 1 m high. Does the ball clear the net? If so, by how much? Ifnot, by how much does it miss?

Solution:

This is a 2D projectile motion problem. We are given the initial height of the ballabove the ground, yi “ 2 m, its initial velocity, vi “ 20.0 m{s, and the angle abovethe horizontal with which the ball is hit, θ “ 5˝. We are also given the horizontaldistance to the net, xf “ 7 m, and the vertical height of the net, yf “ 1 m. We areasked whether the ball clears the net or not.

To solve the problem we first determine the initial velocity of the ball in the horizontal(x) and vertical (y) dimensions:

vxi “ vi cos θ “ p20.0 m{sq ˆ cos p5˝q “ 19.9 m{s, (1)

vyi “ vi sin θ “ p20.0 m{sq ˆ sin p5˝q “ 1.74 m{s. (2)

Next, we use the horizontal kinematic equation of projectile motion to find the timeit takes the ball to travel the distance to the net:

xf “ xi ` vxi∆t (3)

“ vxi∆t (4)

ñ ∆t “ xf

vxi“ 7 m

19.9 m{s “ 0.351 s. (5)

Finally, we use the vertical kinematic equation of projectile motion to find the verticalheight of the ball as it reaches the net:

yf “ yi ` vyi∆t ´ 1

2gp∆tq2, (6)

“ p2.0 mq ` p1.74 m{sq ˆ p0.351 sq ´ 1

2ˆ p9.8 m{s2q ˆ p0.351 sq2 (7)

“ 2.01 m. (8)

So yes, the ball clears the net by about one meter!

2


2. Sand moves without slipping at 6.0 m{s down a conveyer that is tilted at 15˝. The sandenters a pipe 3.0 m below the end of the conveyer belt. What is the horizontal distanced between the conveyer belt and the pipe?

Solution:

This is a 2D projectile motion problem. We are given the initial velocity of the sand,vi “ 6.0 m{s, the angle of the conveyer belt, θ “ 15˝, and the vertical distance to thetop of the pipe, yi “ 3.0 m. We are asked to find the horizontal distance, xf “ d,between the conveyer belt and the pipe.

To solve this problem we must recognize that once the sand leaves the conveyerbelt it is a free-falling projectile. In other words, it maintains the same constanthorizontal velocity it had when it left the belt, while its vertical velocity increasesdue to gravity.

First, we determine the velocity of the sand in the horizontal (x) and vertical (y)dimensions as it leaves the edge of the conveyer belt:

vxi “ vi cos θ “ p6.0 m{sq ˆ cos p15˝q “ 5.80 m{s, (1)

vyi “ ´vi sin θ “ ´p6.0 m{sq ˆ sin p15˝q “ ´1.55 m{s, (2)

where note the negative sign on the vyi velocity. Next, we use the vertical kinematicequation of projectile motion to find the time it takes the sand to fall to the top ofthe pipe (yf “ 0):

yf “ yi ` vyi∆t ´ 1

2gp∆tq2, (3)

0 “ yi ` vyi∆t ´ 1

2gp∆tq2 (4)

This is a quadratic equation for time of the form 0 “ c ` bz ` az2, with solutionz “ p´b ˘

?b2 ´ 4acq{2a. Solving for time, we get:

∆t “´vyi ˘

b

v2yi ´ 4yi ˆ p´g{2q2p´g{2q , (5)

3


“´vyi ˘

b

v2yi ` 2yig

´g , (6)

“ 1.55 m{s ˘a

p´1.55 m{sq2 ` 2 ˆ p9.8 m{s2q ˆ p3.0 mq´9.8 m{s2 , (7)

“ 1.55 m{s ˘ 7.82 m{s´9.8 m{s2 , (8)

“ 0.640 s. (9)

Note that in the last line we had to adopt the “minus” solution to ensure that timewas a positive number.

Finally, we can use the horizontal equation of motion to solve for the final distance,noting that the initial horizontal position xi “ 0:

xf “ d “ xi ` vxi∆t, (10)

“ vxi∆t, (11)

“ p5.80 m{sq ˆ p0.640 sq, (12)

“ 3.7 m. (13)

4


3. A rock stuck in the tread of a 60.0 cm diameter bicycle wheel has a tangential speed of3.00 m{s. When the brakes are applied, the rock’s tangential deceleration is 1.00 m{s2.(a) What are the magnitudes of the rock’s angular velocity and angular acceleration at

t “ 1.50 s?

(b) At what time is the magnitude of the rock’s acceleration equal to g?

Solution:

This is a non-uniform angular acceleration problem. We are given the size of thewheel, r “ 30 cm “ 0.30 m, the initial tangential speed of the rock, vi “ 3.00 m{s, andthe tangential deceleration of the rock when the brakes are applied, aT “ 1.00 m{s2.

