TIME OF COMPLETION - Clayton State University OF COMPLETION_ NAME_ DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 2 Section 1 Version 1 October 30, 2002

TIME OF COMPLETION_______________ NAME_____________________________

DEPARTMENT OF NATURAL SCIENCES

PHYS 1111, Exam 2 Section 1

Version 1 October 30, 2002

Total Weight: 100 points

1. Check your examination for completeness prior to starting. There are a total of ten (10)

problems on eight (8) pages.

2. Authorized references include your calculator with calculator handbook, and the Reference

Data Pamphlet (provided by your instructor).

3. You will have 80 minutes to complete the examination.

4. The total weight of the examination is 100 points.

5. There are six (6) multiple choice and four (4) calculation problems. Work all problems.

Show all work; partial credit will be given for correct work shown.

6. If you have any questions during the examination, see your instructor who will

be located in the classroom.

7. Start: 6:00 p.m.

Stop: 7:20 p.m

PROBLEM

POINTS

CREDIT

1-6

30

7

15

8

15

9

20

10

20

TOTAL

100

PERCENTAGE

CIRCLE THE SINGLE BEST ANSWER FOR ALL MULTIPLE CHOICE QUESTIONS. IN

MULTIPLE CHOICE QUESTIONS WHICH REQUIRE A CALCULATION SHOW WORK FOR

PARTIAL CREDIT.

1. A

a. 27,491 furlongs/fortnight. (5.00 yards/second)(1.00 furlong/220 yards)(14 x 24

x 3600 seconds/1.00 fortnight) =

27,491furlong/fortnight

(5)

b. 13,674 furlongs/fortnight.

c. 6,221 furlongs/fortnight.

d. 2,749 furlongs/fortnight.

2. On the Moon, the acceleration due to gravity is only about 1/6 of that on Earth. An astronaut

whose weight on Earth is 600 N travels to the lunar surface. His mass as measured on the Moon will

be:

a. 600 kg.

b. 100 kg. W = m g m = W/g m = (600 N)/(9.80 m/s2

) = 61.2 kg

(5)

c. 61.2 kg.

d. 9.81 kg.

3. When we subtract a displacement vector from a velocity vector, the result is:

a. A velocity.

b. An acceleration.

(5)

c. Another displacement.

d. Result doesn’t have physical meaning.

4. Which of the following is an example of the type of force which acts at a distance?

a. Gravitational.

b. Magnetic.

(5)

c. Electrical.

d. All of the above.

5. Imagine a book sitting on a horizontal table. One of the forces acting on a book is a normal

force exerted by the table (let’s call it action). What force does form the Newton’s 3rd

law pair

(reaction) ?

e. Weight of the book.

f. Weight of the table.

(5)

g. Normal force exerted on the table by the book.

h. Friction.

6. Take a look at the graph representing one-dimensional motion of a body. Is the instantaneous

velocity of the body at point A:

a. Positive.

(5)

b. Negative.

c. Zero.

d. Cannot be determined from the graph.

7. Coasting due west on your bicycle at 8.00 m/s, you encounter a sandy patch of road 7.20 m across.

When you leave the sandy patch your speed has been reduced to 6.50 m/s.

(15)

a. Assuming the bicycle slows with constant acceleration, what was its acceleration

in the sandy patch? Give both magnitude and direction.

V2 = V0

2 + 2 a ( x- x0) V0 = 8.00 m/s; V = 6.50 m/s; x0 = 0 m; x = 7.20 m

a = (V2 - V0

2)/ 2( x- x0)

a = {(6.50 m/s)2 - (8.00 m/s)

2}/{2 x (7.50 m)} = -1.51 m/s

2 west or 1.51 m/s

2 east

b. What time did it take you to cross the patch?

V = V0 + a t

t = (V - V0)/a

t = (6.50 m/s – 8.00 m/s) / (-1.51 m/s2) = 0.993 s

c. What was your average velocity on the sand?

Vaverage = (x - x0)/(t - t0) = (7.20 m)/(0.993 s) = 7.25 m/s

8. Deep inside an ancient physics text you discover two vectors:

A: 45.0 m @150.0 o

B: 30.0 m @ -15.0 o

Not content with these hoary relics, you are asked to find a new vector R = A - B.

Find the magnitude and direction of vector R

(15)

A: 45.0 m @ 150

o

B: 30.0 m @ -15.0o

Ax = A cos(A) = 45.0m cos(150o

) = -39.0 m

Ay = A sin(A) = 45.0m sin(150o

) = 22.5 m

Bx = B cos(B) = 30.0m cos(-15.0o

) = 29.0 m

By = B sin(B) = 30.0m sin(-15.0o

) = -7.76 m

Rx = Ax - Bx = -39.0 – 29.0 m = -68.0 m

Ry = Ay - By = 22.5 m + 7.76 m = 30.3 m

R = (Rx

2

+ Ry

2

)1/2

= ( (-68.0 m)2

+ (30.3 m)2

)1/2

= 74.4 m

R = tan-1

(Ry /Rx) = tan-1

(30.3 m /-68.0 m) = -24.0o

+ 180o

= 156o

(Rx < 0)

9. A soccer ball is kicked with a speed of 9.50 m/s at an angle of 40.0o above the horizontal

direction. Some time later the ball lands at the same level from which it was kicked.

a. Find the components of the ball’s initial velocity.

Vi: 9.50 m/s @ 40.0

o

Vx0 = V0 cos(0) = 9.50m/s cos(40.0o

) = 7.28 m/s

Vy0 = V0 sin(0) = 9.50m/s sin(40.0o

) = 6.11 m/s

(20)

b. What are the components of the ball’s velocity at the top of the trajectory?

Vx = 7.28 m/s

Vy = 0 m/s

c. How long does it take the ball to reach the top of the trajectory?

Vy = Vy0 - gt

(Vy0 - Vy) /g = t

t = (6.11 m/s – 0 m/s)/9.80 m/s2 = 0.623 s

d. How far does the ball go in the horizontal direction?

x = x0 + Vx0 (t) = 0 m + (7.28 m/s) (1.25 s) = 9.07 m

10. During your winter break, you enter a dogsled race in which students replace the dogs. Wearing

cleats for traction, you begin the race by pulling on a rope attached to the sled with a force of 150 N

at 25o with the horizontal. The mass of the sled is 80.0 kg and there is negligible friction between the

sled and ice.

a. Draw a free-body diagram.

(20)

b. Find the acceleration of the sled.

Ax = A cos() = (150 N) cos(

) = 136 N

Ay = A sin() = (150 N) sin(

) = 63.4 N

nx = n cos(9) = 0

ny = n sin(9) = n

wx = w cos(270) = 0

wy = w sin(270) = – (80.0 kg) (9.80 m/s

2

) = -784 N

Fx = m ax Ax = m ax ax = Ax /m = (136.0 N)/(80.0 kg) = 1.70 m/s2

c. Find the normal force exerted by the surface on the sled.

Fy = m ay ay = 0 ny + wy + Ay = 0

n – 784 N – 63.4 N = 0

n = 721 N

Problem 5

TIME OF COMPLETION_______________

NAME______SOLUTION_______________________






problems on seven (7) pages.









7. Start: 10:30 a.m.

Stop: 11:45 a.m

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

20

10

15

TOTAL

100

PERCENTAGE



PARTIAL CREDIT.

1. Which of the following operations will not change a vector?

a. Translate it parallel to itself.

b. Rotate it.

(5)

c. Multiply it by a constant factor.

d. Add a constant vector to it.

2. Suppose that an object travels from one point in space to another. Make a comparison

between the displacement and the distance traveled.

a. The displacement is either greater than or equal to the distance traveled.

b. The displacement is always equal to the distance traveled.

(5)

c. The displacement is either less than or equal to the distance traveled.

d. The displacement can be either greater than, smaller than, or equal to the distance

traveled.

3. The position, x, of an object is given by the equation x = A + Bt +Ct2, where t refers to time.

What are the dimensions of A, B, and C?

a. Distance, distance, distance.

b. Distance, time, time2.

(5)

c. Distance, distance/time, distance/ time2.

d. Distance/time, distance/ time2, distance/ time

3.

4. In the diagram shown, the unknown vector is

a.

b.

(5)

c.

d. Cannot be determined from the diagram.

5. Which of Newton's laws best explains why motorists should buckle-up?

a. The first law.

b. The second law.

(5)

c. The third law.

d. The law of gravitation.

