Aug18 New tasks rev9-7-08 - Wikispacesdiaznova1.wikispaces.com/file/view/tipers2.pdf · Aug18 New tasks rev9-7-08 nT2C-CCT1: Two Vectors—Vector Subtraction ... Projectile Motion

1/51 Alpha version © nTIPERs nT 9-7-08

Aug18 New tasks rev9-7-08 nT2C-CCT1: Two Vectors—Vector Subtraction.................................................................................... 3 nT2C-CT1: Two Vectors—Vector Addition and Vector Subtraction ..................................................... 4 nT3A-CRT3: Velocity vs. Time Graph—Acceleration vs. Time Graph ................................................. 5 nT3A-CCT2: Velocity vs. Time Graphs of Two Objects I—Displacement............................................ 6 nT3A-CRT4: Velocity vs. Time Graphs of Two Objects I—Velocity Equations ................................... 7 nT3A-WWT3: Velocity vs. Time Graphs of Two Objects I—Velocity Equations ................................. 8 nT3A-TT3: Velocity vs. Time Graphs of Two Objects I—Velocity Equations ...................................... 9 nT3C-CCT1: Position Time Equations—Instantaneous Speed............................................................. 10 nT3C-TT1: Position Time Equations—Instantaneous Speed................................................................ 11 nT3G-CCT1: Position vs. Time Graph—Acceleration ......................................................................... 12 nT3G-CT1: Position vs. Time Graph—Acceleration............................................................................ 13 nT4A-CT2: Motorcycle Trips I—Displacement ................................................................................... 14 nT4A-CCT3: Motorcycle Trips II—Displacement ............................................................................... 15 nT4B-CCT1: Students’ Journeys—Average Velocity .......................................................................... 16 nT4B-CCT2: Students’ Journeys—Average Speed .............................................................................. 17 nT4C-CCT1: Motorcycle on Road Course—Acceleration ................................................................... 18 nT4E-QRT7: Projectile Motion for Two rocks—Velocity and Acceleration Graphs II........................ 19 nT4E-WWT2: Air Pucks—Speed......................................................................................................... 20 nT4E-LMCT1: Dropped Bomb—Landing Point .................................................................................. 21 nT5A-CT1: Curler Pushing Stone—Force on Stone............................................................................. 22 nT5A-CCT1: Curler Pushing Stone—Force on Stone .......................................................................... 23 nT5B-CCT1: Block Moving at Constant Speed—Forces on Block...................................................... 24 nT5B-TT1: Block Speeding Up—Forces on Block .............................................................................. 25 nT5B-RT7: Two Objects in Two Configurations Accelerating Upward—Force.................................. 26 nT5B-RT8: Two Connected Objects Accelerating Upward—Force..................................................... 27 nT5B-RT9: Two Connected Objects Accelerating Downward—Net Force ......................................... 28 nT5B-CT1: Blocks Moving at Constant Speed—Force on Block......................................................... 29 nT5F-CT2: Blocks Moving at Constant Speed—Normal Force on Block ............................................ 30 nT5F-WWT1: Blocks Moving at Constant Speed—Normal Force on Block ....................................... 31 nT5G-RT7: Hanging Mass—String Tension ........................................................................................ 32 nT5I-WBT1: Gravitational Force on Probe near the Sun and Earth—Location Diagram..................... 33 nT5I-WWT2: Two Planetary Objects— Gravitational Force on Each.................................................. 34 nT5I-QRT3: Two Objects—Gravitational Force on Each .................................................................... 35 nT5I-QRT4: Three Objects Exerting Gravitational Forces—Net Force on One .................................. 36 nT5I-QRT5: Two Objects—Gravitational Force on Each .................................................................... 37 nT5I-QRT6: Object in a Uniform Gravitational Field—Motion, Potential Energy, & Force................ 38 nT5I-QRT7: Gravitational Field Equipotential Surfaces—Work, Field, and Force .............................. 39 nT5I-QRT8: Gravitational Field Equipotential Surfaces—Force at Two Points................................... 40 nT6B-CCT3: Identical Toboggans on Horizontal—Speed ................................................................... 41 nT6B-CT1: Toboggans on Horizontal—Speed..................................................................................... 42 nT6D-CT1: Velocity and Acceleration Graphs for Two Falling Rocks—Initial Kinetic Energy.......... 43 nT6D-CRT1: Box Lifted Up Moving Up I—Kinetic Energy & Velocity Graphs ................................ 44 nT6D-CRT2: Box Lifted Up Moving Up II—Kinetic Energy & Velocity Graphs ............................... 45 nT6H-WWT2: Mass on a Compressed Tilted Spring System—Height up Incline ............................... 46 nT7D-CCT5: Colliding Carts that Stick Together—Final Kinetic Energy............................................ 47 nT7D-RT3: Bobs hitting Blocks—Final Speed .................................................................................... 48 nT8C-QRT7: Rotating Cylinder— Direction of Torque and Angular Momentum ............................... 49 nT9B-RT1: Six Pulley Systems—Force to Keep Moving Up............................................................... 50 nT9B-RT2: Six Pulley Systems—Force to Keep Moving Downward .................................................. 51



NT2C-CCT1: TWO VECTORS—VECTOR SUBTRACTION Two vectors labeled

!

r A and

!

r B each have a length six, and each makes a small angle α with the

horizontal as shown. Four students are arguing about the vector difference

!

r C =

r A "

r B .

!

6

!

6A

rB

r

Arlo: “Since we’re subtracting vector B, we flip it around so it points in the same direction as

vector A. The difference will be 12 units long and will point in the same direction as vector A.”

Bob: “We’re subtracting, so the resultant will be smaller than six. Both vectors point down, so the difference will point down as well.”

Celine: “When you flip vector B around to get negative B, it points up and to the left. Then we add it to vector A, we get a long vector pointing horizontally to the right.”

Delbert: “Both vectors are six long, so the difference is zero. It doesn’t point in any direction.”

Which, if any, of these students do you agree with? Arlo _____ Bob _____ Celine _____ Delbert _____ None of them______ Please explain carefully your reasoning.

None of the students is correct, although Celine is close. The difference will be almost 12 units long and will point to the left.