(a) In order to determine the final angular velocity, ωf of the rock after t “ 1.50 s,we first have to determine the initial angular velocity:

wi “ vi

r“ 3.00 m{s

0.30 m“ 10 rad{s. (1)

The angular acceleration is a constant and given by:

α “ aT

r“ ´1.00 m{s2

0.30 m“ ´3.33 rad{s2. (2)

Finally, using the appropriate rotational kinematic equation and substituting, weobtain

ωf “ ωi ` α∆t (3)

“ 10 rad{s ´ p3.33 rad{s2q ˆ p1.5 sq (4)

“ 5.01 rad{s. (5)

5


(b) In order to find the time at which the magnitude of the angular acceleration ofthe rock equals g, we first need to find the linear velocity of the rock at that time.We want |~a| “ g “

a

a2T ` a2r, where aT is given in the problem and ar “ v2{r.Substituting and solving for v we get

g “

d

a2T `ˆ

v2

r

˙2

(6)

g2 “ a2T ` v4

r2(7)

ñ v4 “ r2pg2 ´ a2Tq (8)

v “ r1{2pg2 ´ a2Tq1{4 (9)

“ p0.30 mq1{2 ˆ rp9.8 m{s2q2 ´ p1.00 m{s2q2s1{4 (10)

“ 1.71 m{s. (11)

To find the time at which the rock reaches this velocity, we use the 1D kinematicequation for velocity and solve for the time:

vf “ vi ´ aT∆t (12)

ñ ∆t “ vi ´ vf

aT(13)

“ 3.00 m{s ´ 1.71 m{s1.00 m{s2 (14)

“ 1.29 s. (15)

6


4. You and your friend Peter are putting new shingles on a roof pitched at 25˝. You’resitting on the very top of the roof when Peter, who is at the edge of the roof directlybelow you, 5.0 m away, asks you for the box of nails. Rather than carry the 2.5 kg boxof nails down to Peter, you decide to give the box a push and have it slide down to him.If the coefficient of kinetic friction between the box and the roof is 0.55, with what speedshould you push the box to have it gently come to rest right at the edge of the roof?

Solution:

The pictorial representation and free-body diagram are shown below:

This is a 1D dynamics problem. The relevant forces are gravity, FG, the normalforce, n, and the kinetic friction force, fk. Note that we do not include the initialforce that was applied to the box of nails to get it moving, but we do include thefact that the box has some initial velocity.

The most natural coordinate system is one that is rotated by 25˝ and thereforealigned with the roof. The interaction and free-body diagrams are shown above.The net force along the x- and y-axis is

pFnetqx “ÿ

Fx “ FG sin 25˝ ´ fk “ ma (1)

pFnetqy “ÿ

Fy “ n ´ FG cos 25˝ “ 0 (2)

ñ n “ FG cos 25˝ (3)

In the second line we used the fact that the shingles are not leaping off the roof toset the acceleration in the y-direction equal to zero. The magnitude of the force ofgravity is FG “ mg, and the magnitude of the kinetic force of friction is

fk “ µkn (4)

“ µkFG cos 25˝ (5)

“ µkmg cos 25˝. (6)

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Substituting equation (6) into equation (1) and solving for the acceleration a we get

mg sin 25˝ ´ µkmg cos 25˝ “ ma (7)

ñ a “ psin 25˝ ´ µk cos 25˝qg (8)

“ psin 25˝ ´ 0.55 ˆ cos 25˝q ˆ 9.8 m{s2 (9)

“ ´0.743 m{s2 (10)

where the minus sign indicates the acceleration is directed up the incline. To find theinitial speed, vi, necessary to have the box come to rest (i.e., vf “ 0) after ∆x “ 5.0 mcan be found from the kinematic equation linking velocity and acceleration:

v2f “ v2i ` 2a∆x (11)

ñ vi “?

´2a∆x (12)

“b

´2p´0.743 m{s2qp5.0 mq (13)

“ 2.7 m{s. (14)

8


5. The 1.0 kg block in the figure below is tied to the wall with a rope. It sits on top of the2.0 kg block. The lower block is pulled to the right with a tension force of 20 N. Thecoefficient of kinetic friction at both the lower and upper surfaces of the 2.0 kg block isµk “ 0.40.

(a) What is the tension in the rope holding the 1.0 kg block to the wall?

(b) What is the acceleration of the 2.0 kg block?

Solution:

The free-body diagram for the problem is shown below:

To solve this problem we need to use both Newton’s third and second laws. The sep-arate free-body diagrams for the two blocks show that there are two action/reactionpairs. Notice how the top block (block 1) both pushes down on the bottom block

9


(block 2) with force ~n11, and exerts a retarding friction force ~f2,top on the top surface

of block 2.

(a) Block 1 is in static equilibrium (a1 “ 0 m{s2), but block 2 is accelerating to theright. Newton’s second law for block 1 is

pFnet on 1qx “ f1 ´ Trope “ 0 ñ Trope “ f1 (1)

pFnet on 1qy “ n1 ´ m1g “ 0 ñ n1 “ m1g. (2)

Although block 1 is stationary, there is a kinetic force of friction because there ismotion between block 1 and block 2. The friction model means

f1 “ µkn1 “ µkm1g. (3)

Substituting this result into equation (1) we get the tension of the rope:

Trope “ f1 “ µkm1g (4)

“ p0.40q ˆ p1.0 kgq ˆ p9.8 m{s2q (5)

“ 3.9 N. (6)

(b) Newton’s second law for block 2 is

ax ” a “ pFnet on 2qxm2

“ Tpull ´ f2 top ´ f2 bot

m2

(7)

ay “ pFnet on 2qym2

“ n2 ´ n11 ´ m2g

m2

“ 0 (8)

Forces ~n1 and ~n11 are an action/reaction pair, so ~n1

1 “ ~n1 “ m1g. Substituting intoequation (8) gives

n2 “ pm1 ` m2qg. (9)

This result is not surprising because the combined weight of both objects pressesdown on the surface. The kinetic friction on the bottom surface of block 2 is then

f2 bot “ µkn2 “ µkpm1 ` m2qg. (10)

Next, we recognize that the forces ~f1 and ~f2 top are an action/reaction pair, so

f2 top “ f1 “ µkm1g. (11)

Finally inserting these friction results into equation (7) gives

a “ Tpull ´ µkm1g ´ µkpm1 ` m2qgm2

(12)

“ 20 N ´ p0.40qp1.0 kgqp9.8 m{s2q ´ p0.40qp1.0 kg ` 2.0 kgqp9.8 m{s2q2.0 kg

(13)

“ 2.2 m{s2. (14)

10


6. The lower block in the figure below is pulled on by a rope with a tension force of 20 N.The coefficient of kinetic friction between the lower block and the surface is 0.30. Thecoefficient of kinetic friction between the lower block and the upper block is also 0.30.What is the acceleration of the 2.0 kg block?

