1) The components of vectors B and

C are given as follows:

The angle between vectors

B and

C , in degrees, is closest to:

A) 162 B) 111 C) 69 D) 18 E) 80 Solution

B ⋅C = (−6.1)(−9.8) + (−5.8)(4.6) = 33.1  and  

B = 6.12 + 5.82 = 8.42    

C = 9.82 + 4.62 = 10.8

            cosθ =33.1

8.42 ×10.8= 0.363     θ = 68.7° ≈ 69°

2) Vectors

B and

C are shown in the figure. The length of

vector B is 5.6 and the length of vector

C  is 4.8 Vector

D

is given by D =B −C . The magnitude of

D is closest to

A) 3.2 B) 9.2 C) 8.0 D) 5.3 E) 6.6

Solution Either use the law of cosines

B −C = B2 + C 2 − 2BC cosθ = 5.62 + 4.82 − 2(5.6)(4.8)cos79° = 6.65 ≈ 6.6

or do it the long way:

B = B(cos213o,sin2130 ) = (−4.70,−3.05)C = C(cos292o,sin292o ) = (+1.80,−4.45)

D =B −C = (−6.50,+1.40) ⇒

D = 6.52 +1.42 = 6.65

Bx = −6.1 Cx = −9.8By = −5.8 C y = +4.6

3) Vectors

A,B and

C satisfy the vector equation

A +B =C . Their magnitudes

are related by A −

B =

C . Which of the following is an accurate statement?

A)

B and

C are perpendicular vectors.

B) B and

C are antiparallel vectors.

C) A,B and

C form the sides of an equilateral triangle.

D) A,B and

C form the sides of a right triangle.

E) The angle between A and

B can have any value, in view of the limited

information given. Solution The only case when the length of the sum of two vectors is equal to the difference of their respective lengths is when the two vectors are pointing in opposite directions. 4) Two vectors are given by

A = i + 2 j + 3k and

B = 3i + 2 j + k . Find

A ×B .

A) i + 2 j + 3k

B) 3i + 2 j + k

C) 2i + 2 j + 2k

D) −2i + 2 j − 2k

E) −4i + 8 j − 4 k Solution

A ×B = i + 2 j + 3k( ) × 3i + 2 j + k( ) =

i j k1 2 33 2 1

= 2 − 6( ) i + 9 −1( ) j + 2 − 6( ) k = −4i + 8 j − 4 k

5) What is the average speed of a car that travels first 4.0 km at an average speed of 20 km/h and another 4.0 km at an average speed of 80 km/h?

A) 60 km/h B) 58 km/h C) 50 km/h D) 38 km/h E) 32 km/h

Solution: The average speed is given by v =

ΔxΔt

Δx = Δx1 + Δx2 = 4.0km + 4.0km = 8.0km

Δt1 =Δx1

v1

=4.0km

20km/h= 0.20h; Δt2 =

Δx2

v2

= 4.00km80km/h

= 0.05h

Δt = Δt1 + Δt2 = 0.20h + 0.05h = 0.25h

v =ΔxΔt

=8.0km0.25h

= 32km/h

6) A car moving at a velocity of 20 m/s is behind a truck moving at a constant velocity of 18 m/s. When the car is 50 m behind the front of the truck, the car accelerates uniformly at 1.8 m/s2. The car continues at the same acceleration until it reaches a velocity of 25 m/s, which is the legal speed limit. The car then continues at a constant velocity of 25 m/s, until it passes the front of the truck. The distance the car travels while accelerating is closest to:

A) 54 m B) 50 m C) 58 m D) 66 m E) 62 m

Solution The car is accelerating uniformly (i.e. with constant acceleration) from v0 = 20 m/s to v1 = 25 m/s. The distance it travels during this process is given by

v12 − v0

2 = 2aΔx ⇒ Δx = v12 − v0

2

2a=

252 − 202( )m/s2 ×1.8m/s2 = 62.5m ≈ 62m

7) A ball is thrown straight upward with some initial speed. When it reaches the top of its flight (at a height h), a second ball is thrown straight upward with the same initial speed. Where will the balls cross paths? A) At height h B) Above height h/2 C) At height h/2 D) Below height h/2 but above 0 E) At height 0 Solution The first ball starts at the top with no initial speed. The second ball starts at the bottom with a large initial speed. Since the balls travel the same time until they meet, the second ball will cover more distance in that time, which will carry it over the halfway point before the first ball can reach it. 8) A particle starts from rest has acceleration 4.0m/s2 + 6.0m/s3t . Find the distance covered in 2.0 s.