NT2C-CT1: TWO VECTORS—VECTOR ADDITION AND VECTOR SUBTRACTION Two vectors labeled

!

r A and

!

r B each have a length six, and each makes a small angle α with the

horizontal as shown. Let

!

r C =

r A +

r B and

!

r D =

r A "

r B .

!

6

!

6A

rB

r

Is the magnitude of vector C greater than, less than, or equal to the magnitude of vector D? Please carefully explain your reasoning.

Less than. Vector C points downward and is smaller than six units as shown in the vector sum here. Vector D points to the left and is larger than six units as shown in the vector difference here.


NT3A-CRT3: VELOCITY VS. TIME GRAPH—ACCELERATION VS. TIME GRAPH The time dependence of the velocity of an object is shown.

velocity

time

acceleration Which of the following graphs correctly shows the time dependence of the acceleration?

Answer (d) slope of graph acceleration

time

acceleration

time

time time

time time

(a)

(e)

(d)

(c)

(b)

(f)

accelerationacceleration

accelerationacceleration


NT3A-CCT2: VELOCITY VS. TIME GRAPHS OF TWO OBJECTS I—DISPLACEMENT The graphs below show the velocity of two objects during the same time interval.

v1

time64200

1

2

v2

time64200

1

2

Object 2

Object 1

Three students are discussing the displacements of these objects for this interval.

Andy: “I think Object 2 will have the greater displacement because it gets to a higher speed faster than Object 1.”

Badu: “No, Object 1 will have the greater displacement because it travels for longer than Object 2.”

Coen: “I think Andy has the right answer, but for the wrong reason. The reason Object 2 has the larger displacement is because the area under the graph is more.”

Which, if any, of these three students do you agree with? Andy_____ Badu _____ Coen _____ None of them______

Answer: Coen is correct, because the displacement is determined by the area under a velocity-time graph.


NT3A-CRT4: VELOCITY VS. TIME GRAPHS OF TWO OBJECTS I—VELOCITY EQUATIONS The graphs below show the velocity in m/s of two objects during the same time (in seconds) interval.

v1

time64200

1

2

v2

time64200

1

2

Object 2

Object 1

Write the equations for the velocity in m/s as a function of time in s for each one-second interval for these two motions up to five seconds.

Answer: solution fit to line mt +b where m is the slope which can be read from graph and then determine b by fitting it to one end or the other of the segment (check by testing it against the other end of the segment)

Object 1 Object 2 0 s—1 s v = t v = 2 t 1 s—2 s v = 2 - t v = 2 2 s—3 s v = -4 + 2 t v = 2 3 s—4 s v = 2 v = 8 - 2 t 4 s—5 s v = 10 - 2 t v = 0


NT3A-WWT3: VELOCITY VS. TIME GRAPHS OF TWO OBJECTS I—VELOCITY EQUATIONS The graphs below show the velocity in m/s of two objects during the same time (in seconds) interval.

v1

time64200

1

2

v2

time64200

1

2

Object 2

Object 1

A student writes the following equations for the velocity in m/s as a function of time in s for each one-second interval for the first five seconds for these two motions based on these graphs. Object 1 Object 2 0 s—1 s vf = (1 m/s2)(t)= t vf = (2 m/s2)(t)= 2 t 1 s—2 s vf = 1 m/s + (–1 m/s2)( t)=1 - t vf = 2 m/s= 2 2 s—3 s vf = (2 m/s2)( t)=2 t vf = 2 m/s= 2 3 s—4 s vf = 2 m/s=2 vf = 2 m/s + (–2 m/s2)( t)=2 - 2 t 4 s—5 s vf = 2 m/s + (–2 m/s2)( t)=2 - 2 t vf = 0 m/s= 0 What, if anything, is wrong with the student’s work? If something is wrong, identify it and explain how to correct it.

Answer: There are several problems with the student’s answer. The solution is generated by the fit to line mt +b where m is the slope which can be read from graph and then determine b by fitting it to one end or the other of the segment (check by testing it against the other end of the segment)



NT3A-TT3: VELOCITY VS. TIME GRAPHS OF TWO OBJECTS I—VELOCITY EQUATIONS The graphs below show the velocity in m/s of two objects during the same time (in seconds) interval.

v1

time64200

1

2

v2

time64200

1

2

Object 2

Object 1

A student writes the following equations for the velocity in m/s as a function of time in s for each one-second interval for the first five seconds for these two motions based on these graphs. Object 1 Object 2 0 s—1 s vf = (1 m/s2)(t)= t vf = (2 m/s2)(t)= 2 t 1 s—2 s vf = 1 m/s + (–1 m/s2)( t)=1 - t vf = 2 m/s= 2 2 s—3 s vf = (2 m/s2)( t)=2 t vf = 2 m/s= 2 3 s—4 s vf = 2 m/s=2 vf = 2 m/s + (–2 m/s2)( t)=2 - 2 t 4 s—5 s vf = 2 m/s + (–2 m/s2)( t)=2 - 2 t vf = 0 m/s= 0 There is an error in the student’s work, identify the error and explain how to correct it.

Answer: There are several problems with the student’s answer. The solution is generated by the fit to line mt +b where m is the slope which can be read from graph and then determine b by fitting it to one end or the other of the segment (check by testing it against the other end of the segment)



NT3C-CCT1: POSITION TIME EQUATIONS—Instantaneous Speed Three students are discussing the motion of an object which is described by the following equation with x in meters and t in seconds:

!

x t( ) = "4 " 9t + 2t 2

The students make the following contentions about the motion during the first two seconds:

Arial: “I think this object’s instantaneous speed will increase with time since the acceleration is positive.”

Bonnie: “No, the object will have a decreasing speed since the acceleration is directed opposite to the initial velocity.”

Chad: “I don’t think we can tell what will happen to the speed from this equation since it tells us about the position as a function of time, not about speed or velocity.”

Which, if any, of these three students do you agree with? Arial_____ Bonnie _____ Chad _____ None of them______

Answer: Bonnie is correct since the velocity and acceleration terms have opposite signs.


NT3C-TT1: POSITION TIME EQUATIONS—Instantaneous Speed A student is told that the motion of an object is described by the following equation with x in meters and t in seconds:

!

x t( ) = "4 " 9t + 2t 2 The student makes the following contention: “This object will be moving at 17 m/s two seconds after the start of the motion.” There is an error in the student’s work, identify the error and explain how to correct it.