Solution:

The pictorial representation and free-body diagrams are shown below:

The blocks accelerate with the same magnitude but in opposite directions. Thus theacceleration constraint is a2 “ a “ ´a1, where a will have a positive value becauseof how we have defined the `x direction. There are two real action/reaction pairs.The two tension forces will act as if they are action/reaction pairs because we areassuming a massless rope and a frictionless pulley.

Make sure you understand why the friction forces point in the directions shown inthe free-body diagrams, especially force ~f 1

1 exerted on the bottom block (block 2) bythe top block (block 1).

11


We have quite a few pieces of information to include. First, Newton’s second lawapplied to block 1 gives

pFnet on 1qx “ f1 ´ T1 (1)

“ µkn1 ´ T1 “ m1a1 “ ´m1a (2)

pFnet on 1qy “ n1 ´ m1g “ 0 (3)

ñ n1 “ m1g. (4)

For block 2 we have

pFnet on 2qx “ Tpull ´ f 11 ´ f2 ´ T2 (5)

“ Tpull ´ f 11 ´ µkn2 ´ T2 “ m2a2 “ m2a (6)

pFnet on 2qy “ n2 ´ n11 ´ m2g “ 0 (7)

ñ n2 “ n11 ` m2g. (8)

Note that to simplify the two x-equations above we have already used the kineticfriction model. Next, from Newton’s third law we have three additional constraints:

n11 “ n1 “ m1g (9)

f 11 “ f1 “ µkn1 “ µkm1g (10)

T1 “ T2 “ T. (11)

Knowing n11 we can now use the y-equation for block 2 to find n2. Substituting

all these pieces into the two x-equations, we end up with two equations with twounknowns:

µkm1g ´ T “ ´m1a (12)

Tpull ´ T ´ µkm1g ´ µkpm1 ` m2qg “ m2a. (13)

Subtracting the first equation from the second we get

Tpull ´ T ´ µkm1g ´ µkpm1 ` m2qg ´ µkm1g ` T “ m2a ` m1a (14)

Tpull ´ 3µkm1g ´ µkm2g “ pm2 ` m1qa (15)

Tpull ´ µkp3m1 ` m2qg “ pm2 ` m1qa. (16)

And finally solving for a we get

ñ a “ Tpull ´ µkp3m1 ` m2qgm1 ` m2

(17)

“ 20 N ´ p0.30qp3 ˆ 1.0 kg ` 2.0 kgqp9.8 m{s2q1.0 kg ` 2.0 kg

(18)

“ 1.8 m{s2. (19)

12


7. The 1.0 kg physics book in the figure below is connected by a string to a 500 g coffeecup. The book is given a push up the slope and released with a speed of 3.0 m{s. Thecoefficients of friction are µs “ 0.50 and µk “ 0.20.

(a) How far does the book slide?

(b) At the highest point, does the book stick to the slope, or does it slide back down?

Solution:

The pictorial representation and free-body diagrams are shown below:

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To solve this problem we use the particle model for the book (B) and the coffee cup(C), the models of kinetic and static friction, and the constant-acceleration kinematicequations.

(a) To find the distance x1 the book slides we must find its acceleration. Newton’ssecond law applied to the book gives

ÿ

pFon Bqy “ nB ´ pFGqB cos θ “ 0 (1)

ñ nB “ mBg cos θ (2)ÿ

pFon Bqx “ ´T ´ fk ´ pFGqB sin θ (3)

“ ´T ´ µknB ´ mBg sin θ (4)

“ ´T ´ µkmBg cos θ ´ mBg sin θ “ mBaB. (5)

Similarly, for the coffee cup we haveÿ

pFon Cqy “ T ´ pFGqC “ T ´ mCg “ mCaC . (6)

Equations (5) and (6) can be rewritten using the acceleration constrain aC “ aB “ a

as

´T ´ µkmBg cos θ ´ mBg sin θ “ mBa (7)

T ´ mCg “ mCa. (8)

Adding these two equations and solving for the acceleration gives

´µkmBg cos θ ´ mBg sin θ ´ mCg “ pmB ` mCqa (9)

ñ a “ ´„

mBpµk cos θ ` sin θq ` mC

mB ` mC

g (10)

“ ´„p1.0 kgq ˆ p0.20 cos 20˝ ` sin 20˝q ` 0.5 kg

1.0 kg ` 0.5 kg

ˆ p9.8 m{s2q (11)

“ ´6.73 m{s2. (12)

Finally, to solve for the distance x1 we use the following kinematic equation withv1x “ 0, v0x “ 3.0 m{s, and x0 “ 0:

v21x “ v20x ` 2apx1 ´ x0q (13)

0 “ v20x ` 2ax1 (14)

ñ x1 “ ´v20x2a

(15)

“ ´p3.0 m{sq22 ˆ p´6.73 m{s2q (16)

