practice test for midterm exam - edl test for midterm exam formulas d = vt d t= v o t + ½ at2 d v o...

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South Pasadena • AP Physics Name _______________________________ Period ___ Date ___/___/___ PRACTICE TEST for Midterm Exam FORMULAS d = vt d = vot + ½ at 2 d = vo + v 2 t v = vo + at v 2 = vo 2 + 2ad v = vx 2 + vy 2 = tan 1 vy vx vx = v cos vy = v sin dx = vxt vy = voy + at dy = voyt + ½ at 2 vy 2 = voy 2 + 2ady F = ma W = mg P = F A F = mgsin g = 9.8 m/s 2 freq = rev t v = 2 π r (freq) aC = v 2 r FC = mv 2 r π = 3.14 FG = G mM r 2 FS μS N FK = μK FN G = 6.67 × 10 11 N·m 2 /kg 2 p = mv I = ∆p = m(v v0) = Ft pbefore = pafter KE = ½ mv 2 PE = mgh W = F·d = ∆KE P = W t KE = ½ mv 2 IMA = Fout Fin = din dout AMA = Fout-actual Fin-actual Eff = AMA IMA × 100% PEbefore + KEbefore = PEafter + KEafter Use these terms to identify each description given in terms 1-12 below: Acceleration Centripetal Acceleration Centripetal Force Coefficient of Friction Displacement Distance Energy Force Force of Friction Frequency Gravitational Force Gravitational Potential Energy Impulse Kinetic Energy Mass Momentum Normal Force Position Power Pressure Speed Tension Time Velocity Weight Work

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Page 1: PRACTICE TEST for Midterm Exam - Edl TEST for Midterm Exam FORMULAS d = vt d t= v o t + ½ at2 d v o + v 2 v = v o + at v2 = v o 2 + 2ad v = v x 2 + v y 2 = tan–1 v y v

South Pasadena • AP Physics Name _______________________________

Period ___ Date ___/___/___

PRACTICE TEST for Midterm Exam

FORMULAS

d = vt d = vot + ½ at2 d =

vo + v

2 t v = vo + at v2 = vo

2 + 2ad

v = vx2 + vy

2 = tan–1

vy

vx vx = v cos vy = v sin

dx = vxt vy = voy + at dy = voyt + ½ at2 vy2 = voy

2 + 2ady

F = ma W = mg P = F

A F = mgsin g = −9.8 m/s2

freq = rev

t v = 2 π r (freq) aC =

v2

r FC =

mv2

r π = 3.14

FG = G mM

r2 FS ≤ μS N FK = μK FN G = 6.67 × 10–11

N·m2/kg2

p = mv I = ∆p = m(v – v0) = Ft pbefore = pafter

KE = ½ mv2 PE = mgh W = F·d = ∆KE P = W

t KE = ½ mv2

IMA = Fout

Fin =

din

dout AMA =

Fout-actual

Fin-actual Eff =

AMA

IMA × 100% PEbefore + KEbefore = PEafter + KEafter

Use these terms to identify each description given in terms 1-12 below:

Acceleration Centripetal Acceleration Centripetal Force

Coefficient of Friction Displacement Distance

Energy Force Force of Friction

Frequency Gravitational Force Gravitational Potential Energy

Impulse Kinetic Energy Mass

Momentum Normal Force Position

Power Pressure Speed

Tension Time Velocity

Weight Work

Page 2: PRACTICE TEST for Midterm Exam - Edl TEST for Midterm Exam FORMULAS d = vt d t= v o t + ½ at2 d v o + v 2 v = v o + at v2 = v o 2 + 2ad v = v x 2 + v y 2 = tan–1 v y v

Description Quantity Vector/Scalar Units

1 A push or pull perpendicular to

an object’s motion. Centripetal force Vector N or

kg·m

s2

2 The attraction between two

objects due to their masses. Gravitational Force “Vector” N or

kg·m

s2

3 The change in an object’s

kinetic energy. Work Scalar J or

kg·m2

s2

4 The duration of an object’s

motion. Time Scalar s

5 The force exerted over an area. Pressure Vector Pa or kg

m·s2

6 The force exerted over time. Impulse Vector N·s or kg·m

s

7 The pull exerted on a string or

rope. Tension Vector N or

kg·m

s2

8 The rate at which an object’s

velocity changes. Acceleration Vector

m

s2

9 The rate at which momentum

changes. Force Vector N or

kg·m

s2

10 The rate of change of an

object’s position. Velocity Vector

m

s

11 The stored energy of an object

due to its height.