A) 4.0 m B) 8.0 m C) 12 m D) 16 m E) 20 m

Solution

v t( ) = v0 + a t( ) t0

t

∫ dt = 0 + 4.0m/s2 + 6.0m/s3t( ) 0

t

∫ dt = 4.0m/s2t + 3.0m/s3t 2

x t( ) = x0 + v t( )t0

t

∫ dt = 0 + 4.0m/s2t + 3.0m/s3t 2( )0

t

∫ dt = 2.0m/s2t 2 +1.0m/s3t 3

x 2s( ) = 16m

9) A child standing on a bridge throws a rock straight down. The rock leaves the

child’s hand at t = 0. Which of the following graphs shown here best represents the velocity of the stone as a function of time?

A) B) C)

D) E)

Solution

Since both the initial velocity and the acceleration are downwards, v will start out negative and get even more negative as time goes on. The acceleration is constant so the slope of the line should also be constant.

10) Shown here are the velocity and the acceleration vectors for a car in several different types of motion. In which case is the car slowing down and turning to the right from the driver’s point of view?

A) B) C)

D) E) Solution Slowing down means that

a has a component opposing v . Turning right means

that a also has a component perpendicular to

v in the right direction (from the car’s point of view).

11) A girl shoots an arrow from the top of a cliff. The arrow is initially at a point 19 m above the level field below. The arrow is shot at an angle of 30o above horizontal with a speed of 15.4 m/s. How far out from the base of the cliff will the arrow land?

A) 21.0m B) 27.8m C) 17.8m D) 38.8m E) 28.3m

Solution We first calculate the time the arrow is in the air using the y-component of the velocity. Then we calculate the distance using the x-component. v0y = 15.4m/s( )sin 30o = 7.70m/s, v0x = 15.4m/s( )cos 30o = 13.3m/s

⇒ Δy = v0yt −gt 2

2 ⇒ t = 2.91s

⇒ x = v0xt = 38.8m

12) Shown in the figure to the right are the trajectories of four artillery shells.

Each was fired with the same speed. Which shell was in the air the longest time?

A) A B) B C) C D) D E) All were in the air for the

same time. Solution Shell A reaches the highest point and therefore takes the longest time to fall back to the ground. Since the time it takes going up is the same as the time it takes going down, it takes shell A the longest time for the journey.

13) An object moves at constant speed along a circular path in a horizontal plane, with the center at the origin. When the object is at x = -2 m and y = 0, its velocity is −4 m/s( ) j . What is the object’s acceleration when it is at y = 2 m?

A) 4m/s2( ) j

B) −4m/s2( ) j

C) 8m/s2( ) j D) −8m/s2( ) j

E) 8m/s2( ) i Solution The center of the circle is at the origin, this means that the velocity is along the y direction when y = 0. At x = −2 m velocity is in the negative y direction. Now, we can conclude that the object passes point (x, y) = (−2 m,0), moves in counter-clockwise direction, and the radius of the circle is r = 2 m. The magnitude of the acceleration is a = v2 / r = −4m/s( )2 / 2m = 8m/s2

At point y = 2m we have x = 0, the velocity is −4 m/s( ) i , and the (centripetal) acceleration is directed to the center – in the negative y direction. Finally:

a = − 8m/s2( ) j

14) Snow is falling vertically at a constant speed of 8.0 m/s. At what angle from the vertical do the snowflakes appear to be falling as viewed by the driver of a car traveling on a straight, level road with a speed of 50 km/h?

