1. The angle between the vector !A = 3i ! 2 j ! 5k and the positive y axis, in

degrees, is closest to:

A) 19 B) 71 C) 90 D) 109 E) 161

Solution The dot product between the vector

!A = 3i ! 2 j ! 5k and the unit vector along the

positive y-axis j is given by !A• j =

!A j cos! , where ! , is the angle between the

two vectors and thus cos! =!A• j!A j

. With !A• j = !2 and

!A = 32 + (2)2 + (!5)2 = 38

and j =1 we have cos! =!A• j!A j

=(!2)

38= !0.32 and thus ! = arccos("0.32) = 109!

     2. A rock is thrown directly upward from the edge of the roof of a building that is

34.9 meters tall. The rock misses the building on its way down, and is observed to strike the ground 4.00 seconds after being thrown. Neglect any effects of air resistance. With what speed was the rock thrown?

A) 6.4 m/s B) 12.1 m/s C) 3.6 m/s D) 8.7 m/s E) 10.9 m/s

Solution The vertical position of the rock is given by the equation for free fall:

y(t) = y0 + v0t ! 12 gt

2 . With y(tfinal) = 0, we can solve this for v0: v0 =

12 gt2 ! y0

t final

Putting in y0 = 34.9 m, tfinal = 4.00 s, g = 9.81 m/s, we get v0 = 10.9 m/s.

3. The figure on the right shows the graph of

the position x as a function of time for an object moving in a straight line (along the x-axis). Which of the following graphs best describes the velocity along the x-axis as a function of time for this object?

Solution The object starts with a large positive velocity given by the slope of the x(t) curve. At large t the slope goes to zero. Therefore, the correct graph is B).

Position of motorist as a function of time t after the chase startedxM (t) = vMt

Position of police officer while she is accelerating: xO (t) = 0 + 12

at2

The time it takes her to reach v=70 m/s is v / a = (70m/s)/(2m/s2 ) = 35 sAt t = 35 s, xO = 1225 m and xM = (41.7m/s)(35s) = 1460 m

Distance between police and motorist: 235 mExtra time required for officer to catch motorist = (235m)/(70m/s-41.7m/s) = 8 sTotal time = (35+ 8) s = 43 s  

4. A motorist traveling at a constant speed of 41.7 m/s in a 50 km/h speed zone

passes a police car. Immediately after the car passes, the police car starts off in pursuit. The police officer accelerates at 2 m/s2 up to a speed of 70 m/s, and then continues at this speed until she overtakes the speeding motorist. How long from the time she started does it take the police car to overtake the motorist? The motorist continues at a constant speed during this process.

A) 99 s B) 43 s C) 38 s D) 35 s E) 56 s

Solution

5. Three people are pushing a stalled car up a hill at constant velocity. The net

force on the car is

A) down the hill and greater than the weight of the car. B) up the hill and greater than the weight of the car. C) up the hill and equal to the weight of the car. D) down the hill and equal to the weight of the car. E) zero.

Solution Newton’s first law states that if no net force acts on an object it will move with constant velocity.

6. A girl throws a stone from a bridge. Consider the following ways she might

throw the stone. Case A: Thrown straight up. Case B: Thrown straight down. Case C: Thrown at an angle of 45o above horizontal. Case D: Thrown horizontally. The speed of the stone as it leaves her hand is the same in each case. In which case will the speed of the stone be greatest when it hits the ground below.

A) Case A B) Case B C) Case C D) Case D E) The speed will be the same in all cases.

Solution Assuming the absence of air resistance, the easiest way to analyze this question is with the work-energy theorem. The only force acting is the weight, which is a constant force downwards. So, the net work done is force times the vertical displacement in the downward direction, which is the height of the bridge in all cases. Therefore the same work is done in all cases, leading to the same final kinetic energy and the same landing speed. (However, the direction of the final velocity might be different in different cases.) If air resistance is taken into account the stone with the shortest trajectory will be first hitting the ground. This would be case B.

7. For general projectile motion, when the projectile is at the highest point of its

trajectory

A) its acceleration is zero. B) its velocity and acceleration are both zero. C) the horizontal and vertical components of its velocity are zero. D) the horizontal component of its velocity is zero. E) its velocity is perpendicular to the acceleration.