Answer: There are two problems with the student’s answer. First, the student is taking the acceleration as increasing the speed of the object rather than decreasing it. Second, the acceleration is 4 m/s2, not the 2 m/s2 the student used. So the actual velocity would be -1 m/s.


NT3G-CCT1: POSITION VS. TIME GRAPH—Acceleration The graph below shows the position versus time graph for the motion of an object moving in one dimension. Six points are marked on the parabolic curve.

A

Time (s)

Position (m)

B

C

D

E

F

Three students discussing this graph make the following contentions:

Ahmad: “I think this object has a constant negative acceleration since the curve slopes downward.”

Benita: “No, the object will have an increasing negative acceleration since the slope of the curve is increasing.”

Courtney: “No, you are both wrong. The object is moving with increasing speed, that is what the slope means, so the acceleration has to be positive since the velocity is increasing”

Which, if any, of these three students do you agree with? Ahmad_____ Benita _____ Courtney _____ None of them______

Answer: Ahmad is correct, although his explanation could be better. It would be useful if he had explicitly mentioned in his explanation that it is a parabolic curve.


NT3G-CT1: POSITION VS. TIME GRAPH—Acceleration The graph below shows the position versus time graph for the motion of an object moving in one dimension. Six points are marked on the parabolic curve.

A

Time (s)

Position (m)

B

C

D

E

F

Is this object’s acceleration at D greater than, less than or equal to its acceleration at E? ______________________ Explain.

Answer: The acceleration is the same at both points.


NT4A-CT2: MOTORCYCLE TRIPS I—DISPLACEMENT Shown below are the paths two motorcyclists took on an afternoon ride.

Start

End

Start

End

15 km 19 km

8 km

4 km

Trip A Trip B

Is the displacement for the trip on the left greater than, less than or the same as the displacement for the trip on the right? __________ Explain.

Answer: The displacement is greater for the trip on the right.


NT4A-CCT3: MOTORCYCLE TRIPS II—DISPLACEMENT Shown below are the paths two motorcyclists took on an afternoon ride.

Start

End

15 km

8 kmTrip A

Start

End

15 km

8 kmTrip B

Three physics students are discussing the situation, they make the following contentions.

Ali: “The lengths of the paths the two motorcyclists travel are the same so they have the same

displacement.” Bob: “Well, the displacements are the same size, but their directions are opposite each other so they

are actually different.” Carol: “I agree with Ali that the displacements are the same, but the reason is that the motorcyclists

go between the same two points. What path they follow doesn’t matter.” Which, if any, of these three students do you think is correct?

Ali _____ Bob _____ Carol _____ None of them______

Please explain carefully your reasoning.

Answer: Bob is correct .


NT4B-CCT1: STUDENTS’ JOURNEYS—Average Velocity The two figures below show the paths followed by two students who went to the mall for pizza to celebrate after completing their physics final. Both went directly from their dorm to the nearby mall, but they returned by different routes taking different times. Values for the total distances they traveled and the total times they took to walk their routes are given in the figures.

BA DormMall Mall

Dorm

t = 1800 s

d = 3200 m

t = 4500 s

d = 6000 m Three other students discussing the average velocities for the first two students make the following contentions:

Adrian: “Obviously student A had the lower average velocity since 6000 m/4500 s is less than 3200 m/1800 s.”

Bonnie: “No, you can’t do it that way, since we don’t know how long they stayed at the mall so we

can’t actually find their average velocities.” Carlton: “No, you are both wrong this is one of those trick physics questions and the answer is that

the average velocity is zero for both students.”

Which, if any, of these three students do you think is correct?

Adrian _____ Bonnie _____ Carlton _____ None of them______


Carlton is correct since the displacements for both students are zero.


NT4B-CCT2: STUDENTS’ JOURNEYS—Average Speed The two figures below show the paths followed by two students who went to the mall for pizza to celebrate after completing their physics final. Both went directly from their dorm to the nearby hangout, but they returned by different routes taking different times. Values for the total distances they traveled and the total times they took to walk their routes are given in the figures.

BA DormMall Mall

Dorm

t = 1800 s

d = 3200 m

t = 4500 s

d = 6000 m Three other students are discussing the average speeds for the first two students and make the following contentions:

Adrian: “Obviously student A had the lower average speed since the average speed is the total distance over the total time and 6000 m divided by 4500 s is less than 3200 m divided by 1800 s.”

Bonnie: “No, you can’t do it that way, since we don’t know how long they stayed at the mall so we

can’t actually find their average speeds.” Carlton: “No, you are both wrong since this is one of those trick physics questions and the answer

is that the average speed is zero for both students since the average velocity is zero.”


Adrian _____ Bonnie _____ Carlton _____ None of them______


Adrian is correct since the average speed is the distance traveled, given in the figures, divided by the elapsed time.


NT4C-CCT1: MOTORCYCLE ON ROAD COURSE—ACCELERATION The figure below shows the top view of a road course on which a motorcycle race is being run. The figure shows one of the motorcycles in the race. It is braking at this point to get around the bend. The arrow in the figure represents the motorcycle’s velocity.

Three physics students are discussing the situation and they make the following contentions:

Alexi: “The motorcycle’s acceleration is in the opposite direction to the velocity since it is slowing down.”

Bindi: “No, the acceleration will have two components, one opposite the velocity and the other

toward the center of the curve.” Carlos: “No, you are both wrong this is one of those trick physics questions and the answer is that

the motorcycle doesn’t have an acceleration since it is braking.” Which, if any, of these three students do you think is correct?

Alexi _____ Bindi _____ Carlos _____ None of them______


Answer: Bindi is correct. Both the rate of motion and the direction of motion are changing.


NT4E-QRT7: Projectile Motion for Two rocks—Velocity and Acceleration Graphs II Two students throw two rocks horizontally from a cliff with different velocities. Both rocks hit at the same time but one rock (A) hits farther from the base of the cliff.

Using the system where up is + vertical direction, away from the cliff is + horizontal direction, and the origin is at the top of the cliff at the point of release, sketch velocity vs. time graphs for each of them.