A) 40 B) C) D) E)

Solution

θ = arctan vcar / vsnow( )

θ = arctan 50 km/h8 m/s

⎛⎝⎜

⎞⎠⎟= arctan

50km/h( ) 103m/km( )3.6 ⋅103s/h( )8m/s

⎛

⎝⎜

⎞

⎠⎟ = 60

15) Which of the arrows correctly indicates the direction of the acceleration of a particle that moves clockwise at a constant speed around the path shown in the figure? A) P, R and T B) P, R, T and U C) Q and S D) R and U E) None of the arrows correctly indicates the direction of acceleration. Solution The acceleration points to the inside of the curve (which is not the same as the inside of the shape). Therefore only P, R and T are correct.

16) Three forces FA ,FB ,FC are applied to a mass located at the origin.

FA = −3N( )

i + 4N( )

j , FB = −FB j andFC = 5N . The mass is not moving. GiveFB .

A) 2N B) 4N C) 5N D) 8N E) 9N

Solution From the x coordinate:

FA,x + FB,x + FC ,x = 0 ⇒ FC ,x = 3N

⇒ FC = 5N ⇒ FC ,y = ± FC2 − FC ,x

2 = ±4N

From the y coordinate:

FA,y + FB,y + FC ,y = 0 ⇒ FB,y = −(4 ± 4)N

Thus, FB is either 0N or 8N.

17) A box with a mass of 10 kg is suspended by a cable in an elevator that is moving up with an acceleration of 1.2 m/s². What is the tension in the cable?

A) 86 N B) 98 N C) 110 N D) 120 N E) 130 N

Solution T − mg = ma ⇒ T = m(g + a) = 10kg(9.8 +1.2)m/s2 = 110N 18) A system of blocks and a frictionless and massless pulley is shown in the figure below. Block A has a mass of 8.0 kg and is on a rough surface (µs = 0.40). Block C has a mass of 5.0 kg. An external force P = 14.0 N, applied vertically to block A, maintains the system in static equilibrium. The mass of block B is closest to:

A) 3.8 kg B) 2.9 kg C) 2.6 kg D) 3.5 kg E) 3.2 kg

Solution The system is in static equilibrium. Thus, the net force acting on the knot above B has to be zero. The tensions in the three strings at that knot have to add up to zero. From the vertical coordinate we get TA,y + TB,y + TC ,y = 0 − mBg + mCgcos 40o = 0

⇒ mB = mC cos 40o = 3.83kg

19) A mass, m1 = 100 g, slides on a frictionless inclined plane that is at an angle θ = 30with respect to the horizontal, as shown in the figure. This mass is attached to an ideal string that goes over an ideal massless pulley. The other end of the string is attached to a hanging mass m2 = 200g. What is the acceleration of mass m1?

A) 1.2 m/s2 B) 2.3 m/s2 C) 3.5 m/s2 D) 4.9 m/s2 E) 9.8 m/s2

Solution: From the x-coordinate, which we choose to be along the string, and ignoring the internal force of the system, the tension in the string, we get using Newton’s second law: m1gsinθ − m2g = (m1 + m2 )a

⇒ a = g m1 sinθ − m2

m1 + m2

= 9.8m/s2 0.100kg × 0.5 − 0.200kg0.100kg + 0.200kg

= −4.9m/s2

The acceleration is 4.9 m/s2 in negative x-direction. 20) Three blocks of mass 3m, 2m, and m are connected by strings and pulled with constant acceleration a. Rank tensions from biggest to smallest.

A) T3 > T2 > T1 B) T3 > T1 > T2 C) T2 > T3 >T1 D) T1 = T2 = T3 E) T1 > T2 > T3

Solution: T1 pulls the whole set of blocks along, so it must be the largest. T2 pulls the last two masses, but T3 only pulls the last mass, so T3 < T2 < T1.