Solution At the highest point of the trajectory for projectile motion the projectile’s vertical velocity vy goes from being positive to being negative and is zero. The vertical acceleration is the entire time ay = –9.81 m/s2 and pointing down. Since there’s no acceleration in the horizontal direction, the horizontal velocity vx is equal to its initial value v0x. Thus, at the highest point the velocity

!v has only a horizontal component and is perpendicular to

!a which has only a vertical component. 8. A projectile is fired from point 0 at the edge of a

high cliff, with initial velocity components of v0x = 60.0 m/s and v0y = 175 m/s, as shown in the figure. The projectile rises and then falls into the sea at point P. The time of flight of the projectile is 40.0 s, and it experiences no appreciable air resistance in flight. The height of the cliff is closest to ….

A) 120 m B) 180 m C) 230 m D) 410 m E) 850 m

Solution:

v0 y = 175 m/s t = 40.0 s ay = !9.81 m/s2

"y = v0 yt +12 ayt

2 = !848 m

9. Two particles, A and B, are in uniform circular motion about a common

center. The acceleration of particle A is 5.5 times that of particle B. The period of particle B is 2.7 times the period of particle A. The ratio rA / rB of the radius of the motion of particle A to that of particle B is closest to

A) 2.0 B) 15 C) 4.1 D) 0.75 E) 0.49

Solution: For uniform circular motion the radial acceleration is given by ar =!

2r with

! =2"T

, where r is the radius, T the period and ω the angular speed of the

motion. Thus, rArB=aAaB

!B

!A

!

"#

$

%&

2

=aAaB

TATB

!

"#

$

%&

2

=5.5aBaB

!

"#

$

%&

TA2.7TA

!

"#

$

%&

2

=5.52.72

= 0.75

10. An airplane flies between two points on the ground that are 500 km apart. The

destination is directly north of the origination of the flight. The plane flies with an air speed of 120 m/s. If a constant wind blows at 30.0 m/s due west during the flight, what direction must the plane fly relative to north to arrive at the destination?

A) 75.5° west of north B) 14.5° east of north C) 14.0° west of north D) 14.0° east of north E) 14.5° west of north

Solution We have

!vplane,air = 120 m/s and

!vair ,ground = 30 m/s .

From the sketch on the right, we see

sin! = !vair ,ground / !vplane,air = 30m/s( ) / 120m/s( ) = 0.250

and thus ! = arcsin(0.250) = 14.5! .

11. A man pushes against a rigid, immovable wall. Which of the following is the

most accurate statement concerning this situation?

A) If the man pushes on the wall with a force of 200N, we can be sure that the wall is pushing back with a force of exactly 200N on him.

B) The friction force on the man’s feet is directed to the left. C) Since the wall cannot move, it cannot exert any force on the man. D) The man cannot be in equilibrium since he is exerting a net force on the

wall. E) The man can never exert a force on the wall that exceeds his weight.

Solution Newton’s 3rd law if the man pushes on the wall with a force 200N (action), then the wall must be pushing back on him with an equal and opposite force (reaction). 12. A cylinder slides down a frictionless iceberg as shown in the

figure. Which statement describes the magnitude of the cylinder’s velocity

!v and the cylinder’s acceleration

!a best

while it is sliding down the hill.

A) both !v and

!a remain constant.

B) both !v and

!a increase.

C) !v decreases and

!a increases.

D) !v increases and

!a decreases.

E) both !v and

!a decrease.

Solution: The slope of the hill is negative and there will be a downward component of the weight at any place on the slope. Therefore, the acceleration down the hill, will always be positive and

!v will increase all the time. However, since the slope

flattens at the bottom of the hill, the downward force and the magnitude of the tangent acceleration decreases. However, the total acceleration, the vector sum of tangent and radial acceleration, can increase. For example, if the iceberg slope was circular, the radial acceleration at the bottom would be larger than the tangent acceleration at the top.

13. In the figure a block of mass M hangs in

equilibrium. The rope, which is fastened to the wall is horizontal and has a tension of 27N. The rope, which is fastened to the ceiling has a tension of 83N, and makes an angle θ with the ceiling. The angle θ is

A) 55 o B) 19 o C) 71 o D) 45 o E) 18 o

Solution Consider balance of forces in the horizontal direction: (83 N) cos θ=27 N θ=71 o

14. Dick and Jane stand on a platform of negligible weight, as shown in the figure. Dick weighs 500N and Jane weighs 400N. Jane is supporting some of her weight by holding a rope. Assume that all ropes and pulleys are ideal. What is the downward force Jane is exerting on the platform?