6. A net force F accelerates a mass m with an acceleration a. If the same net force is applied to

mass 3m, then the acceleration will be

a. 3a.

b. 2a.

(5)

c. a/2.

d. a/3.

7. A projectile is fired with an initial speed of 40.2 m/s at an angle of 35.0o above the horizontal on a

long flat firing range. Determine

a. The components of the projectile’s initial velocity.

Vi: 40.2 m/s @ 35.0o

Vx0 = V0 cos(0) = (40.2 m/s) cos(35.0o) = 32.9 m/s

Vy0 = V0 sin(0) = (40.2 m/s) sin(35.0o) = 23.1 m/s

b. The maximum height reached by the projectile.

Vy2= Vy0

2- g (y – y0)

(y – y0) = (Vy2 - Vy0

2)/(-g)

(y – 0) = (02 – (23.1 m/s)

2)/(-9.81 m/s

2)

y = 54.4 m

c. The time it takes to reach the maximum height.

Vy = Vy0 - gt

(Vy0 - Vy) /g = t

t = (23.1 m/s – 0 m/s)/(9.81 m/s2) = 2.35 s

d. The projectile’s range.

x = x0 + Vx0 (2t) = 0 m + (32.9 m/s) (4.70 s) = 155 m

Bonus (5 extra points): Find velocity of the projectile 1.00 s after firing.

Vx = Vx0 = 32.9 m/s

Vy = Vy0 – gt = (23.1 m/s) – (9.81 m/s2)(1.00 s) = 13.3 m/s

V = (Vx2 + Vy

2)1/2

= ( (32.9 m/s)2 + (13.3 m/s)

2)

1/2 = 35.5 m/s

= tan-1

(Vy /Vx) = tan-1

((13.3 m/s) /(32.9 m/s)) = 22.0o

8. A stone is thrown vertically upward with a speed of 18.0 m/s.

a. How fast is it moving when it reaches a height of 7.00 m?

Vy2= Vy0

2- g (y – y0)

Vy2= (18.0 m/s)

2- (9.81 m/s

2)((7.00 m) – 0)

Vy = +/- 16.0 m/s

b. How long is required to reach this height?

Vy = Vy0 - gt

(Vy0 - Vy) /g = t

t1 = (18.0 m/s – 16.0 m/s)/(9.81 m/s2) = 0.204 s

t2 = (18.0 m/s + 16.0 m/s)/(9.81 m/s2) = 3.47 s

c. Why are there two answers to (b)?

Because the stone reaches this point twice: on its way up and on its way down.

9. A student decides to move a box of books into her dormitory room by pushing on the box. She

pushes with a force of 100 N at an angle of 15.0o with the horizontal. The box has a mass of 25.0

kg, and we are going to neglect friction between the box and the floor for simplicity.

a. Draw a free body diagram.

b. Find the acceleration of the box.

Ax = A cos(-15.0o) = (100 N) cos(-15.0

o) = 96.6 N

Ay = A sin(-15.0o) = (100 N) sin(-15.0

o) = -25.9 N

nx = n cos(90.0o) = 0

ny = n sin(90.0o) = n

wx = w cos(-90.0o) = 0

wy = w sin(-90.0o) = – (25.0 kg) (9.81 m/s

2) = -245 N

Fx = m ax

Ax + 0 + 0 = m ax

ax = Ax /m

ax = (96.6 N)/(25.0 kg) = 3.86 m/s2

c. Find the normal force the box exerts on the floor.

Fy = m ay

ay = 0

ny + wy + Ay = 0

n – 245 N – 25.9 N = 0

n = 271 N

Note: this is the normal force that the FLOOR exerts on the BOX, but the third Newton’s Law

states that this force is equal in magnitude and opposite in direction to the force that the BOX

exerts on the FLOOR.

10. Vector A has a magnitude of 127.0 m/s and points in a direction 20.0o below the x axis. A

second vector, B, has a magnitude of 80.0 m/s and points in a direction 95.0o above the x

axis.

Find a new vector R = A + B. Find the magnitude and direction of vector R.

A: 127 m/s @ -20.0o

B: 80.0 m/s @ 95.0o

Ax = A cos(A) = (127 m/s) cos(-20.0o) = 119 m/s

Ay = A sin(A) = (127 m/s) sin(-20.0o) = - 43.4 m/s

Bx = B cos(B) = (80.0 m/s) cos(95.0o) = -6.97 m/s

By = B sin(B) = (80.0 m/s) sin(95.0o) = 79.7 m/s

Rx = Ax + Bx = 119 m/s + (-6.97 m/s) = 112 m/s

Ry = Ay + By = - 43.4 m/s + 79.7 m/s = 36.3 m/s

R = (Rx2 + Ry

2)

1/2 = ( (112 m/s)

2 + (36.3 m/s)

2)1/2

= 118 m/s

qR = tan-1

(Ry /Rx) = tan-1

((36.3 m/s) /(112 m/s)) = 18.0o


NAME______SOLUTION_______________________











5. There are six (6) multiple choice and four (4) calculation problems. Work all calculation

problems and 5 (five) multiple choice. Show all work; partial credit will be given for correct work

shown.



7. Start: 6:00 p.m.

Stop: 7:15 p.m

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT.

1. The metric prefix for one one-hundredth is

a. Milli.

b. Centi.

(4)

c. Kilo.

d. Mega.

2. An object moving in the +x axis experiences an acceleration of 2.0 m/s2. This means the

object is

a. Traveling at 2.0 m in every second.

b. Traveling at 2.0 m/s in every second.

(4)

c. Changing its velocity by 2.0 m/s.

d. Increasing its velocity by 2.0 m/s in every second.

3. When an object is released from rest and falls in the absence of friction, which of the

following is true concerning its motion?

a. The speed of the falling object is proportional to its mass.

b. The speed of the falling object is proportional to its weight.

(4)

c. The speed of the falling object is inversely proportional to its surface area.

d. None of the above is true.

4. Ignoring air resistance, the horizontal component of a projectile's velocity

a. Is zero.

b. Remains constant.

(4)

c. Continuously increases.

d. Continuously decreases.

5. When a football in a field goal attempt reaches its maximum height, how does its speed

compare to its initial speed?

a. It is zero.

b. It is less than its initial speed.

(4)

c. It is equal to its initial speed.

d. It is greater than its initial speed.

6. If you exert a force F on an object, the force which the object exerts on you will

a. Depend on whether or not the object is moving.

b. Depend on whether or not you are moving.

(4)

c. Depend on the relative masses of you and the object.

d. Always be F.

7. Vector 1V

is 6.6 units long and points along the negative x axis. Vector 2V

is 8.5 units long and

points at º45 to the positive x axis.

a. What are the x and y components of each vector?

V1x = V1 cos(1) = (6.60) cos(180o) = - 6.60

V1y = V1 sin(1) = (6.60) sin(180o) = 0

V2x = V2 cos(2) = (8.50) cos(45.0o) = 6.01

V2y = V2 sin(2) = (8.50) sin( 45.0o) = 6.01

b. Determine the sum 21 VV

(magnitude and angle).

Rx = V1x + V2x = (- 6.60) + (6.01) = - 0.59

Ry = V1y + V2y = (0) + (6.01) = 6.01

R = ( Rx2

+ Ry2)1/2

= ( (0.59)2 + (6.01)

2)1/2

= 6.04

R = tan-1

(Ry/Rx) = tan-1

((6.01)/(-0.59)) = -84.4o+ 180

o = 95.6

o

8. A projectile is shot from the edge of a cliff with an initial speed of sm 0.65 at an angle of 37.0º

with the horizontal.

a. Determine the components of the initial velocity of the projectile.

V0x = V0 cos(0) = (65.0 m/s) cos(37.0o) = 51.9 m/s

V0y = V0 sin(0) = (65.0) sin( 37.0o) = 39.1 m/s

b. If it takes 10.5 s for the projectile to reach the base of the cliff, how high is the cliff?

y = y0 + Vy0 t – ½ g t2

0 = y0 + Vy0 t – ½ g t2

y0 = - Vy0 t + ½ g t2

= - (39.1 m/s) (10.5 s) + ½ (9.81 m/s2) (10.5 s)

2

y0 = 130 m

c. Determine the range X of the projectile as measured from the base of the cliff.

x = x0 + Vx0 t

x = (0 m) + (51.9 m/s) (10.5 s) =545 m

Bonus (5 points): Find the maximum height above the cliff top reached by the projectile.