Explain-

Answer

Both rocks hit at the same time but rock A hits farther from the cliff since it travels faster in the horizontal direction.


NT4E-WWT2: AIR PUCKS—Speed The figure below shows a top view of a puck on a very large air hockey table—a table with many tiny air holes so that the puck floats on a cushion of air over the table. The puck is initially sliding from top to bottom near the left side of the table when a stream of air from a hair dryer accelerates it toward the right side of the table. The acceleration lasts until the puck gets to the right side of the table. The trajectory the puck follows is shown in the figure.

A student considering this situation makes the following statement: “The puck will take just as long to get to the bottom of the table in this situation as it would if it had not been accelerated.” What, if anything, is wrong with the student’s statement? If something is wrong, identify it and explain how to correct it.

The student is correct, since the acceleration is perpendicular to the initial velocity, it will not affect the puck’s vertical velocity.


NT4E-LMCT1: DROPPED BOMB—LANDING POINT The figure below shows a military airplane that is flying along at 200 m/s at a height of 1200 m above the ground. The plane drops a bomb that falls to the ground.

200 m/s

1200 m

For each of the numbered changes at the bottom of the page identify, from the lettered choices below, what will happen to the horizontal distance the bomb travels while falling. (a) The horizontal distance will increase. (b) The horizontal distance will decrease. (c) The horizontal distance will not change. (d) This time the bomb will drop straight down vertically from the release point. (e) We cannot determine how this change will affect the horizontal distance.

For each of the following changes only the feature identified is modified from the situation above. 1) The plane’s speed is tripled. _______

A 2) The plane is climbing at a 35° angle at the release point. _______

A 3) The plane is flying in level flight at an altitude of 1100 m. _______

B 4) The mass of the bomb is increased. _______

C 5) The bomb is thrown down from the plane at 15 m/s. _______

B 6) The plane is diving at a 20° angle at a height of 2000m. _______

A 7) The plane’s speed decreases, but it is flying at an altitude of 1800 m. _______

E


NT5A-CT1: CURLER PUSHING STONE—Force on Stone The figures below show two curling stones (the playing pieces in the Winter Olympic sport of curling) that are being pushed by the thrower. For the two stones, which are identical, we are told the instantaneous speed and acceleration at the time shown.

A Bv = 3 m/s

a = 1 m/s2v = 2 m/s

a = 2 m/s2

Is the curler in situation A exerting (a) a larger, (b) smaller or (c) equal force on her stone as the curler in situation B? ____________________ Explain.

Answer: (b) The curler in B is exerting the larger force since the stone has a larger acceleration.


NT5A-CCT1: CURLER PUSHING STONE—Force on Stone The figures below show three curling stones (the playing pieces in the Winter Olympic sport of curling) that are being pushed by the thrower. For each stone, which are all identical, we are told the instantaneous speed and acceleration at the time shown.

A Bv = 3 m/s

a = 1 m/s2v = 2 m/s

a = 2 m/s2

Cv = 5 m/s

a = 1 m/s2

Three students discussing this situation make the following contentions:

Alan: “I think the thrower who has produced the largest speed is the one who has exerted the greater force.”

Benazir: “No, the thrower who is accelerating the stone at the greatest rate is exerting the largest force.”

Colin: “No, you are both wrong. You have to pay attention to both the speed and the acceleration so we multiply the speed by the acceleration”

Which, if any, of these three students do you agree with? Alan_____ Benazir _____ Colin _____ None of them______

Answer: Benazir is correct.


NT5B-CCT1: BLOCK MOVING AT CONSTANT SPEED—FORCES ON BLOCK A student uses a string to pull a block across a table at a constant speed of two meters per second. The string makes an angle θ with the horizontal. A second student makes a free-body diagram of the block, and then uses this free-body diagram to generate a vector sum diagram as shown.

Won block by earth

fon block by table

Non block by tableTon block by string

Non block by table

fon block by table

Won block by earth

Ton block by string

!

Three students are comparing the magnitudes of the forces in the vector sum diagram:

Anja: The vector sum diagram allows us to compare the magnitudes of all four forces: The weight is the largest, then the tension, then friction, then the normal force.

Barb: Well, the weight is definitely greater than the normal force. But there should be a net force to the right in the vector sum, and there isn’t. I don’t think we can use it to rank the other forces.

Cole: I think we can use it to say that the weight is greater than the normal force. Also, the tension is greater than the friction, since the friction is the same length of the dashed line, and this is equal to the tension times the cosine of theta (θ). But we can’t compare the vertical forces with the rest.

Which, if any, of these students do you agree with? Anja _____ Barb _____ Cole _____ None of them______ Please explain carefully your reasoning.

Cole is correct. From the vector sum diagram we can generate two equations: N + Tsinθ = W, and Tcosθ = f. But we cannot compare the vertical forces to the tension or the friction force without more information.

!


NT5B-TT1: BLOCK SPEEDING UP—FORCES ON BLOCK A student uses a string to pull a block across a table. The block is speeding up as it moves to the right. The string makes an angle θ with the horizontal. A second student makes a free-body diagram of the block, and then uses this free-body diagram to generate a vector sum diagram as shown.

Won block by earth

fon block by table

Non block by tableTon block by string

Non block by table

fon block by table

Won block by earth

Ton block by string

!

"F

There is a problem with the vector sum diagram made by the student. Explain what is wrong, and correct the vector sum diagram.

The vectors from the free-body diagram are correctly added head-to-tail, but the net force is pointing in the wrong direction for the forces as drawn. The net force is the resultant, and should point from the tail of the first vector drawn (the weight) and toward the head of the last vector drawn (the friction). We know that the net force must point to the right since the acceleration is in this direction. One possible correct vector sum is shown here.

!


NT5B-RT7: TWO OBJECTS IN TWO CONFIGURATIONS ACCELERATING UPWARD—FORCE Two objects (masses m1 and m2 with m1 < m2) are connected by massless wires in two different orders as shown. The weight of each mass is W1 and W2 and both systems have the same constant acceleration upward. They are pulled upward by forces FA and FB and the tension in the two connecting wires is TA and TB.