A) 0 B) 300 N C) 100 N D) 240 N E) 50 N

Solution Draw the free body diagram for the whole system: the tension in the rope pulls on the system three times (once on Jane’s hand, and twice on the lower pulley) Therefore, 3T = 900N T=300 N. Consider free body diagram for Jane: three forces are acting on her vertically, weight 400 N downward, tension in rope 300 N upward and the force from the platform the platform exerts an upward force of 100 N on her. By Newton’s 3rd law, she is exerting an equal and opposite force on the platform.

15. A worker is dragging a packing crate of mass 100kg across a rough floor

where the coefficient of kinetic friction is 0.40. He exerts a force sufficient to give the crate a positive acceleration. At what angle above the horizontal should his pulling force be directed in order to achieve the maximum acceleration?

A) 34.5 o B) 27.7 o C) 45.0 o D) 21.8 o E) 30.0 o

Solution

F cos! " Ff = ma

FN + F sin! " mg = 0# FN = mg " F sin!

Ff = µk FN # ma = F cos! " µk mg " F sin!( )Taking derivative d(ma)/d! and setting it to zero, we get"F sin! + µk F cos! = 0

Thus, the acceleration is maximum when tan! = µk #! = tan"1 µk = 21.8°

The next two questions pertain to the following scenario: “The Tornado” is a carnival ride that consists of a big vertical cylinder that rotates rapidly about its vertical axis with constant angular speed. As “The Tornado” rotates, the riders are pressed against the inside wall of the cylinder by the rotation. During the ride, the floor of the cylinder drops away. 16. The force preventing the riders from falling downward is …

A) the centripetal force. B) the normal force. C) the centrifugal force. D) the gravitational force. E) the friction force.

Solution The gravitational force pulls downward on the riders. The force preventing them to fall down is the friction force between the riders and the wall of “The Tornado”.

17. While “The Tornado” rotates, the net force acting on a rider …

A) points radially inward. B) points down. C) is zero. D) points in the direction of the riders velocity. E) points radially outward.

Solution There a are several forces acting on a rider. Gravity is pulling down while the friction force between a rider and the wall of the “The Tornado” pulls up. These two are equal in magnitude since the rider neither moves up nor down. The net force is equal to the centripetal force causing the rider to move in a circular trajectory and provided by the normal force of the wall of the “The Tornado”. It points radially inward to the center of “The Tornado”.

18. A 5.00-kg box slides 3.00 m across the floor before coming to rest. What is

the coefficient of kinetic friction between the floor and the box if the box had an initial speed of 3.00 m/s?

A) 0.183 B) 0.245 C) 0.153 D) 0.275 E) 0.214

Solution We use the work-kinetic energy theorem to solve this problem. The work done by the friction force is W = ! fk"x = !µk mg"x and the change in kinetic energy is

!K =

12

mv02 " 0 . Setting these two equal and solving for the coefficient of kinetic

friction gives

W = !K "#µk mg!x = #

12

mv02 " µk =

v02

2g!x=

3.00m/s( )2

2 $ 9.81m/s2 $ 3.00m= 0.153

19. A ball of mass 5.0 kg is suspended by two wires from

a horizontal arm that is attached to a vertical shaft, as shown in the figure. The shaft is in uniform rotation about its axis. The rate of rotation is adjusted until the tensions in the two wires are equal. At that speed, the radial acceleration of the ball is closest to

A) 6.9 m/s2 B) 7.9 m/s2 C) 5.9 m/s2 D) 9.9 m/s2 E) 4.9 m/s2

Solution From the y-coordinate, we get

T1 + T2, y ! mg = 0 and from the x-coordinate we get

T2, y ! marad = 0 . With T1 = T2 and T2,y = 0.6T2 and T2,x = 0.8T2 , we get

T2 + 0.6T2 = mg and 0.8T2 = marad . Division of these two equations gives arad = g/2 = 4.9 m/s2.

20. In order to do work on an object,

A) the object must move. B) the force doing the work must be directed perpendicular to the motion of the object. C) it is necessary that friction not be present. D) the applied force must be greater than the reaction force of the object. E) it is necessary that friction be present.