Vy2 = Vy0

2 -2 g (y-y0)

0 = Vy0

2 -2 g (y-y0)

(y-y0) = Vy02 /(2 g) = 77.9 m

y = 208 m

9.The driver of a truck slams on the brakes when he sees tree blocking the road. The truck slows

down uniformly with an acceleration of -5.60 m/s2 for 4.20 s. If the truck was moving at 27.0

m/s the moment the driver sees the tree,

a. With what speed does the truck hit the tree?

V = V0 + a t

V = (27.0 m/s) + (-5.60 m/s2)(4.20 s) = 3.48 m/s

b. How long are the skid marks?

x = x0 + V0 t + ½ a t2

x = 0 + (27.0 m) (4.20 s) + ½ (-5.60 m/s2) (4.20 s)

2

x = 64.0 m

c. What is the average velocity of the truck over these 4.20 s?

V av = (x - x0)/(t - t0) = (64.0 m)/(4.20 s) = 15.2 m/s

10. At the instant a race began, a 65-kg sprinter exerted a force of 720 N on the starting block at

a 22º angle with respect to the ground.

a. Draw a free body diagram for the sprinter.

b. What was the horizontal acceleration of the sprinter?

Ax = A cos(22.0o) = (720 N) cos(22.0

o) = 668 N

Ay = A sin(22.0o) = (720 N) sin(22.0

o) = 270 N

nx = n cos(90.0o) = 0

ny = n sin(90.0o) = n

wx = w cos(-90.0o) = 0

wy = w sin(-90.0o) = – (65.0 kg) (9.81 m/s

2) = -638 N

Fx = m ax

Ax + 0 + 0 = m ax

ax = Ax /m

ax = (668 N)/(65.0 kg) = 10.3 m/s2

c. If the force was exerted for 0.32 s, with what speed did the sprinter leave the starting

block?

V = V0 + a t

V = (0 m/s) + (10.3 m/s2)(0.32 s) = 3.30 m/s


NAME___SOLUTION__________________________













shown.



7. Start: 6:00 p.m.

Stop: 7:15 p.m

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT.

1. The velocity, V, of an object is given by the equation V = A + Bt, where t refers to time. What

are the dimensions of A, and B?

A) distance/time, distance/time2.

B) distance/time, distance2/time

2.

(4)

C) distance, distance/ time2

D) distance/time2, distance/ time

3

2. Can an object's velocity change direction when its acceleration is constant? Support your

answer with an example.

A) No, this is not possible because it is always speeding up.

B) No, this is not possible because it is always speeding up or always slowing down, but

it can never turn around.

(4)

C) Yes, this is possible, and a rock thrown straight up is an example.

D) Yes, this is possible, and a car that starts from rest, speeds up, slows to a stop, and

then backs up is an example.

3. A ball is thrown straight up, reaches a maximum height, then falls to its initial height. Make a

statement about the direction of the velocity and acceleration as the ball is coming down.

A) Both its velocity and its acceleration point upward.

B) Its velocity points upward and its acceleration points downward.

(4)

C) Its velocity points downward and its acceleration points upward.

D) Both its velocity and its acceleration point downward.

4. An object is moving with constant non-zero acceleration in the +x axis. The position versus

time graph of this object is

A) a horizontal straight line.

B) a vertical straight line.

(4)

C) a straight line making an angle with the time axis.

D) a parabolic curve.

5. Ignoring air resistance, the horizontal component of a projectile's acceleration

A) is zero.

B) remains a non-zero constant.

(4)

C) continuously increases.

D) continuously decreases.

6. A child's toy is suspended from the ceiling by means of a string. The Earth pulls downward

on the toy with its weight force of 8.0 N. If this is the "action force," what is the "reaction

force"?

A) The string pulling upward on the toy with an 8.0-N force.

B) The ceiling pulling upward on the string with an 8.0-N force.

(4)

C) The string pulling downward on the ceiling with an 8.0-N force.

D) The toy pulling upward on the Earth with an 8.0-N force.

7. Vector 1V

is 20.7 m long and points at 55.0o to the positive x axis. Vector 2V

is 15.0 m long and

points at - 15.0o to the positive x axis.


V1x = V1 cos(1) = (20.7 m) cos(55.0o) = 11.9 m

V1y = V1 sin(1) = (20.7 m) sin(55.0o) = 17.0 m

V2x = V2 cos(2) = (15.0 m) cos(-15.0o) = 14.5 m

V2y = V2 sin(2) = (15.0 m) sin( -15.0o) = -3.88 m

b. Determine the vector 21 VV


Rx = V1x - V2x = (11.9 m) - (14.5 m) = -2.60 m

Ry = V1y - V2y = (17.0 m) - (-3.88 m) = 20.9 m

R = ( Rx2

+ Ry2)1/2

= ( (-2.60 m)2 + (20.9 m)

2)1/2

= 21.0 m

R = tan-1

(Ry/Rx) = tan-1

((20.9 m)/(-2.60 m)) = -82.9o+ 180

o = 97.1

o

8. A diver running sm 8.1 dives out from the edge of a vertical cliff at an angle of 25.0o with

the horizontal. He hits the water 5.00 m away from the base of the cliff.

a. What are the components of the initial velocity?

Vx0 = V0 cos(0) = (1.80 m/s) cos(25.0o) = 1.63 m/s

Vy0 = V0 sin(0) = (1.80 m/s) sin( 25.0o) = 0.761 m/s

b. How long was the diver in the air?

x = x0 + Vx0 t

t = (x - x0)/Vx0

t = (5.00 m)/(1.63 m/s) = 3.07 s

c. How high was the cliff ?

y = y0 + Vy0 t – ½ gt2

y0 = y - Vy0 t + ½ gt2

y0 = 0 – (0.761 m/s) (3.07 s) + ½ (9.81 m/s2)(3.07 s)

2 = 43.9 m

Bonus (5 points): How high above the edge of the cliff did the diver rise?

Vy2 = Vy0

2 -2 g (y-y0)

0 = Vy0

2 -2 g (y-y0)

(y-y0) = Vy02 /(2 g)

(y- 43.9 m) = (0.761 m/s)2 /(2 g)

y = 43.93 m (from the ground) or 0.0295 m from the edge of the cliff

9. A world-class sprinter can burst out of the blocks to essentially top speed (of about

sm 5.11 ) in the first 15.0 m of the race.

a. What is the average acceleration of this sprinter?

V2 = V0

2 +2 a (x-x0)

a = (V2 - V0

2)/(2(x-x0))

a = ((11.5 m/s)2 – 0)/(2(15.0 m)) = 4.41 m/s

2

b. How long does it take her to reach that speed?

V = V0 + at

t =(V - V0)/a = ((11.5 m/s) – 0)/( 4.41 m/s2) = 2.61 s

c. What is her average velocity during this time?

Vav = x/t = (15.0 m)/(2.61 s) = 5.75 m/s

10. A box sits at rest on a rough 30º inclined plane. Mass of the box is 10.0 kg.

a. Draw a free body diagram including all the forces acting on the box.

b. Find the magnitude of the normal force exerted on the box by the incline.

fx = f cos(0o) = f

fy = f sin(0o) = 0

nx = n cos(90.0o) = 0

ny = n sin(90.0o) = n

wx = w cos(-120o) = (10.0 kg) (9.81 m/s

2) cos(-120

o) = -13.8 N

wy = w sin(-120o) = (10.0 kg) (9.81 m/s

2) sin(-120

o) = -85.0 N

Fy = m ay

fy+ ny + wy = 0

n - 85.0 N = 0

n = 85.0 N

c. Find the magnitude of the friction force acting on the box.

Fx = m ax

fx+ nx + wx = 0

f -13.8 N = 0

f = 13.8 N


NAME__SOLUTION___________________________



Version 1 February 16, 2005



problems on eight (8) pages.









7. Start: 6:00 p.m.

Stop: 7:15 p.m

PROBLEM

POINTS

CREDIT

1-6

30

7

15

8

15

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT.

2. A furlong is a distance of 220 yards. A fortnight is a time period of 2 weeks. A race horse is

running at a speed of 5 yards per second. What is its speed in furlongs per fortnight?

e. 13,674 furlongs/fortnight.

f. 6,221 furlongs/fortnight.

(5)

g. 2,749 furlongs/fortnight.

h. 27,491 furlongs/fortnight

2. On the Moon, the acceleration due to gravity is only about 1/6 of that on Earth. An astronaut

whose weight on Earth is 600 N travels to the lunar surface. His mass as measured on the Moon will

be:

e. 600 kg.

f. 100 kg.