Gravitational Potential

Energy Scalar J or

kg·m2

s2

12 The total length traveled by an

object. Distance Scalar m

2 Kinematics: Motion in

One-Dimension

1. How long would it take a car, starting from rest

and accelerating uniformly in a straight line at 5

m/s2, to cover a distance of 200 m?

a) 9.0 s c) 12.0 s

b) 10.5 s d) 15.5 s

2. A 0.100 kg rubber ball is thrown downward from

the top of a building 30-m building with a speed of

4.0 m/s. How high above the ground is the ball

after 2.0 s?

a) 2.4 m d) 22.0 m

b) 8.0 m e) 27.6 m

c) 12.2 m

Page 3: PRACTICE TEST for Midterm Exam - Edl TEST for Midterm Exam FORMULAS d = vt d t= v o t + ½ at2 d v o + v 2 v = v o + at v2 = v o 2 + 2ad v = v x 2 + v y 2 = tan–1 v y v

3. Which of the following statements are about

uniformly accelerated motion?

Select two answers.

a) If an object’s acceleration is constant then it

must move in a straight line.

b) If an object’s acceleration is zero, then it’s

speed must remain constant.

c) If an object’s speed remains constant, then its

acceleration must be zero.

d) If the object’s direction of motion is

changing then its acceleration is not zero.

4. On a horizontal number line, a fly is at the

coordinate +6. The fly then flies and lands at the

coordinate of –2. If the time traveled by the fly is

4 s, what is the fly’s velocity?

a) 2.0 units/s

b) 1.0 unit/s

c) –0.50 units/s

d) –1.0 unit/s

e) –2.0 units/s

5. A tennis ball is tossed vertically from the ground

with a speed of 30 m/s. With what speed will the

ball hit the ground?

a) 0 m/s d) 30 m/s

b) 10 m/s e) 40 m/s

c) 20 m/s

6. At a particular time, an object is moving with a

velocity of –20 m/s and an acceleration of 20 m/s2.

Which of the following is true about the object’s

motion?

a) It is not moving.

b) It is speeding up (accelerating).

c) It is slowing down (decelerating).

d) Its speed is not changing.

e) It is both speeding up and slowing down.

3 Kinematics in Two Dimensions

(Projectile Motion & Vectors)

7. A ball was pitched with a speed of 40 m/s at an

angle of 35° to the ground. What is the vertical

component of the velocity of the ball?

a) 23 m/s d) 49 m/s

b) 33 m/s e) 70 m/s

c) 40 m/s

Questions 8-10: A football is kicked with a horizontal

velocity of 20 m/s and a vertical velocity of 15 m/s.

8. What is the speed of the ball when it reaches its

peak?

a) 0 m/s d) 25 m/s

b) 15 m/s e) 35 m/s

c) 20 m/s

9. What is the speed of the ball when it returns to the

ground?

a) 0 m/s d) 25 m/s

b) 15 m/s e) 35 m/s

c) 20 m/s

10. How long does the ball travel in the air?

a) 1.5 s d) 5 s

b) 3 s e) 6 s

c) 4 s

* * * * * * * * * * * * * * * * * * * * * * * *

Page 4: PRACTICE TEST for Midterm Exam - Edl TEST for Midterm Exam FORMULAS d = vt d t= v o t + ½ at2 d v o + v 2 v = v o + at v2 = v o 2 + 2ad v = v x 2 + v y 2 = tan–1 v y v