“ 0.67 m. (17)

14


(b) In order to figure out if the book sticks to the slope or slides back down we haveto determine if the static friction force needed to keep the book in place, fs is largeror smaller than the maximum static friction force

pfsqmax “ µsnB “ µsmBgcosθ (18)

“ p0.50q ˆ p9.8 m{s2q ˆ cos 20˝ (19)

“ 4.60 N. (20)

When the cup is at rest, the string tension is T “ mCg. In this case, Newton’s firstlaw for the book becomes

ÿ

pFon Bqx “ fs ´ T ´ mBg sin θ (21)

“ fs ´ mCg ´ mBg sin θ “ 0 (22)

ñ fs “ pmC ` mB sin θqg (23)

“ p0.5 kg ` 1.0 kg sin 20˝q ˆ p9.8 m{s2q (24)

“ 8.25 N. (25)

Because fs ą pfsqmax, the book slides back down.

15


8. A conical pendulum is formed by attaching a 500 g ball to a 1.0 m-long string, thenallowing the mass to move in a horizontal circle of radius 20 cm. The figure shows thatthe string traces out the surface of a cone, hence the name.

(a) What is the tension in the string?

(b) What is the ball’s angular speed, in rpm?

Solution:

(a) What is the tension in the string?

The forces in the z-direction in the free body diagram are the component of thetension in the z-direction and the force of gravity. To find the z-component ofthe tension,

16


cos θ “ adjacent

hypotenuse“ Tz

T(1)

Tz “ T cos θ (2)

Apply Newton’s second law in the z-directionÿ

Fz “ T cos θ ´ mg “ 0 (3)

T “ mg

cos θ(4)

To calculate cos θ,

cos θ “ adjacent

hypotenuse“

?L2 ´ r2

L“

a

p1mq2 ´ p0.2mq21.0m

“ 0.98 (5)

ñ θ “ 11.5˝ (6)

Plugging this in for tension,

T “ mg

cos θ“ p0.500 kgq ˆ p9.80 m{s2q

0.98(7)

T “ 5.0 N (8)

(b) What is the ball’s angular speed, in rpm?

Referring back to the tension triangle, the radial component of tension is

sin θ “ opposite

hypotenuse“ Tr

T(9)

Tr “ T sin θ (10)

17


Referring back to the triangle for the length of the pendulum,

sin θ “ opposite

hypotenuse“ r

L(11)

r “ L sin θ (12)

Apply Newton’s second law in the r-direction

ÿ

Fr “ Tr “ T sin θ “ mω2r (13)

ñ ω “c

T sin θ

mr(14)

“d

5.0 N ˆ sin 11.5˝

p0.500 kgq ˆ p0.2 mq (15)

“ 3.16 rad{sec (16)

“ 3.16rad

secˆ

ˆ

1 rev

2π rad

˙

ˆˆ

60 sec

1 min

˙

(17)

“ 30 rpm (18)

18


9. A block of mass m slides down a frictionless track, then around the inside of a circularloop-the-loop of radius R. From what minimum height h must the block start to makeit around without falling off? Give your answer as a multiple of R.

Solution:

This is a two-part problem. In the first part, we will find the critical velocity for theblock to go over the top of the loop without falling off. Since there is no friction,the sum of the kinetic and gravitational potential energy is conserved during theblock’s motion. We will use this conservation equation in the second part to find theminimum height the block must start from to make it around the loop.

Place the origin of the coordinate system directly below the block’s starting positionon the frictionless track. From the free-body diagram we have

´Fg ´ n “ ´mv2cR. (1)

For the block to just stay on the track, n “ 0. Therefore, the critical velocity vc is

Fg “ mv2cR

(2)

mg “ mv2cR

(3)

ñ vc “a

gR (4)

We can now use conservation of mechanical energy to find the minimum height h:

Kf ` Uf ` Ki ` Ui (5)

1

2mv2f ` mgyf “ 1

2mv2i ` mgyi. (6)

Using vf “ vc “ ?gR, yf “ 2R, vi “ 0, and yi “ h, we get

1

2gR ` gp2Rq “ 0 ` gh (7)

ñ h “ 2.5R. (8)

19


10. A pendulum is formed from a small ball of mass m on a string of length L. As the figureshows, a peg is height h “ L{3 above the pendulum’s lowest point. From what minimumangle θ must the pendulum be released in order for the ball to go over the top of thepeg without the string going slack?

Solution: This is a two part problem. First, we need to find the velocity for theball to go over the peg without the string going slack. Then we need to find thepotential energy to match that velocity.

(a) Find the velocity of the ball so that the string doesn’t go slack. We’ll set ourorigin on the peg. This means that once the string touches the peg, the ballwill be moving in circular motion with a radius of L{3.When the ball is let go, it moves below the peg and then moves in circularmotion. When it reaches the top of the circle and is directly above the peg, wecan use Newton’s second law and the fact that the ball is moving in circularmotion.

ÿ

F “ ma “ ´mv2

R(1)

´T ´ Fg “ ´mv2

R(2)

(3)

20


The critical speed is just as the tension equals zero. Factoring the negatives andthe masses, and setting Fg “ mg and T “ 0,

0 ` mg “ mv2cR

(4)

vc “a

gR (5)

Setting R “ L{3,

vc “c

gL

3(6)

(b) Find the angle to release the ball. Now that we know the velocity for the stringto remain taut, we’ll use the conservation of energy to find the height to releasethe ball.