(5)

g. 61.2 kg.

h. 9.81 kg.

3. When we subtract a displacement vector from a velocity vector, the result is:

a. A velocity.

b. An acceleration.

(5)

c. Another displacement.

d. Result doesn’t have physical meaning.


a. Gravitational.

b. Magnetic.

(5)

c. Electrical.


5. Imagine a book sitting on a horizontal table. One of the forces acting on a book is a normal

force exerted by the table (let’s call it action). What force does form the Newton’s 3rd

law pair

(reaction)?

e. Normal force exerted on the Earth by the table.

f. Normal force exerted on the table by the Earth.

(5)

g. Normal force exerted on the table by the book.

h. Normal force exerted on the Earth by the book.

6. Take a look at the graph representing one-dimensional motion of a body. The instantaneous

velocity of the body at point A is

e. Positive.

f. Negative.

(5)

g. Zero.

h. Cannot be determined from the graph.

7. Coasting due west on your bicycle at 8.00 m/s, you encounter a sandy patch of road 7.20 m across.

When you leave the sandy patch your speed has been reduced to 6.50 m/s.

a. Assuming the bicycle slows with constant acceleration, what was its acceleration

in the sandy patch? Give both magnitude and direction.

V0= 8.00 m/s

V= 6.50 m/s

x – x0= 7.20 m

V2 = V0

2 + 2 a (x – x0)

a = (V2 - V0

2 )/[2(x – x0)] = -1.51 m/s

2

(15)

b. What time did it take you to cross the patch?

V = V0 + a t

t = (V - V0)/a = 0.993 s

c. What was your average velocity as you were crossing the patch?

Vav = (x – x0)/t = (7.20 m) / (0.993 s) = 7.25 m/s

8. Deep inside an ancient physics text you discover two vectors:

A: 45.0 m @150.0 o

B: 30.0 m @ -15.0 o

Not content with these hoary relics, you are asked to find a new vector R = A - B.

Find the magnitude and direction of vector R

Ax = A cos(A) = (45.0 m) cos(150.0o) = - 40.0 m

Ay = A sin(A) = (45.0 m) sin(150.0o) = 22.5 m

Bx = B cos(B) = (30.0 m/s) cos(-15.0o) = 30.0 m

By = B sin(B) = (30.0 m/s) sin(-15.0o) = -7.76 m

Rx = Ax - Bx = (-40.0 m) - (30.0 m) = -70.0 m

Ry = Ay - By = (22.5 m) - (-7.76 m) = 30.3 m

R = ( Rx2 + Ry

2)1/2

= ( (-70.0 m)2 + (30.3 m)

2)1/2

= 76.3 m

R = tan-1

(Ry/Rx) = = tan-1

((30.3 m)/(-70.0 m)) = -23.4o + 180

o =

157

o

(15)

9. A soccer ball is kicked with a speed of 9.50 m/s at an angle of 40.0o above the horizontal

direction. Some time later the ball lands at the same level from which it was kicked.

a. Find the components of the ball’s initial velocity.

V0: 9.50 m/s @ 40.0o

Vx0 = V0 cos(0) = (9.50 m/s) cos(40o) = 7.28 m/s

Vy0 = V0 sin(0) = (9.50 m/s) sin(40o) = 6.11 m/s

(20)

b. What are the components of the ball’s velocity at the top of the trajectory?

Vx = Vx0 = 7.28 m/s

Vy = 0 m/s

c. How long does it take the ball to reach the top of the trajectory?

Vy = Vy0 - gt

t = - (Vy - Vy0)/g = 0.623 s

d. How far does the ball go in the horizontal direction?

x = x0 + Vx0 t

x = (0 m) + (7.28 m/s) (2 x 0.623 s) = 9.08 m




sled and ice.

a. Draw a free-body diagram.

b. Find the acceleration of the sled.

Ax = A cos(A) = (150 N) cos(25.0o) = 136 N

Ay = A sin(A) = (150 N) sin(25.0o) = 63.4 N

wx = w cos(w) = (80.0 kg)(9.80 m/s2) cos(-90.0

o) = 0 N

wy = w sin(w) = (80.0 kg)(9.80 m/s2) sin(-90.0

o) = -784 N

nx = n cos(n) = n cos(90.0o) = 0 N

ny = n sin(n) = n sin(90.0o) = n

FX = m aX

136 N = (80.0 kg) aX

aX = 1.70 m/s2


FY = m aY = 0

63.4 N -784 N + n = 0

n = 721 N


NAME_SOLUTION___SOLUTION_________________________





1. Check your examination for completeness prior to starting. There are a total of nine (9)






5. There are five (5) multiple choice and four (4) calculation problems. Work all multiple choice

problems and 4 calculation problems. Show all work; partial credit will be given for correct work

shown.



7. Start: 6:00 p.m.

Stop: 7:15 p.m

PROBLEM

POINTS

CREDIT

1-5

20

6

20

7

20

8

20

9

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT.

1. If 1 inch = 2.54 cm, and 1 yd = 36 in., how many meters are in 7.00 yd?

a. 6.40 m.

(7.00 yd)(36 in/1 yd)(0.0254 m/1 in) = 6.40 m

b. 36.3 m.

(4)

c. 640 m.

d. .78 × 103 m.

2. Suppose that an object travels from one point in space to another. Make a comparison

between the displacement and the distance traveled.

a. The displacement is either greater than or equal to the distance traveled.

b. The displacement is always equal to the distance traveled.

(4)

c. The displacement is either less than or equal to the distance traveled.

d. The displacement can be either greater than, smaller than, or equal to the distance

traveled.

3. Ignoring air resistance, the horizontal component of a projectile's velocity

a. Is zero.

b. Remains constant.

(4)

c. Continuously increasing.

d. Continuously decreasing.

4. Ignore this problem.

5. Mass and weight

a. Both measure the same thing.

b. Are exactly equal.

(4)

c. Are two different quantities.

d. Are both measured in kilograms.

6. A person pushes a 14.0-kg lawn mower at constant speed with a force of N 0.88F

directed along the handle, which is at an angle of 45.0º to the horizontal. Making a highly

unrealistic assumption that there is no friction between the mower and the ground,

.

a. Draw the free-body diagram showing all forces acting on the mower.

b. Calculate the acceleration of the mower.

Ax = A cos(-45.0o) = (88.0 N) cos(-45.0

o) = 62.2 N

Ay = A sin(-45.0o) = (88.0 N) sin(-45.0

o) = -62.2 N

nx = n cos(90.0o) = 0

ny = n sin(90.0o) = n

wx = w cos(-90.0o) = 0

wy = w sin(-90.0o) = – (14.0 kg) (9.81 m/s

2) = -137 N

Fx = m ax

Ax + 0 + 0 = m ax

ax = Ax /m

ax = (62.2 N)/(14.0 kg) = 4.44 m/s2

c. Calculate the magnitude of the normal force that the ground exerts on the mower.

Fy = m ay

ay = 0

ny + wy + Ay = 0 n + (-137 N) + (-62.2 N) = 0

n = 199 N

7. A projectile is shot from the edge of a cliff with an initial speed of sm 0.65 at an angle of

37.0º with the horizontal. The projectile lands on the ground 10.0 s later.


V0: 65.0 m/s @ 37.0o

Vx0 = V0 cos(0) = (65.0 m/s) cos(37.0o) = 51.9 m/s

Vy0 = V0 sin(0) = (65.0 m/s) sin(37.0o) = 39.1 m/s

b. How high is the cliff?

y = y0 + Vy0 t – ½ g t2

0 = y0 + Vy0 t – ½ g t2

y0 = ½ g t2 - Vy0 t

y0 = ½ (9.81 m/s2) (10.0 s)

2 - (39.1 m/s) (10.0 s) = 99.5 m

c. Determine the range of the projectile as measured from the base of the cliff.

x = x0 + Vx0 (t) = 0 m + (51.9 m/s) (10.0 s) = 519 m

d. At the instant just before the projectile hits the ground, find the magnitude of the

velocity.

Vx = Vx0 = 51.9 m/s

Vy = Vy0 – g t = (39.1 m/s) - (9.81 m/s2) (10.0 s) = -59.0 m/s

V = (Vx 2 + Vy

2)

1/2 = 78.6 m/s

Bonus (5 points): Find the maximum height above the cliff top reached by the projectile.