W1

W2

FA

TA

W1

W2

FB

TB

Rank the forces (W1, W2, TA , TB, FA and FB) in these systems from greatest to least. Greatest 1 _______ 2 _______ 3 _______ 4 _______ 5 _______ 6 _______ Least

OR, The forces are the same, but not zero. ___

OR, We cannot determine the ranking for the forces. ___


Answer: FA = FB and TA > TB since the acceleration is the same and W2 > W1 along with TA > W2 and TB > W1 so the ranking is FA = FB > TA > W2 > TB >W1


NT5B-RT8: TWO CONNECTED OBJECTS ACCELERATING UPWARD—FORCE Two objects with masses of m1 = 6 kg and m2 = 10 kg are connected by a massless wire. The weights of the two masses are W1 and W2. They are pulled upward by an applied force F. The result is a constant acceleration of 4 m/s2 upward for the two objects. The tension in the wire is T and F1 is the net force on m1 while F2 is the net force on m2.

Rank the forces (W1, W2, F, T , F1 =Fnet on m1 and F2 =Fnet on m2) in this system from greatest to least. Greatest 1 _______ 2 _______ 3 _______ 4 _______ 5 _______ 6 _______ Least



Please explain carefully your reasoning. W1= m1g = 6(9.8)=58.8 N; W2= m2g = 10(9.8)=98N; F1 = m1a = 6(4)=24N and F2 = m2a = 10(4)=40N Now using N2nd law for m2: T - m2g = m2a or T - 98 = 40 or T=138N ; And using N2nd Law for m1: F- m1g-T = m1a or F- 58.8 -138 = 24 or F= 58.8+24+138=220.8N; So F >T > W2 > W1 > F2> F1


NT5B-RT9: TWO CONNECTED OBJECTS ACCELERATING DOWNWARD—NET FORCE Two objects with masses of m1 = 6 kg and m2 = 10 kg are connected by a massless wire. The weights of the two masses are W1 and W2. They are pulled upward by an applied force F. The result is a constant acceleration of 5 m/s2 downward for the two objects. The tension in the wire is T and F1 is the net force on m1 while F2 is the net force on m2.

Rank the forces (W1, W2, F, T , F1 =Fnet on m1 and F2 =Fnet on m2) in these systems from greatest to least. Greatest 1 _______ 2 _______ 3 _______ 4 _______ 5 _______ 6 _______ Least



Please explain carefully your reasoning. W1= m1g = 6(9.8)=58.8 N, W2= m2g = 10(9.8)=98N; F= 76.8N, T = 48N ; F1 = m1a = 30N and F2 = m2a = 50N So W2 > F > W1 > F2 > T> F1


NT5B-CT1: BLOCKS MOVING AT CONSTANT SPEED—FORCE ON BLOCK In both cases shown a block is moving to the right across a table with a constant speed of 2 m/s. The tables and the blocks are identical. In case A the block is pushed with a stick at the angle shown. In case B the block is pulled with a string that makes the same angle with the horizontal.

Will the magnitude of the force on the block by the stick in case A be (a) greater than, (b) less than, or (c) equal to the tension on the block by the string in case B? Please carefully explain your reasoning.

(a) Greater in case A. The force from the stick on the block has a downward component in case A, and for the net force in the vertical direction to be zero the normal force must be greater than the weight. In case B, the tension in the string has an upward vertical component, and since the net vertical force is again zero, the normal force must be less than the weight. Since the weight of both blocks is the same, the normal force on the block in case A must be greater than the normal force on the block in case B. Since the normal force in case A is greater, the friction force on the block by the table is greater in case A than in case B. And since the net horizontal force is zero in both cases, the force on the block by the stick in case A must be greater than the tension on the block by the string in case B.

2 m/sF

Case A

!

2 m/s T

Case B

!


NT5F-CT2: BLOCKS MOVING AT CONSTANT SPEED—NORMAL FORCE ON BLOCK In both cases shown a block is moving to the right across a table with a constant speed of 2 m/s. The tables and the blocks are identical. In case A the block is pushed with a stick at the angle shown. In case B the block is pulled with a string that makes the same angle with the horizontal.

Will the normal force on the block by the table be (a) greater in case A, (b) greater in case B, or (c) the same in both cases? Please carefully explain your reasoning.

Greater in case A. The force from the stick on the block has a downward component in case A, and for the net force in the vertical direction to be zero the normal force must be greater than the weight. In case B, the tension in the string has an upward vertical component, and since the net vertical force is again zero, the normal force must be less than the weight. Since the weight of both blocks is the same, the normal force on the block in case A must be greater than the normal force on the block in case B.

2 m/sF

Case A

!

2 m/s T

Case B

!


NT5F-WWT1: BLOCKS MOVING AT CONSTANT SPEED—Normal FORCE ON BLOCK In both cases shown a block is moving to the right across a table with a constant speed of 2 m/s. The tables and the blocks are identical. In case A the block is pushed with a stick at the angle shown. In case B the block is pulled with a string that makes the same angle with the horizontal. A student makes the following comment about the normal force acting on the block by the table: “The normal force and the weight of the block form a Newton’s third law pair. These forces are equal in magnitude and opposite in direction. Since the blocks are identical, the weights are the same, and since the normal force must equal the weight in both cases, the normal force on the block by the table is the same in both cases.”

What, if anything, is wrong with this statement? If something is wrong, identify it and explain how to correct it.

The statement is incorrect. The normal force and the weight do not form a Newton’s third law pair – the force that forms a third law pair with the weight of the block is the gravitational force on the block by the earth. The force that forms a third law pair with the normal force on the block by the table is the normal force on the table by the block. The normal force on the block by the table is greater in case A. The force from the stick on the block has a downward component in case A, and for the net force in the vertical direction to be zero the normal force must be greater than the weight. In case B, the tension in the string has an upward vertical component, and since the net vertical force is again zero, the normal force must be less than the weight. Since the weight of both blocks is the same, the normal force on the block in case A must be greater than the normal force on the block in case B.

2 m/sF

Case A

!

2 m/s T

Case B

!


NT5G-RT7: HANGING MASS—STRING TENSION In each case below, strings are attached to one or more one kilogram blocks. All of the strings are so light that they can be considered to be massless. In cases A-E the strings pass over pulleys that are frictionless and also massless. All of the masses are at rest.