11. A stone is thrown horizontally with an initial

speed of 30 m/s from a bridge. Find the stone’s

total speed when it enters the water 4 seconds

later. (Ignore air resistance.)

a) 30 m/s c) 50 m/s

b) 40 m/s d) 60 m/s

12. A soccer ball, at rest on the ground, is kicked with

an initial velocity of 10 m/s at a launch angle of

30 . Calculate its total flight time, assuming that

air resistance is negligible.

a) 0.5 s c) 2 s

b) 1 s d) 4 s

13. Which one of the following statements is true

concerning the motion of an ideal projectile

launched at an angle of 45 to the horizontal?

a) The acceleration vector points opposite to the

velocity vector on the way up and in the same

direction as the velocity vector on the way

down.

b) The speed at the top of the trajectory is zero.

c) The object’s total speed remains

constant during the entire flight.

d) The vertical speed decreases on the way

up and increases on the way down.

4 Motion and Force: Dynamics

(Newton’s Three Laws of Motion)

14. A force of x newtons is applied to a crate on a

frictionless surface, and the crate accelerates at 2.0

m/s2. What force (in newtons) should be applied

for the crate to accelerate at 8.0 m/s2?

a) 0.5x

b) x

c) 2x

d) 3x

e) 4x

15. A 100-kg anchor is dropped in the water and falls

at a constant speed of 4.5 m/s. What is the force

of the resistance of the water encountered by the

anchor?

a) 10.2 N d) 530 N

b) 22.2 N e) 980 N

c) 450 N

16. A ball is released and rolls down a rough ramp

with an acceleration of 3.5 m/s2. Which of these

forces is NOT acting on the ball?

a) The weight of the ball.

b) The force of the ramp pushing on the

ball.

c) The force of friction between the ball

and the ramp.

d) The force of the push.

e) All of the above act on the ball.

17. A person standing on a horizontal floor feels two

forces: the downward pull of gravity and the

upward supporting force from the floor. These

two forces

a) have equal magnitudes and form an

action/reaction pair.

b) have equal magnitudes but do not form an

action/reaction pair.

c) have unequal magnitudes and form an

action/reaction pair.

d) have unequal magnitudes and do not form

an action/reaction pair.

18. A person who weighs 800 N steps onto a scale that

is on the floor of an elevator car. If the elevator

accelerates upward at a rate of 5 m/s2, what will

the scale read?

a) 400 N c) 1000 N

b) 800 N d) 1200 N

Page 5: PRACTICE TEST for Midterm Exam - Edl TEST for Midterm Exam FORMULAS d = vt d t= v o t + ½ at2 d v o + v 2 v = v o + at v2 = v o 2 + 2ad v = v x 2 + v y 2 = tan–1 v y v

19. A frictionless inclined plane of length 20 m has a

maximum vertical height of 5 m. If an object of

mass 2 kg is placed on the plane, which of the

following best approximates the net force it feels?

a) 5 N c) 15 N

b) 10 N d) 20 N

20. A 20 N block is being pushed across a horizontal

table by an 18 N force. If the coefficient of kinetic

friction between the block and the table is 0.4, find

the acceleration of the block.

a) 0.5 m/s2 c) 5 m/s2

b) 1 m/s2 d) 7.5 m/s2

21. The coefficient of static friction between a box and

a ramp is 0.5. The ramp’s incline angle is 30. If

the box is placed at rest on the ramp, the box will

do which of the following?

a) Accelerate down the ramp.

b) Accelerate briefly down the ramp but then

slow down and stop.

c) Move with constant velocity down the ramp.

d) Not move.

22. If all of the forces acting on an object balance so

that the net force is zero, then

a) the object must be at rest.

b) the object’s speed will decrease.

c) the object’s direction of motion can

change, but not its speed.

d) None of the above will occur.