Vy2 = Vy0

2 – 2 g (y – y0)

(y – y0) = (Vy2 - Vy0

2)/(-2 g)

(y – 0) = (02 – (39.1 m/s)

2)/(-2 x 9.81 m/s

2)

y = 77.9 m

8. A fugitive tries to hop on a freight train. The fugitive starts from rest and accelerates at

2sm 0.4a to his maximum speed of .sm 0.8

a. How long does it take him to reach the maximum speed?

V = V0 + a t

V = 0 + a t

t = V/a = (8.00 m/s)/(4.00 m/s2) = 2.00 s

b. What is the distance traveled to reach the maximum speed?

x = x0 + V0 t + ½ a t2 = ½ a t

2

x = ½ (4.00 m/s2)(2.00 s)

2 = 8.00 m

c. What is the average velocity of the fugitive during this time interval?

Vav = x/t = (8.00 m)/(2.00 s) = 4.00 m/s

9. A car is driven 215 km 25º north of west and then 85 km southwest. What is the displacement of

the car from the point of origin (magnitude and direction)? Draw a diagram.

A: 215 km @ 155o

B: 85.0 km @ 225o

Ax = A cos(A) = (215 km) cos(155o) = -195 km

Ay = A sin(A) = (215 km) sin(155o) = 90.9 km

Bx = B cos(B) = (85.0 km) cos(225o) = -60.1 km

By = B sin(B) = (85.0 km) sin(225o) = -60.1 km

Rx = Ax + Bx = -195 km + (-60.1 km) = -255 km

Ry = Ay + By = 90.9 km + (-60.1 km) = 30.8 km

R = (Rx2 + Ry

2)

1/2 = ( (-255 km)

2 + (30.8 km)

2)1/2

= 257 km

qR = tan-1

(Ry /Rx) = tan-1

((30.8 km) /(-255 km)) = -6.89o

+180o

= 173o


NAME____SOLUTION_________________________













shown.



7. Start: 6:00 p.m.

Stop: 7:15 p.m

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT.

1. If you are 5'10'' tall, what is your height in meters? (1 in = 2.54 cm.)

a. 1.5 m.

b. 1.6 m.

(4)

c. 1.7 m.

d. 1.8 m.

2. Suppose that a car traveling to the West (-x direction) begins to slow down as it approaches a

traffic light. Make a statement concerning its acceleration.

a. The car’s acceleration is positive.

b. The car’s acceleration is negative.

(4)

c. The acceleration is zero.

d. A statement cannot be made using the information given.

3. A ball is thrown with a velocity of 20.0 m/s at an angle of 60.0° above the horizontal. What is

the horizontal component of its instantaneous velocity at the exact top of its trajectory?

a. 10.0 m/s.

b. 17.0 m/s.

(4)

c. 20.0 m/s.

d. Zero.

4. The acceleration of an object is inversely proportional to

a. The net force acting on it.

b. Its position.

(4)

c. Its velocity.

d. Its mass.

5. Action-reaction forces

a. Sometimes act on the same object.

b. Always act on the same object.

(4)

c. May be at right angles.

d. Always act on different objects.

6. The acceleration due to gravity is lower on the Moon than on Earth. Which of the following

is true about the mass and weight of an astronaut on the Moon's surface, compared to Earth?

a. Mass is less, weight is same.

b. Mass is same, weight is less.

(4)

c. Both mass and weight are less.

d. Both mass and weight are the same.

7. Vector 1V

is 9.60 units long and points along the positive x axis. Vector 2V


points at 75.0o to the positive x axis.


V1: 9.60 units @ 0o

V2: 7.55 units @ 75.0o

V1x = V1 cos(1) = (9.60 units) cos(0o) = 9.60 units

V1y = V1 sin(1) = (9.60 units) sin(0o) = 0

V2x = V2 cos(2) = (7.55 units) cos(75.0o) = 1.95 units

V2y = V2 sin(2) = (7.55 units) sin(75.0o) = 7.29 units



Rx = V1x + V2x = 9.60 units + 1.95 units = 11.6 units

Ry = V1y + V2y = 0 + 7.29 units = 7.29 units

R = (Rx2 + Ry

2)

1/2 = ( (11.6 units)

2 + (7.29 units)

2)1/2

= 13.7 units

R = tan-1

(Ry /Rx) = tan-1

((7.29) /(11.6)) = 32.1o

8. A football is kicked at ground level with a speed of sm 0.18 at an angle of 35.0º to the

horizontal.


V0: 18.0 m/s @ 35.0o

Vx0 = V0 cos(0) = (18.0 m/s) cos(35.0o) = 14.7 m/s

Vy0 = V0 sin(0) = (18.0 m/s) sin(35.0o) = 10.3 m/s

b. How much later does it hit the ground?

y = y0 + Vy0 t – ½ g t2

0 = 0 + Vy0 t – ½ g t2

0 = Vy0 – ½ g t

2 Vy0 /g = t

t = 2.10 s

c. How far does it travel in the horizontal direction ?

x = x0 + Vx0 (t) = 0 m + (14.7 m/s) (2.10 s) = 30.9 m

9. A helicopter is ascending vertically with a speed of .sm 20.5 At a height of 125 m above

the Earth, a package is dropped from a window.

a. How much time does it take for the package to reach the ground? [Hint: The

package’s initial speed equals the helicopter’s.]

y = y0 + Vy0 t – ½ g t2

0 = (125 m) + (5.20 m/s) t – ½ g t2

t = 5.61 s

b. How fast does the package move just before it hits the ground?

V = V0 – g t = (5.20 m/s) - (9.81 m/s2) (5.61 s) = - 49.8 m/s

c. What is the acceleration of the package just before it heats the ground?

Acceleration due to gravity, 9.81 m/s2 downward




sled and ice.

a. Draw a free body diagram for the sled.

(20)

b.Find the acceleration of the sled.



NAME___SOLUTION__________________________













shown.



7. Start: 6:00 p.m.

Stop: 7:15 p.m

PROBLEM

POINTS

CREDIT

1-6

20

7

20

8

20

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT.

1. Which of the following is a scalar quantity?

A) Velocity.

B) Force.

(4)

C) Speed.

D) Acceleration.

2. The density of a solid object is defined as the ratio of the mass of the object to its volume. The

dimension of density is

A) [M]/[L].

B) [L]3/[M].

(4)

C) [M]/[L]3 .

D) [M][L]2.

3. A ball is thrown straight up. When it reaches its highest point,

A) Its velocity is zero.

B) Its acceleration is zero.

(4)

C) Both its velocity and acceleration are zero.

D) Neither is zero.

4. Vector A is along the x-axis and vector B is along the y-axis. Which one of the following

statements is correct?

A) The x-component of vector A is equal to the x-component of vector B.

B) The y-component of vector A is equal to the y-component of vector B.

(4)

C) The x-component of vector A is equal and opposite to the x-component of vector B.

D) Vector A has a zero y-component and vector B has a zero x-component.

5. Under which of the following conditions would a car have a westward acceleration?

A) The car going eastward and its speed is increasing.

B) The car going eastward and its speed is decreasing.

(4)

C) The car going westward and its speed is constant.

D) The car going eastward and its speed is constant.

6. When a parachutist jumps from an airplane, he eventually reaches a constant speed, called the

terminal velocity. This means that

A) The parachutist’s acceleration is equal to g.

B) The force of air resistance is equal to zero.

(4)

C) The effect of gravity has died down.

D) The force of air resistance is equal to the weight of the parachutist.

7. You are a player on a reality TV show. You and two other contestants are brought to the center of

a large, flat field. Each is given a meter stick, a compass, a calculator, a shovel, and following

two displacements:

72.4 m 32.0o east of north,

57.3 m 36.0o south of west.

These two displacements lead to the point where the keys to a new Porsche are buried. Two

players start measuring immediately, but you (knowledge is power!) first calculate where to go

and reach the point first. How far do you go and in what direction?

Ax = A cos(A) = (72.4 m) cos(58.0o) = 38.4 m

Ay = A sin(A) = (72.4 m) sin(58.0o) = 61.4 m

Bx = B cos(B) = (57.3 m) cos(216o) = -46.4 m

By = B sin(B) = (57.3 m) sin(216o) = -33.7 m

Rx = Ax + Bx = (38.4 m) + (-46.4 m) = -8.00 m

Ry = Ay + By = (61.4 m) + (-33.7 m) = 27.7 m

R = ( Rx2 + Ry

2)1/2

= ( (-8.00 m)2 + (27.7 m)

2)1/2

= 28.8 m

R = tan-1

(Ry/Rx) = tan-1

((27.7 m)/(-8.00 m)) = -73.9o +180 =106

o

11. A rock is tossed off a cliff at a speed of 20.0 m/s at an angle of 60.0o to the horizontal. The

rock lands at the base of the cliff 4.03 s later.


