Physics 221 2007S Exam 1 Solutions - Iowa State Universityatwood/221_exam_archive_site/exam1_2007s... · Physics 221 2007S Exam 1 Solutions Page 2 of 21 [3] An elevator has a mass

Physics 221 2007S Exam 1 Solutions

Page 1 of 21

[1]If a 200kg box is subject to a net force of magnitude 100N, what is the magnitude of its acceleration? (A) 0.25 2m/s (B) 0.50 2m/s (C) 1.0 2m/s (D) 2.0 2m/s (E) 4.0 2m/s

Answer (B): Using Newton’s second law 2

100N m0.5200kg s

Fam

= = =

[2] A car is traveling a circular track at a constant speed of 20m/s. If the magnitude of the acceleration of the car is 21 m/s , what is the radius of the track? (A) 50m (B) 100m (C) 200m (D) 400m (E) 800m

Answer (D): Centripetal acceleration is given by 2va

R= so

2

22 (20 ) 400(1 )

ms

ms

vR ma

= = = .


Page 2 of 21

[3] An elevator has a mass of 600kg, not including passengers. The elevator is designed to ascend, at a constant speed, a vertical distance of 20.0m (i.e. five floors) in 16.0s, and is driven by a motor that can provide up to 30kW of power. What is the maximum number of passengers that can ride in the elevator? Assume that an average passenger has a mass of 65kg. (A) 28 (B) 24 (C) 20 (D) 16 (E) 12 Answer (A): The force that the motor must exert on the elevator is equal to the weight of the elevator since it is moving at a constant speed. Since P Fv= the force exerted by the motor and hence the weight of the elevator is

/ (30000 ) /(20 /16 ) 24000W F P v W m s N= = = = The mass of the elevator is thus 2

ms

/ (24000N) /(9.8 ) 2449kgm W g= = = From the problem, the number of passengers n is ( 600kg) / 65kg (2449kg 600kg) / 65kg=28.4n m= − = − The limit is thus 28 passengers. [4] A person pulls a 20-kg block up a ramp of height h = 10 m and angle θ = 20° at constant speed. The rope makes an angle φ = 35° with the ramp as shown and has a tension of 160N . Find the work done by kinetic friction. (A) −661J (B) −740J (C) −980J (D) −1880J (E) −5800J Answer (D): The length of the ramp is

o

10m 29.3sin sin 20

hx mθ

Δ = = =

The work done by the tension is ocos (160N)(29.3m)cos(35 ) 3840JTW T x ϕ= Δ = =

The work done by gravity is

2ms

(20kg)(9.8 )(10m) 1960JgW mgh= − = − = − . The normal force does no work since it is perpendicular to the ramp. By the work energy theorem, the total work on the block is zero because its kinetic energy does not change as the block is moving at a constant speed. Thus

0

( ) (( 3840J) ( 1960J)) 1880Jnet g T k

k T g

W W W W

W W W

= = + +

∴ = − + = − + + − = −


Page 3 of 21

[5] A bullet of mass 1g is fired at a block of wood traveling at a speed of 300 m/s. It comes to rest 60μs after hitting the surface of the wood. If you assume that the wood slows down the bullet exerting a constant force against the bullet’s motion, what is the magnitude of that force? (A) 2500N (B) 5000N (C) 7500N (D) 10000N (E) 25000N Answer (B): The acceleration of the bullet is (change in velocity)/(time interval) so