23. Assuming a frictionless, mass-less pulley,

determine the acceleration of the blocks once they

are released from rest.

m

M

a) m g

M + m

b) M g

m

c) (M + m) g

(M – m)

d) (M − m) g

(M + m)

24. A block of mass m is at rest on a frictionless,

horizontal table placed in a laboratory on the

surface of the Earth. An identical block is at rest

on a frictionless horizontal table placed on the

surface of the Moon. Let F be the net force

necessary to give the Earth – bound block an

acceleration of a across the table. Given that g

Moon is one-sixth of g Earth, the force necessary to

give the Moon – bound block the same

acceleration a across the table is

a) F/6 c) F

b) F/3 d) 6 F

25. A force F of strength 20 N acts on an object of

mass 3 kg as it moves a distance of 4 m. If F is

perpendicular to the 4 m displacement, the work it

does is equal to

a) 0 J c) 80 J

b) 60 J d) 600 J

Page 6: PRACTICE TEST for Midterm Exam - Edl TEST for Midterm Exam FORMULAS d = vt d t= v o t + ½ at2 d v o + v 2 v = v o + at v2 = v o 2 + 2ad v = v x 2 + v y 2 = tan–1 v y v

26. Under the influence of a force, an object of mass 4

kg accelerates from 3 m/s to 6 m/s in 8 s. How

much work was done on the object during this

time?

a) 27 J c) 72 J

b) 54 J d) 96 J

5 Circular Motion and

Law of Gravitation

27. A 0.80-kg rubber stopper is connected to a string

and makes a circular path. Which of the following

requires the greatest tension on the string? (v =

speed of the rubber stopper;

r = radius of the circular path)

a) v = 6.0 m/s r = 0.40 m

b) v = 6.0 m/s r = 0.80 m

c) v = 9.0 m/s r = 0.60 m

d) v = 12.0 m/s r = 0.40 m

e) v = 12.0 m/s r = 0.80 m

28. A person’s weight is the same on Titan (one of

Saturn’s moons) as on Europa (one of Jupiter’s

moons). However, Europa’s radius is half that of

Titan’s. If the Europa has a mass of m kg, what is

Titan’s mass?

a) 0.25m d) 2m

b) 0.50m e) 4m

c) m

29. When a car is banking a curve (assuming circular

motion), which of the following is pointed in the

same direction as the friction between the tires and

the road?

I. Centripetal Acceleration

II. Centripetal Force

III. Instantaneous Velocity

a) I only.

b) II only.

c) III only.

d) I and II only.

e) I, II, and III.

(The Next Five Questions Are From the Chapter 5

Practice Test)

30. A ball is connected to a string and makes a circular

path. The tension on the string (which is the

centripetal force) is 6.0 N. If the radius remains

constant and the velocity doubles, what will be the

new tension on the string?

a) 1.5 N d) 12 N

b) 3.0 N e) 24 N

c) 6.0 N

31. A 0.020 kg rubber stopper is attached to one end of

a string that is passed through a plastic tube. At the

other end of the string weights are attached. A

student whirls the rubber stopper in a horizontal

circular path at constant speed, making ten

revolutions every 4 seconds. If the radius of the

circular path is 0.50 m, what is the centripetal

acceleration acting on the stopper?

a)0.16 m/s2 d) 12.5 m/s2

b) 2.5 m/s2 e) 123 m/s2

c) 5.0 m/s2

32. If an object moves with a frequency of 12

revolutions per minute, how many seconds does it

take to make three revolutions?

a) 9 d) 24

b) 12 e) 36

c) 15

33. A 500 g toy train (as shown below) completes 10

laps of its circular track in 1 minute and 40

seconds. If the diameter of the track is 1 meter,

find the train’s

(a) centripetal acceleration, ac = 0.197 m/s2

(b) centripetal force, Fc = 0.099 N

(c) period, T = 10 s

(d) frequency, f = 0.1 Hz

Page 7: PRACTICE TEST for Midterm Exam - Edl TEST for Midterm Exam FORMULAS d = vt d t= v o t + ½ at2 d v o + v 2 v = v o + at v2 = v o 2 + 2ad v = v x 2 + v y 2 = tan–1 v y v