A B C

ED F

AB C

DE

F

Rank these situations from greatest to least on the magnitude of the tension in the strings at the labeled points.

Greatest 1 _______ 2 _______ 3 _______ 4 _______ 5 _______ 6 _______ Least

OR, The magnitude of the tension is the same for ALL SIX arrangements. ___

OR, The magnitude of the tension is zero for ALL SIX arrangements. ___

OR, We cannot determine the ranking for the magnitude of the tensions in these arrangements. ___

Please explain carefully your reasoning. Answer: The tensions at all of the labeled points are the same. For a massless string, the tension is the same at all points along the string. A free-body diagram of any hanging block will have a tension force upward and a 10-N weight downward, so the tension in all strings is 10 Newtons.


NT5I-WBT1: GRAVITATIONAL FORCE ON PROBE NEAR THE SUN AND EARTH—Location Diagram A student is trying to calculate a location for a scientific probe at a location where it is under the influence of both the earth and the sun, and arrives at the following equation:

!

GmPM

S

(D " x)2

=GmPM

E

x2

Make a diagram of the earth (E), the scientific probe (P), and the sun (S) that is consistent with this equation. Label any appropriate distances. Then explain the significance of the distance x (i.e., why this location is significant).

The equation is for a location where the force on the probe by the sun is exactly equal and opposite to the force on the probe by the earth. The distance D is the earth-sun distance, and the distance x is the distance of this location from the center of the earth.


NT5I-WWT2: TWO PLANETARY OBJECTS— GRAVITATIONAL FORCE ON EACH (1) Two planetary objects each with a mass of m exert a gravitational force on each other as shown below.

m m

Planet A Planet B

Fon A by B Fon B by A

What’s wrong, if anything? Explain.

Answer: The forces are attractive so the directions should be reversed.

(2) Two planetary objects with masses of m and 3m exert gravitational forces on each other as shown.

m 3m

Planet A Planet B

Fon A by B Fon B by A

What’s wrong, if anything? Explain.

Answer: The forces are attractive so the directions should be reversed and they are equal.


NT5I-QRT3: TWO OBJECTS—GRAVITATIONAL FORCE ON EACH Two objects each with a mass of m exert a gravitational force of magnitude F on each other. Now we replace one of the objects with another whose mass is 4m:

(1) If the original magnitude of the gravitational force on the object with a mass of m was F, what is the magnitude of the gravitational force on the object with a mass of m now? (a) 16F (b) 4F (c) F (d) F/4 (e) other

Answer (b) 4F (2) What is the magnitude of the gravitational force on the object with a mass of 4m now? (a) 16F (b) 4F (c) F (d) F/4 (e) other

Answer (b) 4F (3) If the original magnitude of the gravitational force on the object with a mass of m was F and we move the two objects with masses m and 4m to be 3 times as far apart as they were, now what is the magnitude of the gravitational force on the object with a mass of 4m?

m 4m

(a) F/9 (b) F/3 (c) 4F/9 (d) 4F/3 (e) other

Answer (c) 4F/9 (4) and what is the magnitude of the gravitational force on the object with a mass of m? (a) F/9 (b) F/3 (c) 4F/9 (d) 4F/3 (e) other

Answer (c) 4F/9


NT5I-QRT4: THREE OBJECTS EXERTING GRAVITATIONAL FORCES—NET FORCE ON ONE Three objects each with a mass of M exert a gravitational force on each other. Which of the arrows below is in the direction of the net force on mass B?

(a) (b) (c) (d) (e) none of these Answer (a) vector addition of the force on B by A(left) and the force on B by C(down)


NT5I-QRT5: TWO OBJECTS—GRAVITATIONAL FORCE ON EACH The picture below shows an object (labeled B) that has mass of +1 unit. Several centimeters to the left is another object (labeled A) that has a mass of 2 units. Choose the pair of force vectors (the arrows) that correctly compare the gravitational force on A (caused by B) with the gravitational force on B (caused by A).

2m m

A B

gravitational force on A by B

Fon A by B gravitational force on B by A

Fon B by A

(a)

(b)

(c)

(d)

(e)

(f) Answer (b) attractive and equal, Newton’s Third Law


NT5I-QRT6: OBJECT IN A UNIFORM GRAVITATIONAL FIELD—MOTION, POTENTIAL ENERGY, & FORCE

A mass is placed at rest at the center of a region of space in which there is a uniform, three-dimensional gravitational field. (A uniform field is one whose strength and direction are the same at all points within the region.) (1) When the mass is released from rest in the uniform gravitational field, what will its subsequent motion be? ______ Explain

Answer (c) since the force is constant producing constant acceleration (a) It will move at a constant speed. (b) It will move at a constant velocity. (c) It will move at a constant acceleration. (d) It will move with a linearly changing acceleration. (e) It will remain at rest in its initial position. (2) What happens to the gravitational potential energy of the system, after the mass is released from rest in the uniform gravitational field? ______ Explain

Answer (e) (a) It will remain constant because the gravitational field is uniform. (b) It will remain constant because the mass remains at rest. (c) It will increase because the mass will move in the direction of the gravitational field. (d) It will decrease because the mass will move in the opposite direction of the gravitational field. (e) It will decrease because the mass will move in the direction of the gravitational field.

A mass is placed at one of two different locations in a region where there is a uniform gravitational field, as shown below.

1 2

(3) How do the gravitational forces on the mass at positions 1 and 2 compare? ______ Explain

Answer (d) (a) Force on the mass is greater at 1. (b) Force on the mass is greater at 2. (c) Force at both positions is zero. (d) Force at both positions is the same but not zero. (e) Force at both positions has the same magnitude but is in opposite directions. (4) A mass is placed at a position on the x-axis where the gravitational potential energy is + 10 J. Which idea below best describes the future motion of the mass? (a) The mass will move down (-y) since the potential is positive (b) The mass will move up (+y) since the potential is positive (c) The mass will move left (-x) since the potential is positive. (d) The mass will move right (+x) since the potential is positive. (e) The motion cannot be predicted with the information given. ______ Explain

Answer (e) since you need more information since you do not know the direction of increasing or decreasing potential energy or force.