62(300 ) /(60 ) 5.0 10m ma s

s sμ= = × in the direction against the motion of the bullet. From

Newton’s second law, the force on the bullet is 3 62

m(10 kg)(5.0 10 ) 5000Ns

F ma −= = × =

[6]A lady bug and a gentleman bug are sitting on a circular turntable with radius 50cm which is rotating clockwise at a constant angular velocity. The ladybug is at the rim of the turntable while the gentleman bug is half way between the center and the rim. If the lady bug is moving at a speed of 0.10m/s, what is the magnitude of the angular velocity of the gentleman bug? (A) 0.05 rad/s (B) 0.10 rad/s (C) 0.20 rad/s (D) 0.30 rad/s (E) 0.40 rad/s Answer (C): The angular velocity of the gentleman bug is the same as that of the lady bug because they are on the same turn table. This angular velocity is

ms/ (0.10 ) /(0.50m) 0.20rad/sv Rω = = =


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[7]Sally runs 1 km north at 6 m/s and then runs 2 km south at 4 m/s. What is the magnitude of Sally’s average velocity for this trip? (A) 0.0 m/s (B) 1.5 m/s (C) 2.0 m/s (D) 3.0 m/s (E) 5.0 m/s Answer (B): The total displacement of Sally’s trip is 1km south. The first part of her trip takes an amount of time 1 1000m /(6m/s) 167st = = . The second part of her trip takes an amount of time 2 2000m /(4m/s) 500st = = . The total time is thus 1 2 667stott t t= + = . The magnitude of the average velocity is thus (1000m)/(667m/s)=1.5m/s. [8]An object’s height as a function of time is 2 2( ) (120 m) (80 m/s) (20 m/s )y t t t= + − . What is the highest point this object reaches? (A) 100m (B) 150m (C) 300m (D) 400m (E) None of the above Answer (E): The extremum is given by y’=0. Taking the derivative,

2

2m ms s

' (80 ) (40 )

solving for ' 0(80 ) /(40 ) 2s

m ms s

y t

yt

= −

== =

This is in fact a maximum because 2'' 40 ms

y = − is negative. Plugging this time back into the expression for y we get the height at the maximum:

2 2max

2 2

(120 m) (80 m/s) (20 m/s )

(120 m) (80 m/s)(2s) (20 m/s )(2s)200m

y t t= + −

= + −=

(Which is not one of the choices A-D)


Page 5 of 21

[9]A 100kg person stands on top of a box that is inside the cabin of an elevator. The elevator has an upward acceleration of 23 m/s . What is the magnitude of the net force acting on the person? (A) 100N (B) 300N (C) 680N (D) 980N (E) 1280N Answer (B): By Newton’s second law 2(100 )(3 ) 300m

net sF ma kg N= = = .

[10] The figure shows the overhead view of two stones that travel in circles over a frictionless horizontal surface. Each stone is tied to a cord whose opposite end is anchored at the center of the circle. The radius of the circle of stone B is 2 times the radius of the circle of stone A. The tension in the long cord is 2 times the tension in the short cord. The period of stone B is 4 times the period of stone A. What is the ratio

(mass of stone B):(mass of stone A) (A) 16:1 (B) 4:1 (C) 1:1 (D) 1:4 (E) 1:16

Answer (A): The centripetal force is 2

2 22 2

144

mR FTF mR mT R

ω ππ

= = ∴ =

Thus ( )2

22

12 4 162

B B B A

A A A B

m F T Rm F T R

⎛ ⎞= = ⋅ =⎜ ⎟⎝ ⎠

A B


Page 6 of 21

[11]A rubber ball is tossed horizontally from a window 5m above a parking lot. When it strikes the ground it rebounds. Which of the following graphs best represents the y-component of the ball’s velocity as a function of time? The +y direction is up and neglect air resistance.

Answer (C): While the ball is in freefall, before and after the bounce, it is accelerating downwards so the velocity graph should have a constant negative slope. During the bounce the velocity will change rapidly from downwards to upwards. Only graph (C) has these features.

A B C

D E

Time Time Time Time

Time Time

vy vy vy

vy vy


Page 7 of 21

[12]Robin Hood fires an arrow at velocity v0=30m/s at an angle of θ=30o above the horizontal from the top of a cliff of height 20m. The sheriff of Nottingham stands at distance d from the base of the cliff. What is the speed of the arrow when it strikes the sheriff? Neglect air resistance. (A) v=36 m/s (B) v=42 m/s (C) v=46 m/s (D) v=50 m/s (E) Cannot determine the speed without knowing d. Answer (A): Using the v-squared equation in three dimensions:

2 20 2v v a r− = ⋅Δ

Taking the x-axis horizontal and y-axis vertical, 2(0, 9.8 )mas

= − and ( , 20m)r dΔ = − so

22 20 2 (30 ) 2( 9.8 )( 20 ) 36m m m

s ssv v a r m= + ⋅Δ = + − − =

Note that the distance d is determined by the angle θ but to get the speed of impact we don’t have to know either of these quantities.