34. Alan makes 38 complete revolutions on the

playground “Round-A-Bout” in 30 seconds. If the

radius of the Round-A-Bout is 1meter, determine

the

(a) Period of the motion, T = 0.789 s

(b) Frequency of the motion, f = 1.27 Hz

(c) Linear speed at which Alan revolves = 7.96 m/s

(d) Centripetal force on 40 kg Alan = 2530 N

6 Work and Energy

35. A 2.5-kg box is pushed across a surface with a

force of 2.0 N at a constant speed of 1.5 m/s. What

is the coefficient of kinetic friction between the

box and the surface?

a\ 0.080

b) 0.53

c) 0.80

d) 1.3

e) 1.9

36. A box of mass m slides down a frictionless

inclined plane of length L and vertical height h.

What is the change in its gravitational potential

energy?

a) − m g L c) − m g L/h

b) − m g h d) − m g h/L

37. While a person lifts a book of mass 2 kg from the

floor to a tabletop, 1.5 m above the floor, how

much work does the gravitational force do on the

book?

a) − 30 J c) 0 J

b) − 15 J d) 15 J

38. A block of mass 3.5 kg slides down a frictionless

inclined plane of length 6.4 m that makes an angle

of 30 with the horizontal. If the block is released

from rest at the top of the incline, what is its speed

at the bottom?

a) 5.0 m/s c) 8.0 m/s

b) 6.4 m/s d) 10 m/s

39. An astronaut drops a rock from the top of a crater

on the Moon. When the rock is halfway down to

the bottom of the crater, its speed is what fraction

of its final impact speed?

a) 1/4 c) 1/2

b) 1/2√2 d) 1/√2

40. A force of 200 N is required to keep an object

sliding at a constant speed of 2 m/s across a rough

floor. How much power is being expended to

maintain this motion?

a) 50 W c) 200 W

b) 100 W d) 400 W

41. A 2.5-kg box is pushed across a surface with a

force of 2.0 N at a constant speed of 1.5 m/s. What

is the coefficient of kinetic friction between the

box and the surface?

a) 0.080 d) 1.3

b) 0.53 e) 1.9

c) 0.80

Page 8: PRACTICE TEST for Midterm Exam - Edl TEST for Midterm Exam FORMULAS d = vt d t= v o t + ½ at2 d v o + v 2 v = v o + at v2 = v o 2 + 2ad v = v x 2 + v y 2 = tan–1 v y v

42. Two identical cars (they have the same mass) are

driving on a freeway. One is traveling at 72 mph,

while the other one is traveling at 32 mph. What

is the ratio of the kinetic energy of the first car to

that of the second car?

a) 0.444 d) 2.25

b) 0.667 e) 5.0

c) 1.50

43. A delivery man carries a 5.0-kg box up a flight of

stairs 4.0 m above the ground. How much work

was done on the box by the man?

a) 0 J d) 98 J

b) 20 J e) 196 J

c) 49 J

44. A 57-g tennis ball was dropped from a height of

2.0 m. If it bounces back up to a height of 1.8 m,

how much energy was transferred when it hit the

ground?

a) 0 J d) 0.56 J

b) 0.011 J e) 1.0 J

c) 0.11 J

45. A pulley system lifts a 20-kg box 1.5 m into the

air. If 6.0 m of rope was pulled, what was the

tension on the string?

a) 5.0 N d) 196 N

b) 49 N e) 784 N

c) 80 N

7 Linear Momentum

Chapter 7 will have the greatest number of questions

since no test on Chapter 7 will be given before the

Midterm and so more practice is needed.

46. A hammer that hits a nail with a 150 N force

delivers an impulse of 0.75 N • s. How long did

the hammer make contact with the nail?

a) 0.0050 s d) 2.0 s

b) 0.020 s e) 200 s

c) 0.50 s

47. A force of 15 N was applied to a 5.0 kg ball so it

accelerates from 2.0 m/s to 8.0 m/s. How long

was this force applied?

a) 0.50 s d) 3.0 s

b) 2.0 s e) 6.0 s

c) 2.5 s

48. Ball A has a mass of 2.0 kg and is rolling across a

smooth surface. If Ball B is moving twice as fast

as Ball A, but has the same momentum, what is

Ball B’s mass?

a) 0.50 kg

b) 1.0 kg

c) 2.0 kg

d) 4.0 kg

e) Not enough information to determine.