V0: 20.0 m/s @ 60.0o

Vx0 = V0 cos(0) = (20.0 m/s) cos(60.0o) = 10.0 m/s

Vy0 = V0 sin(0) = (20.0 m/s) sin(60.0o) = 17.3 m/s

b. How high is the cliff?

y = y0 + Vy0 t – ½ gt2

0 = y0 + Vy0 t – ½ gt2

y0 = - Vy0 t + ½ gt2

y0 = - (17.3 m/s)(4.03 s) – ½ (9.80 m/s2)(4.03 s)

2 = 9.94 m

c. What is the angle of impact?

Vx = Vx0 = 10.0 m

Vy = Vy0 - gt

Vy = (17.3 m/s) – (9.80 m/s2)(4.03 s) = -22.2 s

= tan-1

(Vy/Vx) = tan-1

((-22.2 m/s)/(10.0 m/s)) = -65.8o

Bonus (5 points): How high above the base of the cliff does the rock rise?

Vy2 = Vy0

2 - 2 g (y – y0)

(y – y0) = (Vy2 - Vy0

2)/(- 2 g)

Vy = 0 m/s

(y – y0) = (Vy02)/(2 g) = 15.3 m

12. A student throws a water balloon vertically downward from the top of the building. The

ballooh leaves the thrower’s hand with a speed of 6.00 m/s. Air resistance may be

ignored, so the water balloon is in free fall after it leaves the trower’s hand.

a.What is its speed after falling for 2.00 s?

(20)

b. How far does it fall in 2.00 s?

c. What is its velocity after falling 10.0 m?

13. A stockroom worker pushes a box with mass 11.2 kg on a horizontal surface with a constant

speed of 3.50 m/s. The worker’s force makes an angle of 30.0o with the horizontal. There is a

constant friction force acting on the box with a magnitude of 22.0 N.

d. Draw a free body diagram including all the forces acting on the box.

(20)

e. What is the magnitude of the force the worker must apply to maintain the motion? .

f. Find the magnitude of the normal force the floor exerts on the box.


NAME___SOLUTION__________________________



Version 1 June 10, 2004









problems multiple choice problems. Show all work; partial credit will be given for correct work

shown.



7. Start: 10:15 a.m.

Stop: 11:35 a.m

PROBLEM

POINTS

CREDIT

1-6

30

7

20

8

15

9

15

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT.

1. If L stands for length, T for time, and M for mass, the dimensions of acceleration are?

i. ML2.

(5)

j. ML/T.

k. L/T2.

l. L/M.

2. Which one of the following statements is NOT true in case of 2-dimentional motion?

a. If the acceleration is zero, the speed must be constant.

b. If the speed is constant, the acceleration must be zero.

(5)

c. If the acceleration is zero, the direction of motion must be constant.

d. If the speed and direction of the motion are constant, the acceleration must be

zero.

3. Figure below represents the parabolic trajectory of a projectile going from point A to point C.

What is the direction of the projectile’s acceleration at point B?

i. Up and to the right.

(5)

j. Down and to the left.

k. Straight up.

l. Straight down.

4. You have two vectors, A (6.00; -3.00) and B (-3.00; 4.00). Which one of the following vectors is

A + B?

a. (9.00; -7.00).

b. (3.00; 1.00).

(5)

c. (-18.0; -12.0).

d. (3.00; -7.00).

5.Imagine a physics textbook laying on a horizontal table. One of the forces acting on a textbook is a

normal force exerted by the table (let’s call it action). What force does form the Newton’s 3rd

law

pair (reaction) ?

a. Gravitational force exerted on the textbook by the Earth.

b. Gravitational force exerted on the table by the textbook.

(5)

c. Normal force exerted on the table by the textbook.

d. Normal force exerted by the Earth on the table.

6. Take a look at the graph representing one-dimensional motion of a body. Is the instantaneous

velocity of the body at point A:

i. Positive.

(5)

j. Negative.

k. Zero.

l. Cannot be determined from the graph.

7. A joyful physics student throws his cap into the air with an initial velocity of 24.5 m/s at 36.9o

from the horizontal. Another physics student catches the cap at the same height it was thrown.

a. What are the components of the cap’s initial velocity?

V0: 24.5 m/s @ 36.9

o

V0x = V0 cos(0) = 24.5 m/s cos(36.9o

) = 19.6 m/s

V0y = V0 sin(0) = 24.5 m/s sin(36.9o

) = 14.7 m/s

b. How long is the cap in the air?

y = y0 + V0yt – ½ g t2

0 = 0 + V0yt – ½ g t2

0 = (V0y – ½ g t) t

(V0y – ½ g t) = 0

t = 2 V0y /g = 2 (14.7 m/s)/ (9.80 m/s2) = 3.00 s

c. What is the maximum height of the cap’s trajectory?

y = y0 + V0yt – ½ g t2

y = (0 m) + (14.7 m/s)(1.50 s) – ½ (9.80 m/s2) (1.50)

2 = 11.0 m

d. How far does the cap travel in horizontal direction?

x = x0 + V0x (t) = 0 m + (19.6 m/s) (3.00 s) = 58.8 m

8. A bullet traveling at 350 m/s strikes a telephone pole and penetrates a distance of 12.0 cm

before stopping.

a. Assuming that acceleration is constant, find the acceleration of the bullet

inside the pole.

V2 = V0

2 + 2 a ( x- x0) V = 0 m/s; V0 = 350 m/s; x- x0= 0.120 m

a = (V2 - V0

2 )/ (2 ( x- x0))

a = -510,417 m/s2

b. How long does it take for the bullet to stop?

V = V0 + a t

t = (V - V0)/a

t = (0 m/s – 350 m/s) / (-510,417 m/s2) = 6.84 x 10

-4 s

c. What is the average velocity of the bullet inside the pole?

Vaverage = (x - x0)/(t - t0) = (0.120 m)/(6.84 x 10-4

s) = 175 m/s

9. Two cats pull horizontally on the ropes attached to a toy mouse. Cat A exerts a force of 130 N at

20O north of east and cat B exerts a force of 90 N due west. Find the magnitude of the resultant force

on the toy and its direction.

A: 130 N @ 20

o

B: 90.0 N @ 180.0o

Ax = A cos(A) = (130 N) cos(20o

) = 122 N

Ay = A sin(A) = (130 N) sin(20o

) = 44.5 N

Bx = B cos(B) = (90.0 N) cos(180o

) = -90.0 N

By = B sin(B) = (90.0 N) sin(180o

) = 0 N

Rx = Ax - Bx = 122 N – 90.0 N = 32.0 N

Ry = Ay - By = 44.5 N – 0 N = 44.5 N

R = (Rx

2

+ Ry

2

)1/2

= ( (32.0 N)2

+ (44.5 N)2

)1/2

= 54.8 N

R = tan-1

(Ry /Rx) = tan-1

(44.5 N /32.0 N) = 54.3o

10. A 30.0 kg sled is coasting with constant velocity of 5.00 m/s over perfectly smooth, level ice.

It enters a rough stretch of ice 20.0 m long in which the force of friction exerted by the ice on the

sled is 12.0 N.

a .Draw a free body diagram indicating all forces acting on the sled while on rough stretch.

b. Calculate the magnitude of the normal force the ice exerts on the sled.

nx = n cos(9) = 0

ny = n sin(9) = n

fx = f cos(180) = -f = -12.0 N

fy = f sin(180) = 0

wx = w cos(270) = 0

wy = w sin(270) = – (30.0 kg) (9.80 m/s

2

) = -294 N

Fy = m ay = 0

- 294 N + n = 0

n = 294 N

c. Calculate the acceleration of the sled while on rough stretch.

Fx = m ax

-12.0 N = (30.0 kg) ax

ax = -0.400 m/s2

Bonus (Worth 5 points): What is the coefficient of kinetic friction between ice and the sled?

f = n

= f/n = (12.0 N)/ (294 N) = 0.0408


NAME____SOLUTION_________________________















7. Start: 11:25 a.m.

Stop: 12:50 p.m

PROBLEM

POINTS

CREDIT

1-6

30

7

15

8

15

9

20

10

20

TOTAL

100

PERCENTAGE



PARTIAL CREDIT.