Kf ` Uf ` Ki ` Ui (7)

1

2mv2f ` mgyf “ 1

2mv2i ` mgyi (8)

1

2v2c ` gyf “ 0 ` gyi (9)

Plugging in yf “ L{3 and v2c “ gL{3,gL

6` gL

3“ gyi (10)

L

6` L

3“ yi (11)

yi “ L

2(12)

This result tells us that we need to release the ball at a height of L/2 above thepeg.

To find the angle, consider the third figure in the three-part figure above. Weneed to find the distance of the ball below the point where the string is attached.We can use the string length and result above.

h “ L ´ L

2´ L

3“ L

6(13)

To find the angle, we can use cosine,

cos θ “ L{6L

“ 1

6(14)

ñ θ “ 80.4˝ (15)

21


11. A sled starts from rest at the top of the frictionless, hemispherical, snow-covered hillshown below.

(a) Find an expression for the sled’s speed when it is at angle φ.

(b) Use Newton’s laws to find the maximum speed the sled can have at angle φ withoutleaving the surface.

(c) At what angle φmax does the sled “fly off” the hill?

Solution:

(a) When the sled is at a position that relates to angle φ, the height is yf “ R cosφ.Using conservation of energy with yi “ R, we get

Kf ` Uf “ Ki ` Ui (1)

1

2mv2f ` mgyf “ 1

2mv2i ` mgyi (2)

1

2v2f ` gR cosφ “ 0 ` gyi (3)

ñ v2f “ 2gR ´ 2gR cosφ (4)

“ 2gRp1 ´ cosφq (5)

ñ vf “a

2gRp1 ´ cosφq (6)

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(b) To find the maximum speed we use the free-body diagram:

ÿ

F “ ma “ ´mv2

R(7)

FN ´ Fg cosφ “ ´mv2

R(8)

The critical speed is just as the normal force equals zero. Factoring the negativesand the masses, and setting Fg “ mg and FN “ 0,

g cosφ “ v2cR

(9)

vc “a

gR cosφ (10)

(c) Find the angle when the sled gets air! We can set the speeds from the twoprevious parts equal to each other to find φ.

a

gR cosφ “a

2gRp1 ´ cosφq (11)

gR cosφ “ 2gRp1 ´ cosφq (12)

cosφ “ 2 ´ 2 cosφ (13)

3 cosφ “ 2 (14)

ñ φ “ cos´1

ˆ

2

3

˙

“ 48˝. (15)

23


12. Truck brakes can fail if they get too hot. In some mountainous areas, ramps of loosegravel are constructed to stop runaway trucks that have lost their brakes. The combi-nation of a slight upward slope and a large coefficient of rolling resistance as the trucktires sink into the gravel brings the truck safely to a halt. Suppose a gravel ramp slopesupward at 6.0˝ and the coefficient of rolling friction is 0.40. Use work and energy to findthe length of a ramp that will stop a 15, 000 kg truck that enters the ramp at 35 m/s(« 75 mph).

Solution:

We’ll identify the truck and the loose gravel as the system. We need the gravel insidethe system because friction increases the temperature of the truck and the gravel.We will also use the model of kinetic friction and the conservation of energy equation.

Kf ` Uf ` ∆Eth “ Ki ` Ui ` Wext (1)

0 ` Uf ` ∆Eth “ Ki ` 0 ` 0 (2)

The thermal energy created by friction is

∆Eth “ fkp∆xq (3)

“ pµkFNqp∆xq (4)

“ pµkmg cos θqp∆xq (5)

“ pµkmg cos θqp∆xq (6)

Using geometry, the final gravitationl potential energy of the truck is

Uf “ mgyf (7)

“ mgp∆xq sin θ (8)

Finally, the initial kinetic energy is simply

Ki “ 1

2mv2i (9)

24


Plugging everything into the conservation of energy equation, we get

mgp∆xq sin θ ` µkmg cos θp∆xq “ 1

2mv2i (10)

g∆xpsin θ ` µk cos θq “ 1

2v2i (11)

ñ ∆x “ v2i2gpsin θ ` µk cos θq (12)

“ p35 m{sq22 ˆ p9.8 m{s2q ˆ psin 6˝ ` 0.4 ˆ cos 6˝q (13)

“ 124 m (14)

“ 0.12 km. (15)

25


13. The spring shown in the figure below is compressed 50 cm and used to launch a 100 kgphysics student. The track is frictionless until it starts up the incline. The student’scoefficient of kinetic friction on the 30˝ incline is 0.15.

(a) What is the student’s speed just after losing contact with the spring?

(b) How far up the incline does the student go?

Solution:

(a) This is a conservation of energy problem. We want to know the initial velocityimmediately after the spring is no longer in contact with the student.

Kf ` Ugf ` Usf ` ∆Eth “ Ki ` Ugi ` Usi ` Wext (1)

1

2mv2f ` mgyf ` 1

2kp∆xf q2 ` 0 “ 1

2mv2i ` mgyi ` 1

2kp∆xiq2 ` 0 (2)

1

2mv2f “ 1

2kp∆xiq2 (3)

26


ñ vf “c

k

m∆xi “ 14 m{s (4)

(b) Using the conservation of energy equation again, we can how far the studenttravels up the incline. Let’s define the y-height when the student reaches thehighest point as yf “ ∆s sin 30˝. Then we have:

Kf ` Ugf ` Usf ` ∆Eth “ Ki ` Ugi ` Usi ` Wext (5)