49. A car of a train moves at a constant speed of 20 m/s

toward another identical car that is at rest. If the

two cars lock together when they collide, what will

be the speed that the two cars will move?

a) 0 m/s d) 30 m/s

b) 10 m/s e) 40 m/s

c) 20 m/s

50. A boy delivers a 0.72 N·s impulse on a 1.2-kg toy

train by pushing it for 5.0 s. If the toy train’s initial

velocity was 0.60 m/s, what’s the final momentum

of the train?

a) 0.60 N·s d) 1.44 N·s

b) 0.72 N·s e) 3.00 N·s

c) 1.20 N·s

51. An object rolling at a constant speed has a

momentum of 60 N • s. If the mass of the object is

doubled and the speed is tripled, what is the new

momentum?

a) 10 N • s d) 120 N • s

b) 30 N • s e) 360 N • s

c) 60 N • s

Page 9: PRACTICE TEST for Midterm Exam - Edl TEST for Midterm Exam FORMULAS d = vt d t= v o t + ½ at2 d v o + v 2 v = v o + at v2 = v o 2 + 2ad v = v x 2 + v y 2 = tan–1 v y v

52. When a porcelain bowl is dropped 1 m on a

padded mat, it does not break. Which of these

is smaller compared to dropping it on the hard

ground?

I. The impulse.

II. The speed of the bowl.

III. The force of impact.

a) I only.

b) II only.

c) III only.

d) I and III only.

e) I, II, and III.

53. Runner A is moving at 3 m/s and Runner B is

moving at 6 m/s. If the two runners have the

same momentum, and Runner A has a mass of

x kg, what is the mass of Runner B in kg?

a) 0.33x

b) 0.50x

c) 2.0x

d) 3.0x

e) 6.0x

54. A particle is moving at a constant speed in a

circular path. The momentum vector is

pointing:

a) Curved along the path of the circle.

b) Inward, toward the center of the circle.

c) Outward, away from the center of the circle.

d) Tangent to the circle, in the direction of

the motion.

e) Tangent to the circle opposite to the

direction of motion.

55. The units for impulse are:

a) kg • m • s

b) kg • m • s–1

c) kg • m • s–2

d) kg • m–1 • s–2

e) kg • m2 • s–2

56. When a force of 30 N is applied to each of the

following, which will require the most time to

stop?

a) m = 30 kg v = 12 m/s

b) m = 30 kg v = 24 m/s

c) m = 45 kg v = 18 m/s

d) m = 60 kg v = 12 m/s

e) m = 60 kg v = 24 m/s

57. A force is applied to a block over a period of 6

seconds. The graph is shown below. What is

the impulse from t = 0 s to t = 3 s?

a) 3.3 N • s d) 30 N • s

b) 10 N • s e) 45 N • s

c) 15 N • s

58. A 20-N force is applied for 3 s against a 15-kg

ball moving at 5 m/s to slow it down. What is

the final speed of the ball?

a) 0.50 m/s d) 2.0 m/s

b) 1.0 m/s e) 2.5 m/s

c) 1.5 m/s

Fo

rce

(N)

Time (s)

5

10

15

0 0 2 4 6

Page 10: PRACTICE TEST for Midterm Exam - Edl TEST for Midterm Exam FORMULAS d = vt d t= v o t + ½ at2 d v o + v 2 v = v o + at v2 = v o 2 + 2ad v = v x 2 + v y 2 = tan–1 v y v

59. A 5.0-kg ball is dropped from the top of a

balcony for 3.0 s. Approximately what is the

impulse on the ball over this time?

a) 15 N • s

b) 30 N • s

c) 50 N • s

d) 75 N • s

e) 150 N • s

60. Which of the following is true for an inelastic

collision?