3. If a is acceleration, v is velocity, x is position, and t is time, then which equation is NOT

dimensionally correct?

m. t = x/v.

n. a = v2/x.

(5)

o. v = a/t.

p. t2 = 2x /a

2. Which of the following expresses a principle initially stated by Galileo and later incorporated into

Newton’s laws of motion?

i. An object’s acceleration is inversely proportional to its mass.

j. For every action is an equal but opposite reaction.

(5)

k. The natural condition for a moving object is to remain in motion.

l. The natural condition for a moving object is to come to rest..

3. A 7.00 kg bowling ball experiences a net force of 5.00 N. What is its acceleration?

m. 35.0 m/s2.

n. 7.00 m/s2.

(5) a = F/m = (5.00 N)/(7.00 kg)

o. 1.40 m/s2.

p. 0.714 m/s2.


a. Strong nuclear.

b. Weak nuclear.

(5)

c. Electromagnetic.


5. Imagine a box sitting on a horizontal table. One of the forces acting on a box is a normal force

exerted by the table (let’s call it action). What force does form the Newton’s 3rd

law pair (reaction)?

q. Normal force exerted on the Earth by the table.

r. Normal force exerted on the table by the Earth.

(5)

s. Normal force exerted on the table by the box.

t. Normal force exerted on the Earth by the box.

6. Vector A points north and vector B points east. If C = B – A, then vector C points:

m. North of east.

n. South of east.

(5)

o. North of west.

p. South of west.

7. A projectile is fired with an initial speed of 51.2 m/s at an angle of 44.5o above the horizontal on a

long flat firing range. Determine

a. The components of the projectile’s initial velocity.

V0: 51.2 m/s @ 44.5o

Vx0 = V0 cos(0) = (51.2 m/s) cos(44.5o) = 36.5 m/s

Vy0 = V0 sin(0) = (51.2 m/s) sin(44.5o) = 35.9 m/s

b. The maximum height reached by the projectile.

Vy2 = Vy0

2 - 2 g (y – y0)

(y – y0) = (Vy2 - Vy0

2)/(- 2 g)

Vy = 0 m/s

(y – y0) = (Vy02)/(2 g) = 65.8 m

c. The total time in the air.

Time to reach the top:

Vy = Vy0 – gt1/2

t1/2 = Vy0/g = 3.66 s

Total time: t = 2 t ½

t = 7.33 s

d. The projectile’s range.

x = x0 + Vx0 t

x = (0 m) + (36.5 m/s) (7.33 s) = 267 m

Bonus (5 extra points): Find velocity of the projectile 1.50 s after firing.

Vx = Vx0 = 36.5 m/s

Vy = Vy0 – gt = (35.9 m/s)- (9.80 m/s2)(1.50 s) = 21.2 m/s

V =((36.5 m/s)2 + (21.2 m/s)

2)1/2

= 42.2 m/s

= tan-1

(Vy/Vx) = = tan-1

((21.2 m/s)/(36.5 m/s)) = 30.0o

9. A stone is thrown vertically upward with a speed of 10.0 m/s from the edge of the cliff

5.0 m high.

a. How much later does it reach the bottom of the cliff?

y = y0 + Vy0 t – ½ gt2

y = 5.00 m

y0 = 0 m

Vy0 = +10.0 m/s

0 = 5 + 10 t – 4.9 t2

4.9 t2 - 10 t - 5 = 0

t = 2.46 s

b. What is its speed just before hitting?

Vy = Vy0 - gt

Vy = (10.0 m/s) – (9.80 m/s2)(2.46 s) = -14.1 s

|Vy| = 14.1 s

c. What total distance did it travel?

To the top:

Vy2 = Vy0

2 - 2 g (y – y0)

(y – y0) = (Vy2 - Vy0

2)/(- 2 g)

Vy = 0 m/s

y - y0 = (Vy02)/(2 g) = 5.10 m

Total distance traveled:

5.10 m + 5.10 m + 5.00 m = 15.2 m

10. Jets at JFK International airport accelerate from rest at one end of a runaway, and must

attain takeoff speed before reaching the other end of the runaway. A plane has

acceleration of 11.0 m/s2 and takeoff velocity of 90.0 m/s.

a. What is the minimum runaway required for this airplane?

V0= 0 m/s

V= 90.0 m/s

a = 11.0 m/s2

V2 = V0

2 + 2 a (x – x0)

(x – x0) = (V2 - V0

2)/(2 a)

(x – x0) = 368 m

b. How long time does it take the plane to attain the takeoff speed?

V = V0 + a t

t = (V - V0)/a = 8.18 s

c. What is the average velocity of the plane in the process?

Vav = (x – x0)/t = (368 m) / (8.18 s) = 45.0 m/s

10. Vector A has a magnitude of 50.0 m and points in a direction 20.0o below the x axis. A

second vector, B, has a magnitude of 70.0 m and points in a direction 50.0o above the x axis.

Find a new vector R = A + B. Find the magnitude and direction of vector R.

Ax = A cos(A) = (50.0 m) cos(-20.0o) = 47.0 m

Ay = A sin(A) = (50.0 m) sin(-20.0o) = -17.1 m

Bx = B cos(B) = (70.0 m) cos(50.0o) = 45.0 m

By = B sin(B) = (70.0 m) sin( 50.0o) = 53.6 m

Rx = Ax + Bx = (47.0 m) + (45.0 m) = 92.0 m

Ry = Ay + By = (-17.1 m) + (53.6 m) = 36.5 m

R = ( Rx2 + Ry

2)1/2

= ( (92.0 m)2 + (36.5 m)

2)1/2

= 99.0 m

R = tan-1

(Ry/Rx) = tan-1

((36.5 m)/(92.0 m)) = 21.6o


NAME__SOLUTION___________________________













shown.



7. Start: 1:30 p.m.

Stop: 2:45 p.m

PROBLEM

POINTS

CREDIT

1-6

30

7

30

8

30

9

30

10

30

TOTAL

150

PERCENTAGE



PARTIAL CREDIT.

1. Which of the following is a vector quantity?

a. Time

b. Mass.

(6)

c. Displacement.

d. Speed.

PHYS 1111 Exam 1, Version 1

Fall 2002 69

2. Suppose that a car traveling to the East (+ x direction) begins to slow down as it approaches a

traffic light. Make a statement concerning its acceleration.

a. The car’s acceleration is positive.

b. The car’s acceleration is negative.

(6)

c. The acceleration is zero.

d. A statement cannot be made using the information given.

3. If the position versus time graph of an object is a horizontal line, the object is

a. Moving with constant non-zero velocity.

b. Moving with constant non-zero acceleration.

(6)

c. At rest.

d. Moving with infinite velocity.

4. The acceleration of an object is linearly proportional to

a. The net force acting on it.

b. Its position.

(6)

c. Its velocity.

d. Its mass.

5. A ball is thrown straight up. When it reaches its highest point,


Fall 2002 70

a. Both its velocity and its acceleration are zero.

b. Its velocity is zero and its acceleration is not zero.

(6)

c. Its velocity is not zero and its acceleration is zero.

d. Neither its velocity nor acceleration is zero.

6. A ball is thrown straight into the air. Ignore air resistance. At the highest point of its

trajectory, the net force acting on it is

a. Equal to its weight.

b. Greater than its weight.

(6)

c. Less than its weight.

d. Zero.

7. Vector 1V

is 12.5 units long and points along the negative x axis. Vector 2V


points at 25.0o to the positive x axis.


(30)

V1x = V1 cos(1) = (12.5) cos(180o) = - 12.5

V1y = V1 sin(1) = (12.5) sin(180o) = 0

V2x = V2 cos(2) = (4.35) cos(25.0o) = 3.94

V2y = V2 sin(2) = (4.35) sin( 25.0o) = 1.84


Fall 2002 71



Rx = V1x + V2x = (- 12.5) + (3.94) = - 8.56

Ry = V1y + V2y = (0) + (1.84) = 1.84

R = ( Rx2

+ Ry2)1/2

= ( (-8.56)2 + (1.84)

2)

1/2 = 8.76

R = tan-1

(Ry/Rx) = tan-1

((1.84)/(-8.56)) = -12.1o+ 180

o = 168

o

14. A ball is thrown from the top of a 100-m building with an initial speed of 15.0 m/s at an angle

of 40.0º to the horizontal. The ball hits the ground 5.60 s later.