1

2mv2f ` mgyf ` 1

2kp∆xf q2 ` 0J “ 1

2mv2i ` mgyi ` 1

2kp∆xiq2 ` 0J (6)

mg∆s sin 30˝ ` µkFn∆s “ mgyi ` 1

2p∆xq2 (7)

(8)

From the sum of the forces in the y-direction on the incline, the normal force isFn “ mg cos 30˝. Plugging this in, we get

mg∆s sin 30˝ ` µkmg cos 30˝∆s “ mgyi ` 1

2p∆xq2 (9)

∆smgpsin 30˝ ` µk cos 30˝q “ mgyi ` 1

2p∆xq2 (10)

ñ ∆s “ mgyi ` 12kp∆xq2

mgpsin 30˝ ` µk cos 30˝q (11)

∆s “ 32.1 m (12)

27


14. A hollow sphere is rolling along a horizontal floor at 5.0 m{s when it comes to a 30˝

incline. How far up the incline does it roll before reversing direction?

Solution:

Assume that the hollow sphere is a rigid rolling body and that the sphere rolls upthe incline without slipping. Also assume that the coefficient of rolling friction iszero.

The initial kinetic energy, which is a combination of rotational and translationalenergy, is transformed in gravitational potential energy. Choose the bottom of theincline as the zero of the gravitational potential energy.

Starting from conservation of energy, we have

Kf ` Ugf “ Ki ` Ugi (1)

1

2Mv21 ` 1

2Iω2

1 ` Mgy1 “ 1

2Mv20 ` 1

2Iω2

0 ` Mgy0. (2)

Substituting v1 “ 0, ω1 “ 0, y0 “ 0, I “ 2{3MR2 (appropriate for a hollow sphere),ω0 “ v0{R, and solving for the final height, y1, we get

0 ` 0 ` Mgy1 “ 1

2Mv20 ` 1

2

ˆ

2

3MR2

˙

ω20 ` 0 (3)

gy1 “ 1

2v20 ` 1

3R2

ˆ

v20R2

˙

(4)

ñ y1 “ 5v206g

(5)

“ 5 ˆ p5.0 m{sq26 ˆ p9.8 m{s2q (6)

“ 2.126 m. (7)

28


The distance traveled along the incline is

s “ y1

sin 30˝(8)

“ 2.126 m

0.5(9)

“ 4.3 m. (10)

29


15. A 10 g bullet traveling at 400 m{s strikes a 10 kg, 1.0-m wide door at the edge oppositethe hinge. The bullet embeds itself in the door, causing the door to swing open. Whatis the angular velocity of the door just after impact?

Solution:

In order to solve this problem, we use the fact that angular momentum is conservedin the collision for the (bullet+door) system. As the bullet hits the door, its velocity~v is perpendicular to ~r. Consequently, the initial angular momentum about therotation axis, with r “ L, is

Li “ mBvBL. (1)

After the collision, with the bullet in the door, the moment of inertia about thehinges is

I “ Idoor ` Ibullet (2)

“ 1

3mDL

2 ` mBL2. (3)

Therefore, the final angular momentum is

Lf “ Iω (4)

“ˆ

1

3mDL

2 ` mBL2

˙

ω. (5)

30


Equating the initial and final angular momentum and solving for ω we get

Lf “ Li (6)ˆ

1

3mDL

2 ` mBL2

˙

ω “ mBvBL (7)

ñ ω “ 3mBvB

LpmD ` 3mBq (8)

“ 3 ˆ 0.010 kg ˆ 400 m{s1.0 m ˆ p10 kg ` 3 ˆ 0.010 kgq (9)

“ 1.2 rad{s. (10)

31


16. A 355 mL soda can is 6.2 cm in diameter and has a mass of 20 g. Such a soda can halffull of water is floating upright in water. What length of the can is above the waterlevel?

Solution:

The buoyant force, FB, on the can is given by Archimedes’ principle.

Let the length of the can above the water level be d, the total length of the can beL, and the cross-sectional area of the can be A. The can is in static equilibrium, sofrom the free-body diagram sketched above we have

ÿ

Fy “ FB ´ FG,can ´ FG,water “ 0 (1)

ρwaterApL ´ dqg “ pmcan ` mwaterqg (2)

L ´ d “ pmcan ` mwaterqAρwater

(3)

ñ d “ L ´ pmcan ` mwaterqAρwater

(4)

Recalling that one liter equals 10´3 m3 and the density of water is ρwater “ 1000 kg m´3,the mass of the water in the can is

mwater “ ρwater

ˆ

Vcan

2

˙

(5)

“ p1000 kg m´3q ˆˆ

355 ˆ 10´6 m3

2

˙

(6)

“ 0.1775 kg. (7)

To find the length of the can, L, we have:

Vcan “ AL (8)

ñ L “ 355 ˆ 10´6 m3

πp0.031 mq2 (9)

“ 0.1176 m. (10)

32


Finally, substituting everything into equation (4), we get:

d “ 0.1176 m ´ p0.020 kg ` 0.1775 kgqπ ˆ p0.031 mq2 ˆ p1000 kg m´3q (11)

“ 0.0522 m (12)

“ 5.2 cm. (13)

33


17. Water flowing out of a 16 mm diameter faucet fills a 2.0 L bottle in 10 s. At whatdistance below the faucet has the water stream narrowed to 10 mm in diameter?

Solution:

We treat the water as an ideal fluid obeying Bernoulli’s equation. The pressure atpoint 1 is p1 and the pressure at point 1 is p2. Both p1 and p2 are atmosphericpressure. The velocity and the area at point 1 are v1 and A1, and at point 2 theyare v2 and A2. Let d be the distance of point 2 below point 1.