I. After the collision, the collided objects

move together as one object.

II. There is a loss of energy for the system

during the collision.

III. There is a loss of momentum for the

system during the collision.

a) I only.

b) II only.

c) III only.

d) I and II only.

e) I, II, and III.

61. Ball A, which has a mass of 1.0 kg, moves with

a velocity of +1.5 m/s. It collides elastically

with Ball B, which has a mass of 0.50 kg and is

at rest. If Ball A comes to a rest after the

collision, what is the final velocity of Ball B?

a) 0 m/s

b) 0.75 m/s

c) 1.0 m/s

d) 2.0 m/s

e) 3.0 m/s

62. Jon, who has a mass of 60 kg, runs 1.0 m/s

eastward and collides with Peter, who has a

mass of 80 kg and is at rest. After the inelastic

collision,

a) Both Jon and Peter will move eastward at

a speed less than 1.0 m/s.

b) Both Jon and Peter will move eastward at a

speed of 1.0 m/s.

c) Both Jon and Peter will move eastward at a

speed greater than 1.0 m/s.

d) Jon will stop moving, while Peter will move

eastward at a speed less than 1.0 m/s.

e) Jon will fall westward, while Peter will

move eastward at 1.0 m/s.

63. A 0.050 kg bullet is shot from a 5.0 kg rifle. If

the bullet travels at 80 m/s, what is the recoil

velocity of the rifle?

a) 0.0030 m/s

b) 0.010 m/s

c) 0.060 m/s

d) 0.80 m/s

e) 10 m/s

64. Ball A has a mass of 2.0 kg and Ball B has a

mass of 4.0 kg. When the two balls collide

elastically, which of the following is NOT true?

a) The force of Ball A hitting Ball B is equal to

in magnitude as the force of the Ball B

hitting Ball A.

b) The impulse of the collision is zero.

c) The net force on the system is zero.

d) The sum of the velocities of the balls

before the collision is equal to that after

the collision.

e) The total momentum of both balls before the

collision is equal to that after the collision.

Page 11: PRACTICE TEST for Midterm Exam - Edl TEST for Midterm Exam FORMULAS d = vt d t= v o t + ½ at2 d v o + v 2 v = v o + at v2 = v o 2 + 2ad v = v x 2 + v y 2 = tan–1 v y v

The Midterm Exam will not have

any Free Response Questions,

however, solving these Free

Response Questions that follow

will be helpful to your studying!

Free Response

Please show all work neatly with units for

computational problems Answers should be

recorded with the appropriate number of

significant figures, the correct direction (+ or –,

if applicable), and units. Any responses

without adequate work would not be given

full credit for a Unit Exam.

Free Response Answers

Please show all work neatly for computational

problems (I.E.S.A.). Answers should be

recorded with the appropriate number of

significant figures, the correct direction (+ or –,

if applicable), and units. Any responses

without adequate work would not be given

full credit.

Chapter 2: Motion in 1D

1. An object is moving in along a straight line

according to velocity vs. time graph below.

Every interval along the x-axis represents 1

s, and every interval along the y-axis

represents 1 m/s.

Describe the object’s motion from t = 0 s to

t = 6 s. (2 points)

From t = 0 to t = 2, the object is moving at

a constant speed of 3 m/s.

From t = 2 to t = 6, the object is moving

with constant acceleration from 3 m/s to 6

m/s.

2. A foam ball is launched from the ground

straight up with a velocity of 40 m/s. What

is the ball’s velocity after 6.2 s? (4 points)

t = 6.2 s v = v0 +

at = (40 m/s) + (–9.8 m/s2)(6.2 s) = –20.8

m/s

v0 = 40 m/s

v = ?

a = –9.8 m/s2

Chapter 3: Motion in 2D

3. A cannon ball is fired with a horizontal

velocity of 6.0 m/s and a vertical velocity of

2.5 m/s.

(a) What is the initial velocity with which

the cannonball was fired? (2 points)

vx = 6.0 m/s v =

vx2 + vy

2 = (6.0 m/s)2 + (2.5 m/s)2 = 6.5

m/s

vy = 2.5 m/s

v = ?