θ=30o v0=30m/s

Robin Hood

Sheriff of Nottingham

h

d

v=?

20m

x

y


Page 8 of 21

[13] Vectors A and B start at a corner of a cube of edge length 1. Vector A lies along the edge of the cube. Vector B goes to the opposite corner of the cube. What is the angle between and A B ?

(A) 45° (B) 55° (C) 60° (D) 70° (E) 90°

Answer (B): Let us take the origin to be the starting part of the vectors in the picture and the x axis parallel to the vector A. In this coordinate system the vectors are

(1,0,0)

(1,1,1)

A

B

=

=

thus the cosine of the angle between the vectors is given by:

o

(1,0,0) (1,1,1) 1cos| (1,0,0) || (1,1,1) | 3

1arccos 553

A BAB

θ

θ

⋅ ⋅= = =

∴ = =

[14] A particle of mass m = 10.0 kg moves along the x-axis with time dependent acceleration

( )4 23.00 m/sxa t= . The velocity of the particle at t = 0 is -3.00 m/s in the +x direction. What is the velocity of the particle at t = 2.00s? (A) 0 m/s (B) 5 m/s (C) 8 m/s (D) 9 m/s (E) 21 m/s Answer (B): Acceleration is the derivative of velocity thus:

4

4

00

2.002

023

0

3.00 (3 )

3.00 (3 ) 3.00 8.003

5

finalt

sm ms s

s

m m m ms s ss

ms

v v a dt

t dt

t

= +

= − +

⎡ ⎤= − + = − +⎢ ⎥

⎣ ⎦=

∫

∫

1


Page 9 of 21

[15] Two cars move along parallel tracks on a straight road. Shown below are the snapshots taken every second. Assume that the motion between snapshots is what one would assume from the image, nothing strange. Which of the following best describe the direction of the velocity of the rectangular car relative to the square car at t = 2 s and at t = 3 s? (A) ← , → (B) ← , ← (C) → , → (D) ← , 0 (E) → , 0

Answer (B): At 2 and 3 seconds it is clear that square car is faster than the rectangular car and both moving to the right. Since we want the velocity of the rectangular car relative to the square car, using eqn. 3.36 from the text:

/ / /

/ / /

/ /( )

rec ground rec squ squ ground

rec squ rec ground squ ground

rec ground squ ground

v v v

v v v

v v

= +

∴ = −

+ −

Drawing the vectors (the same diagram applies to both t=2s and t=3s):

Therefore both arrows are to the left and so the answer must be (B)

1 s 2 s 3 s 4 s

1 s

2 s

3 s

4 s

= +/( )squ groundv−/rec squv /rec groundv


Page 10 of 21

The following information applies to questions [16] and [17]: In the figures below, the force applied in both cases on the system has the same magnitude and is horizontal. Block A has twice the mass of block B. There is no friction between the blocks and the surface underneath them.

[16]What is the magnitude of the net force on block A in figure 1 and figure 2, respectively?

(A) , 3

2(B) , 3 3

2(C) , 3 32 2(D) , 3 3

(E) , 3 3

FF

F F

F F

F F

F F

[17] For the same figures in the previous problem, which of the following is TRUE? (please read carefully)

(A) The force on A by B is smaller in figure 1 than in figure 2. (B) The force on A by B is the same in both figures. (C) In figure 1, the force on A by B is smaller than the force on B by A. (D) In figure 2, the force on A by B is smaller than the force on B by A. (E) None of the above.