NT5I-QRT7: GRAVITATIONAL FIELD EQUIPOTENTIAL SURFACES—WORK, FIELD, AND FORCE In the figures below, the dotted lines show equipotential surfaces (lines in two-dimensional drawing) of three different gravitational fields. (A mass moving along a line of equal potential line has constant gravitational potential energy.) An object is moved directly from point A to point B. The mass of the object is 1 kg.

10J

40J20J

50J30J

A B

Region I

10J 40J20J 50J30J

A B

Region IIIRegion II

10J

40J20J

50J30J

A B

(1) How does the amount of work needed to move this mass compare for these three cases? (a) Most work required in Region I. (b) Most work required in Region II. (c) Most work required in Region III. (d) Region I and II require the same amount of work but less than Region III. (e) All three would require the same amount of work. ______ Explain

Answer (e) since the potential difference is the same for all three regions (2) How does the magnitude of the gravitational field at B compare for these three cases? (a) Region I > Region III > Region II (b) Region I > Region II > Region III (c) Region III > Region I > Region II (d) Region II > Region I > Region III (e) Region I = Region II = Region III ______ Explain

Answer (d) since the potential difference spacing smallest for II meaning the field is larger and the spacing between lines is largest for III

(3) For Region III, what is the direction of the gravitational force exerted by the field on a 1 kg object when at A and when at B? (a) left at A and left at B (b) right at A and right at B (c) left at A and right at B (d) right at A and left at B (e) no gravitational force at either point. ______ Explain

Answer (a) since the potential is decreasing in that direction


NT5I-QRT8: GRAVITATIONAL FIELD EQUIPOTENTIAL SURFACES—FORCE AT TWO POINTS A mass is first placed at rest in two different gravitational fields at two positions A and B as shown. In these regions, the gravitational field is described by gravitational equipotential surfaces (lines in two-dimensional drawing) shown as the dotted lines. All the points on an equipotential line have the same constant gravitational potential energy with values as labeled.

10J

40J20J

50J30J

A B

Region I

10J

40J20J

50J30J

A B

Region II

(I) Which set of arrows below best describes the relative magnitudes and directions of the gravitational force exerted on a 1 kg mass when at positions A and B for Region I?

gravitational force at A in Region I

gravitational force at B in Region I

(a) (b) (c) (d) (e) (f)

_________ Explain? Answer (e) since the potential difference is decreasing in that direction at a constant rate

(II) Which set of arrows below best describes the relative magnitudes and directions of the gravitational force exerted on a 1 kg mass when at positions A and B for Region II?

gravitational force at A in Region II

gravitational force at B in Region II

(a) (b) (c) (d) (e) (f)

_________ Explain? Answer (c) since the potential difference is decreasing in that direction at a larger rate at B


NT6B-CCT3: IDENTICAL TOBOGGANS ON HORIZONTAL—Speed The figures below represent two identical toboggans that have traveled down a snowy hill. The toboggans both have the same speed at the bottom of the hill. Assume that the horizontal surfaces that they travel along are frictionless except for the shaded areas, where the coefficient of friction is given. These shaded areas have different lengths as shown.

A B

P

6 m

µ = 0.2 P

3 m

µ = 0.5

Three students discussing these toboggans make the following contentions. Alicia: “The toboggan in case B will have the higher speed at P since it goes through a shorter rough

stretch.” Bo: “No, actually toboggan A will have the higher speed at P because the coefficient of friction is

less for the rough patch it travels over.” Colin: “Well I agree with Bo about A going faster but he is wrong about the reason. You need to take

the product of the distance and the coefficient of friction, and that is less for A than B.” Which, if any, of these three students do you think is correct?

Alicia _____ Bo _____ Colin _____ None of them______


Answer: Colin is correct .


NT6B-CT1: TOBOGGANS ON HORIZONTAL—Speed The figures below represent two toboggans that have traveled down a snowy hill. The toboggans both have the same speed at the bottom of the hill, but toboggan A has three times the mass of toboggan B. Assume that the horizontal surfaces that they travel along are frictionless except for the shaded areas, where the coefficient of friction is given. These shaded areas have different lengths as shown.

A B

P

6 m

µ = 0.2 P

3 m

µ = 0.5

3 M M

Will the speed of toboggan A be greater than, less than or the same as the speed of toboggan B at point P? __________ Explain.

Answer: Toboggan A will be going faster at point P. The masses of the toboggans are irrelevant here. There will be more work done in case A than case B, but the toboggan in case A has more KINETIC ENERGY at the bottom, since they have the same speed, but A has more mass. But if you take the initial kinetic energy, subtract the work by friction and equate that to the final kinetic energy the mass drops out of the equation.


NT6D-CT1: Velocity and Acceleration Graphs for Two Falling Rocks—Initial Kinetic Energy

One rock is dropped from the top of a cliff at the same instant a second rock is thrown from the cliff. The following graphs describe part of the motion of the rocks. Use a coordinate system in which up is the positive vertical direction and the positive horizontal direction is away from the cliff with the origin at the point the balls were released.

v (horizontal)

time

time

v (horizontal)

time

v (vertical)

time

Rock A Rock B

v (vertical)

Is the kinetic energy of Rock A at the start of the motion (t=0) (a) greater than, (b) less than, or (c) the same as the kinetic energy of Rock B? Explain.

(a) Rock A is greater since Rock B has zero initial kinetic energy.


NT6D-CRT1: BOX LIFTED UP MOVING UP I—KINETIC ENERGY & VELOCITY GRAPHS A 100 N box is initially moving upward at 4 m/s. A woman is applying a vertical force of 80 N with her hand to the box as shown.

100 N or 10 kg

80 N

Sketch a graph of the velocity and kinetic energy of the box as a function of time.

velocity

time

kinetic energy

time

Explain.

Answer

The kinetic energy is decreasing since the hand is doing less work on the box than the gravitational force and thus the velocity is decreasing at a constant rate.


NT6D-CRT2: BOX LIFTED UP MOVING UP II—KINETIC ENERGY & VELOCITY GRAPHS A 100 N box is initially moving upward at 4 m/s. A woman is applying a constant upward vertical force of 120 N to the box with her hand as shown.

100 N or 10 kg

120 N

Sketch a graph of the velocity and kinetic energy of the box as a function of time.

velocity

time

kinetic energy

time

Explain the graphs.