T3 T2 T1 3m 2m m

a

21) Block A of mass 3 kg and block X are attached to a rope, which passes over a pulley. A force P = 20 N is applied horizontally to block A, keeping it in contact with a rough vertical face. The coefficients of static and kinetic friction are µs = 0.40 and µk = 0.30. The pulley is light and frictionless. The mass of block X is set so that block A descends at constant velocity when it is set into motion. The mass of block X is closest to:  

A) 3.3 kg B) 2.4 kg C) 3.6 kg D) 3.0 kg E) 2.7 kg

Solution Block A moves at constant velocity, that means that there is no acceleration and that the net force on each object is zero. From the free body diagram of block X we see that the tension T is equal to the weight of block X. Block A is moving down and thus the friction force fk opposing the motion is pointing upward. From the free body diagram of block A, we see that fk + T − mag = 0 ⇒ µkP + T = mag ⇒ T = mag − µkP = (3.0kg × 9.8m/s2 ) − 0.30 × 20N = 23.4N

We calculate the mass of block X from the tension T:

⇒ mX = T / g = ma −µkPg

= 2.4kg

22) As shown in the figure, a constant external force P = 170 N is applied to a 20 kg box, which is on a rough horizontal surface. The force pushes the box a distance of 8.0 m, in a time interval of 6.0 s, and the speed changes from v1 = 0.3 m/s to v2 = 2.5 m/s. The work done by the external force P is closest to:

A) 1180 J B) 810 J C) 1060 J D) 940 J E) 680 J

Solution W = Pd cos300 = (170N)(8.0m) × 0.866 = 1180J

23) A ball of mass m = 40 g is tied to a string and is being whirled around in a circle of radius R = 10 cm with constant speed v = 5.0 cm/s. What is the work done by tension as the ball makes one and a half revolution?

A) 1.0 ×102π( )J

B) 3.0 ×102π( )J

C) 1.0 ×10−4π( )J

D) 3.0 ×10−4π( )J E) 0

Solution Since the ball is moving at constant speed its kinetic energy doesn’t change. Therefore, no work is done on the ball. Also, at any time the tension is perpendicular to the velocity and thus the displacement. It’s the centripetal force. The work done by the centripetal force, which is perpendicular to the displacement is zero.

24) In order to do work on an object,

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

the object. D) It is necessary that friction be present. E) The object must move.

Solution The work W done on an object by a force

F over a displacement Δ

r is given by W =

FiΔr = FΔr cosθ , where θ is the angle between

F and Δ

r . So, if there is no displacement because the object does not move, no work is done on the object.

25) A car traveling at 33.0 mph skids to a stop in 39 meters from the point where the brakes were applied. In approximately what distance would the car stop had it been going at 66.0 mph?

A) 39 m B) 110 m C) 55 m D) 78 m E) 156 m

Solution The kinetic energy of the car is four times larger when the speed doubles. The kinetic friction of the car with the ground brings the car to a stop. The kinetic friction µkmg is the same. Thus, the distance to stop the car Δx =W / fK where the work is given by the kinetic energy of the car: W = ΔK . So, the car would stop in four times the original distance 39m × 4 = 156m . Also, from v2 − v0

2 = 2aΔx with v = 0. Using Newton’s second law we get

−v02 = −2 fK

mΔx = −

2µKNΔxm

= −2µKgΔx . Thus with two times larger initial speed v0

we get a four times larger displacement Δx. 26) A book is at rest on a table. Which of the following forces forms a Newton’s third law pair with the book’s weight? A) The normal force of the table on the book

B) The force of the book on the table C) The gravitational force of the earth on the book D) The gravitational force of the book on the earth E) None of the above

Solution The weight of the book is the gravitational force of the earth on the book. This force forms a Newton’s third law pair with the gravitational force of the book on the earth:

Fearth,book = −

Fbook ,earth

27) A sand mover at a quarry lifts 2,000 kg of sand per minute a vertical distance of 12.0 meters. The sand is initially at rest and is discharged at the top of the sand mover with speed 5.00 m/s into a loading chute. At what minimum rate must power be supplied to the sand mover?

A) 1.13 kW B) 6.65 kW C) 0.524 kW D) 4.34 kW E) 3.92 kW

Solution The work done on the sand is given by the work needed to lift the sand and the work needed to cause the observed change in kinetic energy W = mgh + 1

2 mv2

= m(gh + 12 v

2 ) = 2000kg (9.80m/s2 ×12.0m) + 12 (5.00m/s)

2( )= 260,200J

The power is the work done per time interval P = 260,000J / 60s = 4340W