Answer (A): The force between the blocks in figure 1 is 1AB BF m a= since this is the only horizontal force on block B. In figure 2 the force between the blocks is

2 2AB A BF m a m a= = since this is the only horizontal force on A. Thus option (A) is true hence (B) is false. (C) and (D) are false due to Newton’s 3rd law because this law implies

AB BAF F= no matter what.

mA mB

F

Figure 1

mA mB

F

Figure 2

Answer (D): Consider first Figure 1. The acceleration of the two blocks is the same as the acceleration of the whole system. Using Newton’s 2nd law on the whole system , one can determine the acceleration. This acceleration is /(3 )Ba F m= . Using Newtons 2nd law on

block A, 223 3A A B

B

F FF m a mm

= = = . In Figure 2 the

magnitude of the net force on the whole system is the same as Figure 1 so the above argument applies and the net force on A is the same.


Page 11 of 21

[18] A block of mass 2kg moving along the x-axis is subject to a position dependent force given by 2 ˆF Ax i= where 22 kg/(ms )A = . Initially the block is at x=0m and has a velocity of ˆ(10 m/s) i . When the block reaches x=4m, what is its kinetic energy? (A) 43J (B) 57J (C) 100J (D) 143J (E) 228J Answer (D): By the work energy theorem, the change in kinetic energy is equal to the work done on the block. In this case,

2

4m2 20

0

4mkg2 3m 1s 3ms 0

12

1 (2kg)(10 ) (2 )2143J

f iK K W mv Ax dx

x

= + = +

⎡ ⎤= + ⎣ ⎦

=

∫


Page 12 of 21

[19] A 10-kg weight hangs from two ideal massless ropes as shown below. Find the tension on the left rope.

(A) 33 N (B) 49 N (C) 85 N (D) 98 N (E) 113 N

Answer (B): This question may be simplified considerably by noting that the two strings intersect at a right angle. If we take the coordinate system shown the y-component of the weight is sinmg θ where θ=30° as shown. Applying Newton’s 2nd law in the y-direction we find that therefore the net force in the y direction is y sinnetF T mg θ= − therefore sin 0yT mg maθ− = = because the 10kg object is not accelerating. Thus

2om

ssin (10kg)(9.8 )sin 30 49NT mg θ= = =

This problem can also be done using the coordinate system with the x-axis horizontal.

30° 60°

T=?

10kg

x y

mg

=θ

mg sin θ mg cos θ


Page 13 of 21

[20] City B is 1000km east of city A. A light plane can fly at a speed of 400km/hr with respect to the air. If there is a constant wind at a speed of 300km/hr blowing in a direction 30° north of east, what is the time it takes for the plane to make the trip from A to B? (A) 2.65 hr (B) 2.20 hr (C) 1.59 hr (D) 1.46 hr (E) The trip as described is not possible. Answer (C): The velocity of the plane with respect to the ground by

, , ,plane ground plane air air groundv v v= + . This relationship is shown in the figure below. We know the magnitude and direction of ,air groundv and just the magnitude of ,plane airv . From the diagram, we see that we can think of the resultant vector ,plane groundv as being composed of

two pieces, P ,Q . The height of the triangle is , sinair groundv θ so

,

2 2, ,

,

2 2, , ,

o 2 o 2km km kmhr hr hr

km km kmhr hr hr

cos

( sin )

cos ( sin )

(300 )cos30 (400 ) (300 sin 30 )

260 371 631

air ground

plain air air ground

plane ground

air ground plain air air ground

P v

Q v v

v P Q

v v v

θ

θ

θ θ

=

= −

= +

= + −

= + −

= + =

The time of the trip is thus km

hr1000km/(631 )=1.59hr . You can also solve this triangle using law of cosines, expressing the vectors in components or other methods

,plane groundv

,plane airv,air groundv

P Qθ=30°

, sinair groundv θ300km/hr

400km/hr

260km/hr 371km/hr

631km/hr


Page 14 of 21

[21] For the vectors shown below, which of the following statements is TRUE? (A)

(B)

(C)

(D)

(E)

A C A D

A B A C

C B C D

A B A D

B C C B

⋅ = ⋅

⋅ > ⋅

⋅ = ⋅

⋅ = − ⋅

⋅ = − ⋅

Answer (D): Taking each square to be one unit, the vectors in component form are:

(6,0)

(3,3)

( 3,3)

(3,0)

A

B

C

D

=

=

= −

=

Thus 18 and 18A B A D⋅ = ⋅ = − hence equation (D) is true. The other equations are all false.