Answer

The kinetic energy is increasing since the hand is doing more work on the box than the gravitational force and thus the velocity is increasing at a constant rate since the net force is constant.


NT6H-WWT2: MASS ON A COMPRESSED TILTED SPRING SYSTEM—HEIGHT UP INCLINE The figures below show blocks moving on identical, tilted, frictionless surfaces. The blocks are just about to lose contact with the end of a spring that had been compressed. In each system, the springs were compressed by the distance indicated in the figure. The masses of the blocks and force constants of the springs are also given for each system. A student makes the following statement about how far the blocks will slide up the inclines.

.

1 kg

1 N/m

0.5 m

5 kg

1 N/m

0.5 m

Case A Case B

Steve: “I think they will both travel up the same distance on the inclines. The kinetic energy at the point shown in the diagram is equal to the initial elastic energy stored in the compressed spring. This is the same for both cases since they both are compressed the same distance and have the same spring constants. The kinetic energy at the point shown is also equal to the gravitational potential energy at the top or mv2/2 = mgh. Thus, the mass cancels out leaving the same heights for each case.”

What, if anything, is wrong with Steve’s statement? If something is wrong, identify it and explain how to correct it.

Answer: The Steve’s answer is incorrect since he did not take into account that the gravitational potential energy will be different because the masses are different.


NT7D-CCT5: COLLIDING CARTS THAT STICK TOGETHER—Final Kinetic Energy In each of the figures below, two carts traveling in opposite directions are about to collide. The carts are identical in size and shape, but they carry different loads and are initially traveling at different speeds. The carts stick together after the collision, and the system, i.e., the two carts and the surface, is frictionless.

20 cm/s

8 kg 4 kg

40 cm/s

Three physics students discussing this situation make the following contentions.

Alex: “These carts will both be at rest after the collision since the initial momentum of the

system is zero, and the final momentum has to be zero also.” Belinda: “No, that would mean that they would have zero kinetic energy after the collision and that

would violate conservation of energy. Since the right hand cart has more kinetic energy, the combined carts will be moving slowly to the left.”

Chano: “No, you are both wrong the pair of carts will be traveling left at 20 m/s. That way conservation of momentum and conservation of energy are both satisfied”


Alex _____ Belinda _____ Chano _____ None of them______


Answer: Alex is correct .


NT7D-RT3: BOBS HITTING BLOCKS—FINAL SPEED The figures below show six situations where a bob on the end of a string is going to swing down and hit a block sitting on a frictionless surface. All of the bobs start from the same height, but three (A, B and D) are made of rubber and will bounce off the block with the same speed as they hit it, while the other three (C, E and F) are made of clay and will stick to the block. The masses of the bobs vary and the values are given in the figures. All of the blocks have a mass of 1000 g.

A

200 g

B

400 g

C

100 g

E

300 g

D

100 g

F

200 g

Rank these situations from greatest to least on the impulse the bobs impart to the blocks. Greatest 1 _______ 2 _______ 3 _______ 4 _______ 5 _______ 6 _______ Least

OR, The impulse is the same for ALL SIX arrangements. ___

OR, The impulse is zero for ALL SIX arrangements. ___

OR, We cannot determine the ranking for the impulses in these arrangements. ___

Please explain carefully your reasoning. Answer: B > A > E > D = F > C; the impulse is the change in momentum of the bobs.


NT8C-QRT7: ROTATING CYLINDER— DIRECTION OF TORQUE AND ANGULAR MOMENTUM A cylinder that started at rest is rotating as shown. (1) Which arrow best represents the direction of the angular momentum of the cylinder? Explain.

(a)

(b)

(c)

(d)

(e) None of these

Answer-(c) (2) Which arrow best represents the direction of the torque on the cylinder that caused this rotation? Explain.

(a)

(b)

(c)

(d)

(e) None of these

Answer-(c) (3) Which arrow best represents the direction of the torque on the cylinder that will cause this rotation to stop? Explain.

(a)

(b)

(c)

(d)

(e) None of these

Answer-(a)


NT9B-RT1: SIX PULLEY SYSTEMS—FORCE TO KEEP MOVING UP Shown are six frictionless pulley systems that are connected to various masses. The pulleys are massless. There is a vertical force for each system that is applied at the end of a support rope that keeps the masses moving upward at a constant velocity of 4 or 6 m/s.

A

FA

B

5 N

FB

C

FC

4 m/s

3 N5 N

6 m/s 6 m/s

D E F

FFFEFD

7 N 7 N3 N

4 m/s 4 m/s6 m/s

Rank, from greatest to least, the forces that keep these systems moving upward at 4 or 6 m/s. Greatest 1 _______ 2 _______ 3 _______ 4 _______ 5 _______ 6 _______ Least

OR, The forces on all these systems will be the same. ___

OR, We cannot determine the ranking for forces on all these systems. ___

Please explain carefully your reasoning. E = F > A = B >C= D. This ranking is the same for holding the systems at rest, namely equal to the weight of the hanging mass since there is no acceleration.


NT9B-RT2: SIX PULLEY SYSTEMS—FORCE TO KEEP MOVING DOWNWARD Shown are six frictionless pulley systems that are connected to various masses. The pulleys are massless. There is a vertical force for each system that is applied at the end of a support rope that keeps the systems moving downward at a constant velocity of 4 or 6 m/s.

A

FA

B

5 N

FB

C

FC

4 m/s

3 N5 N

6 m/s 6 m/s

D E F

FFFEFD

7 N 7 N3 N

4 m/s 4 m/s6 m/s

Rank, from greatest to least, the forces that keep these systems moving downward at 4 or 6 m/s. Greatest 1 _______ 2 _______ 3 _______ 4 _______ 5 _______ 6 _______ Least

OR, The forces on all these systems will be the same. ___

OR, We cannot determine the ranking for forces on all these systems. ___

Please explain carefully your reasoning. E = F > A = B >C= D. This ranking is the same for holding the systems at rest, namely equal to the weight of the hanging mass since there is no acceleration.

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Aug18 New tasks rev9-7-08 - Wikispacesdiaznova1.wikispaces.com/file/view/tipers2.pdf · Aug18 New tasks rev9-7-08 nT2C-CCT1: Two Vectors—Vector Subtraction ... Projectile Motion