Vx0 = V0 cos(0) = (15.0 m/s) cos(40.0o) = 11.5 m/s

Vy0 = V0 sin(0) = (15.0 m/s) sin( 40.0o) = 9.64 m/s

b. How far from the base of the building does the ball land?

x = x0 + Vx0 t

x = (0 m) + (11.5 m/s) (5.60 s) =64.4 m

c. What is the maximum height that the ball reaches?

Vy2 = Vy0

2 -2 g (y-y0)

0 = Vy0

2 -2 g (y-y0)

(y-y0) = Vy02 /(2 g)

(y- 100 m) = (9.64 m/s)2 /(2 g)

y = 105 m (from the ground) or 4.74 m from the top of the building


Fall 2002 72

Bonus (8 points): Find the angle of the impact.

Vx = Vx0 = 11.5 m/s

Vy = Vy0 - gt = (9.64 m/s) – (9.81 m/s2)(5.60 s) = -45.3 m/s

= tan-1

(Vy/Vx) = tan-1

((-45.3 m/s)/(11.5 m/s)) = -75.8o

15. A bullet traveling at 350 m/s strikes a telephone pole and penetrates a distance of 12.0 cm

before stopping.

a. What is the acceleration of the bullet assuming it is constant?

V2 = V0

2 +2 a (x-x0)

a = (V2 - V0

2)/(2(x-x0))

a = (02 – (350 m/s)

2)/(2(0.120 m)) = - 510417 m/s

2 = - 5.10 x 10

5 m/s

2

b. How long did it take for the bullet to stop?

V = V0 + at

t =(V - V0)/a = (0 – (350 m/s))/( - 5.10 x 105 m/s

2) = 6.86 x 10

-4 s

10. Dr. K is trying to move a box full of physics textbooks (total mass of 30 kg) up the incline by

exerting a constant force of 200 N parallel to the surface of the incline. Luckily for her, the surface of

the incline is frictionless. The surface is inclined to the horizontal at an angle of 35o.

g. Draw a free body diagram including all the forces acting on the box.


Fall 2002 73

h. Find the magnitude of the normal force exerted on the box by the incline.

Ax = A cos(0o) = (200 N) cos(0

o) = 200 N

Ay = A sin(0o) = (200 N) sin(0

o) = 0 N

nx = n cos(90.0o) = 0

ny = n sin(90.0o) = n

wx = w cos(-125.0o) = (30.0 kg) (9.81 m/s

2) cos(-125.0

o) = -169 N

wy = w sin(-125.0o) = (30.0 kg) (9.81 m/s

2) sin(-125.0

o) = -241 N

Fy = m ay

Ay+ ny + wy = 0

n -241 N = 0

n = 241 N

i. Find the acceleration of the box.


Fall 2002 74

Fx = m ax

Ax + 0 + wx = m ax

ax = (Ax + wx)/m

ax = (200 N – 169 N)/(30.0 kg) = 1.03 m/s2


NAME___SOLUTION__________________________




Fall 2002 75



1. Check your examination for completeness prior to starting. There are a total of nine (9)

problems on six (6) pages.





5. There are six (6) multiple choice and three (3) calculation problems. Work all calculation


shown.



7. Start: 5:00 p.m.

Stop: 6:15 p.m.

PROBLEM

POINTS

CREDIT

1-6

25

7

25

8

25

9

25

TOTAL

100

PERCENTAGE



PARTIAL CREDIT.

1. A bird flies 3.00 km due west and then 2.00 km due north. What is the magnitude of the


Fall 2002 76

bird’s displacement?

u. 2.00 km.

v. 3.00 km.

(5)

w. 3.61 km.

x. 6.20 km.

2. The gas pedal in a car is sometimes referred to as “the accelerator”. Which other controls on

the vehicle can be used to produce acceleration?

m. The breaks.

n. The steering wheel.

(5)

o. The gear shift.

p. All of the above.

3.Two vectors appear as in figure below. Which combination points directly to the left?

a. QP

b. QP

(5)


Fall 2002 77

c. PQ

d. PQ

4. A ball fired from a canon at point 1 follows a trajectory shown below. Air resistance may be

neglected. Four possible vectors are also shown in the figure. Which vector best represents the

ball’s velocity at point 2?

a. A

.

b. B

.

(5)

c. C

.

d. D

.

5. A ball fired from a canon at point 1 follows a trajectory shown above. Air resistance may be

neglected. Four possible vectors are also shown in the figure. Which vector best represents the

ball’s acceleration at point 4?

a. A

.

b. B

.

(5)

c. C

.

d. D

.


Fall 2002 78

6. A pilot drops a bomb from a plane flying horizontally at a constant speed. Neglecting air

resistance, when the bomb hits the ground, the horizontal location of the plane will

a. Be behind the bomb.

b. Be directly over the bomb.

(5)

c. Be in front of the bomb.

d. Depend on the speed of the plane when the bomb was released.

7. A box is dropped from a spacecraft moving horizontally at a speed of 27.0 m/s, 100 m above the

surface of the Earth.

a. How long does it take the box to reach the surface?

V0: 27.0 m/s @ 0o

Vx0 = V0 cos(0) = (27.0 m/s) cos(0o) = 27.0 m/s

Vy0 = V0 sin(0) = (27.0 m/s) sin(0o) = 0 m/s

y = y0 + Vy0 t – ½ g t2

y - y0 = – ½ g t2

t = sqrt(2 (y0 -y) /g) = 4.51 s

b. What is the total horizontal displacement of the box?

x = x0 + Vx0 t

x = (0 m) + (27.0 m/s) (4.51 s) =122 m

c. What is the speed of the box when it strikes the surface?

Vx = Vx0 = 27.0 m/s

Vy = Vyo –g t = -44.0 m/s


Fall 2002 79

V = sqrt(Vx2 + Vy

2) = 51.9 m/s

16. A hot-air balloonist, rising vertically with a constant velocity of magnitude 5.00 m/s, releases

a sandbag. After it is released, the sandbag is in free fall.

a. If it takes the sandbag 10.0 seconds to reach the ground, how high above the ground

was the balloon when the sandbag was released?

y = y0 + Vy0 t – ½ g t2

0 = y0 + Vy0 t – ½ g t2

y0 = -Vy0 t + ½ g t2

y0 = - (5.00 m/s) (10.0 s) + ½ (9.80 m/s2) (10.0 s)

2 = 440 m

b.How high above the ground is the sandbag 5.00 s after its release?

y = y0 + Vy0 t – ½ g t2

y = 440 m + (5.00 m/s) (5.00 s) – ½ (9.80 m/s2) (5.00 s)

2 =

343 m

b. How long does it take the sandbag to reach its highest point?

Vy = Vyo – g t

0 = Vyo – g t

Vyo = g t

t = Vyo / g = 0.510 s

d. What is the acceleration of the sandbag just before it heats the ground?

Acceleration due to gravity, 9.80 m/s2 downward.


Fall 2002 80

9. Vector 1V

is 21.3 m long and points at 35.0o to the positive x axis. Vector 2V

is 17.2 m long and

points at - 55.0o to the positive x axis.


V1x = V1 cos(1) = (21.3 m) cos(35.0o) = 17.4 m

V1y = V1 sin(1) = (21.3 m) sin(35.0o) = 12.2 m

V2x = V2 cos(2) = (17.2 m) cos(-55.0o) = 9.86 m

V2y = V2 sin(2) = (17.2 m) sin( -55.0o) = -14.1 m

b. Determine the vector 21 VV


Rx = V1x - V2x = (17.4 m) - (9.86 m) = 7.58 m

Ry = V1y - V2y = (12.2 m) - (-14.1 m) = 26.3 m

R = ( Rx2

+ Ry2)1/2

= ( (7.58 m)2 + (26.3 m)

2)1/2

= 27.4 m

R = tan-1

(Ry/Rx) = tan-1

((26.3 m)/(7.58 m)) = 73.9o


Fall 2002 81

Documents

TIME OF COMPLETION - Clayton State University OF COMPLETION_____ NAME_____ DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 2 Section 1 Version 1 October 30, 2002

TIME OF COMPLETION - Clayton State University OF COMPLETION_ NAME_ DEPARTMENT OF NATURAL SCIENCES PHYS 1111, Exam 2 Section 1 Version 1 October 30, 2002