Solution Work done W =

!F ! "!r "!r = 0#W = 0

21. A system comprising blocks, a

massless frictionless pulley, a frictionless incline, and connecting massless ropes is shown. The 9-kg block accelerates downward when the system is released from rest. The tension in the rope connecting the 6-kg block and the 4-kg block is closest to

A) 39 N B) 30 N C) 36 N D) 42 N E) 33 N

Solution  

To solve problem, we begin by solving for the acceleration of the systemm1 = 6 kg, m2 = 4 kg, m3 = 9 kg

Write down equations of motion for m3 and m1 + m2 :

m3g ! T2 = m3aT2 ! (m1 + m2 )g sin" = (m1 + m2 )aAdding the 2 eqns together: m3g ! (m1 + m2 )g sin" = (m1 + m2 + m3)a

# a =m3g ! (m1 + m2 )g sin"

(m1 + m2 + m3)= 2.065 m/s2

Now consider free body diagram for m1

T1 ! m1g sin" = m1a# T1 = m1a + m1g sin" = 42 N

F  m1   m2  

22. Which of the free-body diagrams below represents a block sliding down a

frictionless inclined surface?

Solution Answer D represents a block sliding down a frictionless inclined surface. Weight is directed down, and the normal force is directed perpendicular to the inclined surface. These are the only two forces present. 23. Two blocks with masses m1 = 3.0 kg and m2 = 2.0 kg rest on a horizontal,

frictionless table. A horizontal force F is applied on block 1 from the left, as shown. The magnitude of the force exerted by block 1 on block 2 is 30 N. What is the magnitude of F?

A) 30 N B) 45 N C) 50 N D) 65 N E) 75 N

Solution For block 2, Newton’s second law gives

F2,1 = m2a , where

F2,1 is the normal force

on block 2 by block 1. The acceleration is all the same for block 2, block 1 and the system consisting of blocks 1 and 2. Thus, for the total force F that accelerates the 2-block system, we have

F = (m1 + m2 )a =

m1 + m2

m2

F2,1 =3.0kg+2.0kg

2.0kg(30N)=75N

24. A conical pendulum made with a bead of

mass 4.0 kg and a 1.00 m long massless rope completes 1.2 turns every second. Determine the angle θ between the string and the vertical.

A) 40 o B) 50 o C) 60 o D) 70 o

E) 80 o Solution

x :T sin! = marad = m" 2 R = m" 2 Lsin!

y :T cos! # mg = 0$ T =mg

cos!

$ mgsin!cos!

= m" 2 Lsin!

$ cos! =g

" 2 L= 0.17

where we have used ! = 1.2 rev/s=1.2

2" radrev

revs

=7.5rads

25. The work performed by an engine as a function of time for a process is given

by W = at3, where a = 2.4 J/s3. What is the instantaneous power output of the engine at t = 1.4 s?

A) 7 W B) 10 W C) 14 W D) 20 W E) 30 W

Solution

The instantaneous power of the engine is given by P =

dWdt

=ddt

at3( ) = 3at2 .

At t = 1.4 s, we have an instantaneous power of P = 3 2.4J/s3( ) 1.4s( )2

=14W .

26. When a parachutist jumps from an airplane, she eventually reaches a constant speed, called the terminal speed. Which statement describes the jump correctly?

A) Once the parachutist has reached terminal speed the force of air drag on her is zero. B) Once the parachutist has reached terminal speed the force of air drag on her is equal to her weight. C) Once the parachutist has reached terminal speed the net force on her is equal to her weight. D) When the parachutist jumps out of the plane her acceleration vector points upward. E) Airplanes cannot fly high enough for a parachutist to reach terminal speed.

Solution When the parachutist has reached (constant) terminal speed she does not accelerate anymore and thus the net force on her is zero. Her weight is pulling her towards earth, while the force of air drag opposes the motion of the fall. Since the net force on the parachutist is zero once she has reached terminal speed, the force of air drag on her is equal to her weight. 27. A 4.00-kg box sits atop a 10.0-kg box on a horizontal table. The coefficient of

kinetic friction between the two boxes and between the lower box and the table is 0.600, while the coefficient of static friction between these same surfaces is 0.800. A horizontal pull to the right is exerted on the lower box, as shown in the figure, and the boxes move together. What is the friction force on the upper box?

A) 19.3 N to the left B) 31.4 N to the left C) 19.3 N to the right D) 31.4 N to the right E) 23.5 N to the right

Solution Newton’s second law for the system of two boxes in the x-direction is given by

F ! fk = (m + M )aF ! µk (m + M )g = (m + M )a

" a = Fm + M

! µk g = 150N4kg+10kg

! 0.6(9.8m/s2 ) = 4.83m/s2

For the top box, we get fs =ma = 4kg!4.83m/s2 =19.3N