(b) After how many seconds will the ball

return to the ground? (4 points)

t = ?

dy = v0yt + ½ at2

dy = 0 m (0 m) =

(2.5 m/s)(t) + ½ (–9.8 m/s2)(t2)

v0y = 2.5 m/s t = 0.51

s

a = –9.8 m/s2

velo

city

time

Page 12: PRACTICE TEST for Midterm Exam - Edl TEST for Midterm Exam FORMULAS d = vt d t= v o t + ½ at2 d v o + v 2 v = v o + at v2 = v o 2 + 2ad v = v x 2 + v y 2 = tan–1 v y v

Chapter 4: Forces

4. In the setup shown, box A has a mass of

0.30 kg and box B has a mass of 0.025 kg.

Assume no friction.

(a) Draw the free body diagram for this

system. (2 points)

(b) What is the acceleration with which this

system moves? (4 points)

mA = 0.30 kg

Fnet =

msys asys

mB = 0.025 kg

WB – T

+ T = (mA + mB)(asys)

WA = mAg = (0.30 kg)(9.8 m/s2) = 2.94 N

(0.245 N) = (0.30 kg + 0.025 kg)(asys)

WB = mBg = (0.025 kg)(9.8 m/s2) = 0.245 N

a = 0.754 m/s2

Chapter 5: Circular Motion

5. A 0.150-kg mass is tied to a 0.60-m string

and swung to make a circular orbit.

(a) If the tension on the string is 20.0 N,

what is the speed with which the mass

travels? (4 points)

FC = 20.0 N FC =

mv2

r

m = 0.150 kg (20.0

N) = (0.150 kg)v2

(0.60 m)

v = ?

v = 8.9 m/s

r = 0.60 m

(b) Draw the vector that represents the path

of the mass if the string was let go. (2

points)

A

B

N

WA

WB

T

T

Page 13: PRACTICE TEST for Midterm Exam - Edl TEST for Midterm Exam FORMULAS d = vt d t= v o t + ½ at2 d v o + v 2 v = v o + at v2 = v o 2 + 2ad v = v x 2 + v y 2 = tan–1 v y v

Chapter 6: Energy

6. A 0.400-kg soccer ball was kicked from the

ground with an initial speed of 32.5 m/s.

(a) Find the maximum height if it is moving

at 12.5 m/s at that point. (4 points)

Before After

PEbefore + KEbefore = PEafter +

KEafter

m = 0.400 kg m = 0.400 kg

1

2 mv2 = mgh

h = 0 m h = ?

1

2 (0.400 kg)(32.5 m/s)2 =

(0.400 kg)(9.8 m/s2)h

v = 32.5 m/s v = 0 m/s

h = 53.9 m

(b) Why is the ball traveling slower at the

height of its path? Explain briefly using

energy arguments.

(2 points)

At the height of the path, all the kinetic

energy at the beginning has been

converted to potential energy. With less

kinetic energy, it travels at a slower

speed.

Chapter 7: Momentum

7. A 1500-kg car traveling at 15 m/s collides

inelastically with a 3500-kg pick-up truck at

rest.

(a) With what speed to the vehicles move

after the collision? (4 points)

Before After

pbefore = pafter

m1 = 1500 kg m1 = 1500 kg

m1v1 + m2v2 = (m1 + m2)v

v1 = 15 m/s m2 = 3500 kg

(1500 kg)(15 m/s) + (3500 kg)(0 m/s) =

(1500 kg + 3500 kg)v

m2 = 3500 kg V = ?

v = 4.5 m/s

v2 = 0 m/s

(b) If time of impact of the collision was 2.0

s, what force was experienced by the

pick-up truck? (4 points)

m = 3500 kg J = ∆p

= m(v – v0) = Ft

v0 = 0 m/s (3500

kg)(4.5 m/s – 0 m/s) = (F)(2.0 s)

v = 4.5 m/s F =

7875 N

F = ?

t = 2.0 s