First, we use the time it takes to fill a 2.0 L bottle to compute the flow rate, Q, therate at which water is flowing out of the faucet, remembering that one liter equals10´3 m3:

Q “ p2.0 Lq ˆ p10´3 m3{Lq10 s

(1)

“ 2.0 ˆ 10´4 m3{s. (2)

Next, we use this result to find the velocity, v1, with which the water leaves thefaucet through the D1 “ 16 ˆ 10´3 m diameter aperture (at point 1):

Q “ v1A1 (3)

ñ v1 “ Q

A1

(4)

“ Q

πpD1{2q2 (5)

“ 2.0 ˆ 10´4 m3{sπ ˆ p8 ˆ 10´3 mq2 (6)

“ 1.0 m{s. (7)

34


Now, from the continuity equation we have

v2A2 “ v1A1 (8)

ñ v2 “ v1A1

A2

(9)

“ v1πpD1{2q2πpD2{2q2 (10)

“ v1

ˆ

D1

D2

˙2

(11)

“ 1.0 m{sˆ

16 ˆ 10´3 m

10 ˆ 10´3 m

˙2

(12)

“ 2.56 m{s. (13)

Finally, we turn to Bernoulli’s equation to find the height, d:

p1 ` 1

2ρv21 ` ρgy1 “ p2 ` 1

2ρv22 ` ρgy2 (14)

ρgpy1 ´ y2q “ 1

2ρpv22 ´ v21q (15)

gd “ 1

2pv22 ´ v21q (16)

d “ pv22 ´ v21q2g

(17)

“ p2.56 m{sq2 ´ p1.0 m{sq22 ˆ 9.8 m{s2 (18)

“ 0.283 m (19)

« 28 cm. (20)

35


18. A 100 g ice cube at ´10˝ C is placed in an aluminum cup whose initial temperature is70˝ C. The system comes to an equilibrium temperature of 20˝ C. What is the mass ofthe cup?

Solution:

There are two interacting systems: aluminum and ice. The system comes to thermalequilibrium in four steps: (1) the ice temperature increases from ´10 ˝C to ´0 ˝C;(2) the ice becomes water at 0 ˝C; (3) the water temperature increases from 0 ˝C to20 ˝C; and (4) the cup temperature decreases from 70 ˝C to 20 ˝C.

Since the aluminum and ice form a closed system, we have

Q “ Q1 ` Q2 ` Q3 ` Q4 “ 0. (1)

Each term is as follows:

Q1 “ Mice cice ∆T (2)

“ p0.100 kgq ˆ r2090 J{pkg Kqs ˆ p10 Kq“ 2090 J.

Q2 “ Mice Lf (3)

“ p0.100 kgq ˆ p3.33 ˆ 105 J{kgq“ 33, 300 J.

Q3 “ Mice cwater∆T (4)

“ p0.100 kgq ˆ r4190 J{pkg Kqs ˆ p20 Kq“ 8380 J.

Q4 “ MAl cAl∆T (5)

“ MAl ˆ r900 J{pkg Kqs ˆ p´50 Kq“ ´p45, 000 J{kgqMAl.

Inserting everything into equation (1), we get

43, 770 J ´ p45, 000 J{kgqMAl “ 0

ñ MAl “ 0.97 kg. (6)

36


19. A group of rebels wants to invade the castle and they made a trebuchet as their first lineof attack. The group optimized their trebuchet so that their projectile can be fired froma great distance to hit the castle wall. Their trebuchet includes a counterweight mass of931 kg, a moment arm, R2 “ 2 m, a projectile mass of 7 kg, a moment arm R1 “ 10 m,an arm mass of 25 kg, and a release angle of φ “ 90˝ due to a stopper on the trebuchetbase that stops the projectile arm from going any further. The pivot point is 8 m abovethe ground. Find the distance from the castle wall that the trebuchet needs to be placedso that the projectile hits the wall.

Solution:

This problem uses the torque created by the difference in the force/moment-armratios of the counterweight and projectile to generate an angular acceleration.

There are two torque terms for the trebuchet. The counterweight will create aclockwise rotation (which by convention is negative) and the projectile will create apositive counterclockwise rotation.

Because the projectile is in a cup attached to the arm, the force and the momentarm are perpendicular to each other. For that torque term, R2Fp sinpangleq “ R2Fp.However, the counterweight and its moment arm are not perpendicular to each other.The figure below shows the angle between them. The torque component for thecounterweight is R1FCW sinψ.

Just like theř

F “ ma for linear motion,ř

τ “ Iα for rotational motion where I isthe moment of inertia of the R2 which we’ll consider to be a rod rotating about it’send. Alpha is the angular acceleration.