A

B

C

D


Page 15 of 21

[22] A block of mass 10kg lies at rest on an incline ramp tilted at an angle of 10º to the horizontal. The coefficient of static friction between the block and the ramp is μS=0.4. The coefficient of kinetic friction between the block and the ramp is μK=0.3. What is the magnitude of the frictional force on the block? (A) 0N (B) 17N (C) 29N (D) 39N (E) 49N Answer (B): First we need to see if the block slides or is held by static friction. Assume first that the block is static. The free body diagram in this scenario is as shown below where the block has 0 acceleration therefore the net force on it is 0. Let us take the x-axis parallel to the ramp. From Newton’s 2nd law in the y direction, the normal force is cosN mg θ= where θ is the angle of the ramp. The static friction limit is fmax=μsN=μsmgcosθ. Form Newton’s 2nd law in the x direction, the force of static friction must be

sinsf mg θ= because it must cancel the down-ramp component of the block’s weight. This scenario is consistent only if:

max

o

sin costan

tan10 0.18 0.4 which is true

s

s

s

f fmg mgθ μ θ

θ μ

<

⇔ <⇔ <

⇔ = < ⇔

Thus, given the numbers in the problem this scenario is consistent. The force of static friction is thus given as above: 2

osin (10 )(9.8 )sin10 17ms s

f mg kg Nθ= = = Note in particular that 39N which is the static friction limit is not the correct answer.

θ=10°

N fs

mg

mg sinθ

mg cosθ

Newton’s 2nd Law x-component: sin 0smg fθ − = y-component: cos 0N mg θ− =

x-axis

y-axis


Page 16 of 21

[23]A 2kg cannon ball is fired from ground level at an angle of 45° with respect to the horizontal. The ball hits a castle on a cliff which is 600m away horizontally and 200m above the ground. What was the initial speed of the cannon ball? Neglect air resistance. (A) 43 m/s (B) 63 m/s (C) 66 m/s (D) 94 m/s (E) 112 m/s Answer (D): Method 1: If there were no gravity, the cannon ball would pass over the castle at an altitude of 600m which is 400m above the cliff. The time it takes to hit the castle must therefore be the same as the time it takes for something to fall a distance of d=400m.

Thus 2 9.0dt sg

= = . Horizontally, the cannon ball covers 600m in that time so the

horizontal speed is ms(600m) /(9.0s) 66.4xv = = . The initial speed is thus

o ms/ cos(45 ) 93.9xv v= = .

d=400m

θ=45°

200m

600m

No gravity trajectory

Actual trajectory


Page 17 of 21

Method 2: Let t be the time it takes to hit the castle. Taking the origin to be the point where the cannon is fired, the x and the y components at time this time are:

212

cossin

x vty vt gt

θ

θ

=

= −

The two unknown quantities are t and v but we have two equations so we can solve the system. Starting by solving the first equation for t:

cosxt

v θ=

plugging this back into the second equation:

2

3

2 2

212 2 2

2

2 2

22 2

2

2

2ms

2 o o

2 6m ms s

ms

tancos

tan2 cos

2 costan

2cos ( tan )

(9.8 )(600m)2cos 45 ((600m) tan 45 200m)

(9.8 )(600m) 3.53 10(600m 200m) 400m

93.9

xy x gv

gx x yv

gxvx y

gxvx y

θθ

θθ

θθ

θ θ

= −

∴ = −

∴ =−

∴ =−

=−

×= =

−

=


Page 18 of 21

[24] Let A and B be two vectors such that 2A B+ = and 4A B− = . What is the value

of A · B ? (A) 0 (B) 1/ 2− (C) 1− (D) 2− (E) 3− Answer (E): Expanding:

2 2 2

2 2 2

( ) 2

( ) 2

A B A B A B

A B A B A B

+ = + + ⋅

− = + − ⋅

Therefore 2 2( ) ( ) 4A B A B A B+ − − = ⋅ Thus

2 2142 21

4

[( ) ( ) ]

[(2) (4) ]3

A B A B A B⋅ = + − −

= −

= −


Page 19 of 21

[25] Consider the system shown below made of three identical blocks of mass m, two ideal, massless strings and an ideal, massless pulley. The friction between the blocks and the table is negligible. A force of F=mg is applied to the end of the string as shown.

Let T1 and T2 be the tension on each string, as shown in the figure. Which of the following is true?

(A) T1 = T2 = mg (B) 2T1 = T2 = mg (C)2T1 = T2 = 2

3 mg (D)T1 = T2 = 2

3 mg (E)T1 = T2 = 32 mg

Answer (C): The net force on the three block system of mass 3m is mg to the right. The acceleration is therefore / 3a g= . The only horizontal force on the left block is 1T . This block accelerates at the rate a so by Newton’s 2nd law this tension must be

11 3T ma mg= = . The only net horizontal force on the system consisting of the two left

blocks is 2T and the mass of this system is 2m so again, this force is 1 2

2 3 3(2 )( )T m g mg= = . Thus (C) is the correct answer.

m m T2 T1

m

F=mg


Page 20 of 21

[26] Ball A is dropped from the top of a building of height h at the same instant that ball B is thrown vertically upward from the ground. When the balls collide, they are moving in opposite directions, and the speed of A is two times the speed of B. How far above the ground does the collision occur? (A) h/6 (B) h/3 (C) h/2 (D) 2h/3 (E) 5h/6 Answer (D): Let t be the time after the balls are released when the collision occurs. Since the balls collide: ( ) ( )A By t y t= and from the condition on the velocities at collision: ( ) 2 ( )A Bv t v t= − Using the constant acceleration equations we can expand the positions and velocities in the above as follows:

21

2

021

0 2

A

A

B B

B B

v gt

y h gtv v gt

y v t gt

= −

= −

= −

= −

Plugging these into the above we obtain two equations in the two unknowns vB0 and t:

2 21 1

0 02 2

0 02 2 2 3 2A B B B

A B B B

y y h gt v t gt h v tv V gt v gt gt v= ⇒ − = − ⇒ =

− = ⇒ = − ⇒ =

Solving this system of two equations in the two unknowns we obtain

0

2 233 2

hB t

h ht g

v

gt t

=

∴ = ⇒ =

plugging this into the equation for yA:

212

212 3

23

( )A

hg

y h gth g

h

= −

= −

=

The height of the collision above the ground is therefore 23y h= .


Page 21 of 21

[27] [EXTRA CREDIT PROBLEM] A crate is dragged at constant speed along a horizontal floor with a cable at an angle with the horizontal (see figure). Let these symbols represent the magnitude of the following forces on the crate: T = tension in the cable f = friction force N = normal force by the floor W = weight of the crate

Which of the following is TRUE? (A) T > f ; N < W (B) T < f ; N = W (C) T < f ; N < W (D) T > f ; N = W (E) T > f ; N > W

T

Tx

Ty

N

W

fk

Answer (A): Consider the free body diagram of the system. The block moves at a constant speed so the net force on it is 0. In the vertical direction, N+Ty=W so that N<W. In the x direction

cosk yf T T Tθ= = < since cosθ is always less than 1.

θ

T

Physics 221 2007S Exam 1 Solutions - Iowa State Universityatwood/221_exam_archive_site/exam1_2007s... · Physics 221 2007S Exam 1 Solutions Page 2 of 21 [3] An elevator has a mass

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