37


ÿ

τ “ Iα (1)

FpR2 ´ FCWR1sinψ “ Iα (2)

(3)

The moment of inertia for a rod rotating about its end is I “ 13ML2. We’re interested

in finding the speed of the projectile as it leaves the the moment arm, R2. So, we’lluse the length of R2 for L and the mass of that portion of the rod in our calculation.Let the total mass of the rod be M where the mass of the projectile portion of therod is N1 and the mass of the counterweight portion of the rod is N2. To find N1,

N1

R1

“ M

L(4)

N1 “ “ R1M

L“ 10m ˆ 25kg

12m“ 20.8kg (5)

(6)

ÿ

τ “ Iα (7)

FpR2 ´ FCWR1sin psi “ Iα (8)

m2gR2 ´ mCWgR1sin psi “ 1

3MR2α (9)

(10)

Calculate the angular acceleration, α,

α “ 3gpm2R2 ´ m1R1sinψqN1R2

2 (11)

38


α “ p3 ˆ 9.8m{s2qp7kg ˆ 10m ´ 931kg ˆ 2m sin 143˝q20.8kg ˆ p10mq2 (12)

α “ 14.8rad

s2(13)

(14)

To find the angular velocity, ωf when the projectile leaves the trebuchet, we need touse kinematics. We’re told in the problem, that the projectile moment arm travelsθ “ 90˝ “ π

2radians before the projectile is released.

pωf q2 “ pωiq2 ` 2αθ (15)

pωf q2 “ 2αθ “ 2αθ (16)

ωf “?2αθ “

c

2 ˆ 14.8rad

s2ˆ π

2(17)

ωf “ 6.8rad

s2(18)

(19)

So far, our calculations have been for circular motion as the projectile is moved bythe trebuchet. Once the projectile leaves the trebuchet, we need to switch to linearmotion.

To find the linear velocity with which the projectile leaves the trebuchet, we convertthe angular velocity to linear velocity.

v “ ωfR2 (20)

39


v “ 6.8rad

s2ˆ 10m (21)

v “ 68m{s (22)

(23)

The original description tells us that the projectile travels through an angle of90˝while on the trebuchet. Using this information, we can find the angle at whichthe projectile leaves the trebuchet and its height.

We need to use the kinematic equations for projectile motion to find the time it takesfor the projectile to hit the ground. Then use this time, to calculate the horizontalrange of the trebuchet.

vy “ v sin θ “ 68m{s ˆ sin 37˝ “ 41m{s (24)

(25)

We’re able to calculate the starting height of the projectile using the figure above.Because the trebuchet travels through 90˝, we can use the same triangle for rightbefore the trebuchet is fired and right after the projectile is released. Because thesetriangles are the same, the height at which the projectile leaves the trebuchet is 16m.

yf “ yi ` vy∆t ´ 1

2g∆t (26)

0 “ 16m ` 41m{s ˆ ∆t ´ 1

29.8m{s2∆t (27)

(28)

40


Rearranging this equation into the form of a quadratic equation,

4.9m{s2∆t ´ 41m{s ˆ ∆t ´ 16m “ 0 (29)

∆t “ 8.7 sec (30)

(31)

To find the range, we plug this time into the range equation (i.e. no accelerationin the x-direction). We also need to use the initial velocity of the projectile in thex-direction. From the figure above, we can see that the velocity in the x-direction isvx “ v cos 37˝.

xf “ xi ` vxi∆t “ v cos 37˝∆t (32)

xf “ 68m{s ˆ cos 37˝ ˆ 8.7 sec (33)

xf “ 472m (34)

(35)

41


Kinematics and Mechanics

xf “ xi ` vxit ` 1

2axt

2

vxf “ vxi ` axt

v2xf “ v2xi ` 2axpxf ´ xiq

yf “ yi ` vyit ` 1

2ayt

2

vyf “ vyi ` ayt

v2yf “ v2yi ` 2aypyf ´ yiq

θf “ θi ` ωi∆t ` 1

2α∆t2

ωf “ ωi ` α∆t

ωf2 “ ωi

2 ` 2α∆θ

s “ rθ

c “ 2πr

vt “ ωr

ar “ v2

r“ ω2r

v “ 2πr

T

Forces

~Fnet “ Σ~F “ ma

~Fnet “ Σ~Fr “ ma “ mv2

r

Fg “ mg

0 ă fs ă“ µsFN

fk “ µkFN

~FAonB “ ´~FBonA

Momentum

~pi “ ~pf

~p “ m~v

pvfx q1 “ m1 ´ m2

m1 ` m2

pvix q1

pvfx q2 “ 2m1

m1 ` m2

pvix q1

Energy

Kf ` Ugf “ Ki ` Ugi

∆K ` ∆U ` ∆Eth “ ∆Emech ` ∆Eth “ ∆Esys “ Wext

Kf ` Uf ` ∆Eth “ Ki ` Ui ` Wext

K “ 1

2mv2

U “ mgy

∆Eth “ fk∆s

W “ ~F ¨ ~dP “ ~F ¨ ~vP “ ∆E{∆t

Fluids and Thermal Energy

P “ F

A

ρ “ m

V

Q “ vA

v1A1 “ v2A2

p1 ` 1

2ρv2

1 ` ρgy1 “ p2 ` 1

2ρv2

2 ` ρgy2

Qnet “ Q1 ` Q2 ` ... “ 0

Q “ MLf for freezing/melting

Q “ Mc∆T

I “ÿ

mr2

~τ “ ~r ˆ ~F “ Iα

~L “ m~v ˆ ~r

~A ˆ ~B “ ABsinθ

Krot “ 1

2Iω2

Idisk “ 1

2mr2

Ihollow sphere “ 2

3mr2

I door “ 1

3mr2 about its hinges

I rod “ 1

12mr2 about its end

42


I rod “ 1

3mr2 about its center

Constants

g “ 9.8 m{s2

ρwater “ 1000 kg{m3

ρair “ 1 .28 kg{m3

cice “ 2090J

kg K

cwater “ 4190J

kg K

cAl “ 900J

kg K

Lf “ 333 , 000J

kgice to water

43

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Final Exam - Siena Science - Please Discover the School …mmccolgan/GP130F12/Exams/final_all.pdfequation of projectile motion to ﬁnd the time it takes the sand to fall to the top