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AP PHYSICS B Linear Momentum & Impulse

Teacher Packet

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Linear Momentum & Impulse

Objective To review the student on the concepts, processes and problem solving strategies necessary to successfully answer questions on linear momentum and impulse.

Standards Linear momentum and impulse are addressed in the topic outline of the College Board AP* Physics Course Description Guide as described below.

I. Newtonian Mechanics D. Systems of particles, linear momentum 1. Center of mass 2. Impulse and momentum 3. Conservation of linear momentum, collisions

AP Physics Exam Connections Topics relating to linear momentum and impulse are tested every year on the multiple choice and in most years on the free response portion of the exam in conjunction with work and energy. The list below identifies free response questions that have been previously asked over linear momentum and impulse. These questions are available from the College Board and can be downloaded free of charge from AP Central. http://apcentral.collegeboard.com.

2008 2005 2002 2001 1999

Free Response Questions Question 1 2008 Form B Question 2 2006 Form B Question 1 2005 Form B Question 2 2002 Form B Question 1 (ex a)

Question 1 Question 2 Question 2 Question 1

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Linear Momentum & Impulse

What I Absolutely Have to Know to Survive the AP* Exam

Momentum (p) is a vector quantity which has the same direction as the velocity. Momentum can be thought of as inertia in motion. It is the product of mass times velocity. To have momentum an object must be moving. Impulse (J) equals the change in momentum of an object and is a vector that has the same direction as the net Force. It is derived from Newton’s second law. For a collision, the area under a force versus time graph yields the impulse applied during the collision. There are two types of collisions, elastic and inelastic. In both types of collisions momentum is conserved. In elastic collisions, kinetic energy is conserved as well.

Key Formulas and Relationships

p = mv Momentum Units: kg ⋅ m / s Impulse J = FΔt Units: N ⋅ s Note: kg ⋅ m / s and N ⋅ s are equivalent units Impulse – momentum theorem J = Δp FΔt = mv f − mv i Conservation of linear momentum pa + pb = p'a + p'b

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Linear Momentum & Impulse

Important Concepts

• •

Momentum (p) is a vector quantity which has the same direction as the velocity. Momentum can be thought of as inertia in motion. To have momentum an object must be moving. Impulse (J) equals the change in momentum of an object and is a vector that has the same direction as the net Force. It is derived from Newton’s second law. F = ma

F = m( v f − v i )

Δt FΔt = m( v f − v i ) FΔt = mv f − mv i FΔt = Δp J = FΔt is used on the AP exam for impulse. The impulse is the applied force multiplied by the time over which it acts. Impulse is equal to the change in the momentum of a system. Hence, if we apply a large force for a short time we can generate a momentum change of the same magnitude as having a small force for a long time.

F •

=

t

F

t

The area under a Force vs. Time curve = the change in momentum of the object

Force (N)

Time (s) •

The impulse—momentum theorem says that Impulse (J) is the product of the average force acting on an object and the time interval during which the force acts. This impulse produces a change in an object’s momentum J = Ft = Δp and can be used to derive the law of conservation of linear momentum.

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Linear Momentum & Impulse

Derivation: Imagine two objects traveling in opposite directions that collide. According to Newton’s F1 = − F2 third law they exert equal and opposite forces upon each other Since the forces act upon each other for the same time, F1t = − F2t Since Ft = Δp we can write Δp1 = −Δp 2 p1' − p1 = −( p '2 − p 2 )

therefore m1 v1' − m1 v1 = − m2 v '2 + m2 v 2

this gives us the law of conservation of momentum m1 v1 + m2 v 2 = m1 v1' + m2 v '2

•

•

When no external forces act on a system, the total momentum remains the same. o Example 1: Two objects collide. The momentum of object one before the collision plus the momentum of object two before the collision equals the momentum of objects one and two after the collision. m1v1 + m2v2 = m1 v1' + m2 v2' o Example 2: A gun is fired and recoils. The sum of the momentum of the gun and bullet before it is fired (which is 0) equals the sum of the momentum of the gun and bullet after it is fired (also 0). The gun kicks and has a negative velocity and therefore a negative momentum. The bullet has a positive velocity and momentum, and when added together, the resulting momentum is zero. 0 + 0 = − p a + pb = 0 When objects collide, there are two types of collisions that can occur-- elastic and inelastic. o Elastic: In this type both kinetic energy and momentum are conserved.

If an equation is classified as elastic, both the conservation of linear momentum and conservation of KE equations can be used pa + pb = p'a + p'b

KEa + KEb = KEa' + KEb' When two objects collide and bounce off each other, there are two possible energy interactions. Kinetic energy can be conserved or not conserved. If Kinetic energy is conserved, then we have an elastic collision. Momentum will always be conserved. o Inelastic: In these collisions only momentum is conserved

When two objects collide and bounce off each other, some of the kinetic energy is

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Linear Momentum & Impulse

transformed into other forms of energy. Real world collisions are inelastic. On the AP exam, you may be asked whether kinetic energy is conserved and, if it isn’t, then what happened to that energy or how much kinetic energy is transferred elsewhere. When two objects collide and they stick together after they collide, then this is called a perfectly inelastic collision. The key thing to remember here is that after the collision both objects stick together and have the same velocity. Only momentum is conserved in this type of collision. A third example of an inelastic collision on the AP Exam is the two body explosion. The idea of an explosion is that you have two bodies that are at rest – this means that they have no momentum. After the explosion takes place, the bodies end up moving away from each other. Because momentum is conserved, their final momentum, when added together, must still equal zero (the momentum before the explosion). Remember momentum is a vector.

•

Momentum is conserved in two (and three) dimensions as well. Whenever we analyze a collision in two dimensions the momentum has to be split up into components. The sum of the momenta of all components in the x and y directions before a collision must equal the sum of the momenta after the collision. ∑ px = ∑ px' and ∑ p y = ∑ p 'y

o The diagram below shows the momentum vector of ball A colliding with ball B which is at rest. Ball A has momentum in the x direction but none in the y while ball B has no momentum at all. Ball A

Ball B

After the collision: Ball A

Ball B

The components of the x and y momentum vectors after the collision must equal the components before the collision.

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Linear Momentum & Impulse

=

In the x direction

=

+

In the y direction since the momentum before the collision was zero, the y components of the momentum after the collision must be equal to zero. Since they act in opposite directions, their magnitudes must be equal to each other. =

On the AP Exam, collisions in two dimensions are frequent. Momentum still has to be conserved, but the problems can become fairly complex. Note: keep track of what’s going on by breaking things into x and y components: momentum is a vector, not a scalar.

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Linear Momentum & Impulse

Free Response

Question 1 (15 pts) A 15.0 kg wagon with frictionless wheels travels to the right at 5.0 m/s on a horizontal frictionless surface as shown in the diagram above. When it reaches point A, a 5.0 kg mass is placed upon it. v = 5.0 m/s

5.0 kg mass

A A. Determine the magnitude of the momentum of the wagon before the mass is placed upon it. 1 point for the correct expression for momentum

(2 points max) p=mv p= (15kg)(5m/s)= 75

1 point for the correct answer including correct units and reasonable number of significant digits

kgm s

B. Determine the speed of the wagon after the mass is placed upon it. 1 point for any statement that momentum is conserved

(3 points max) Pwagon+ pmass=pwagon+mass

1 point for adding the masses 75 + 0 = (15 kg + 5kg)v 1 point for the correct answer including correct units and reasonable number of significant digits

m v = 3.75 s

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Linear Momentum & Impulse

The wagon and mass continue moving until reaching another frictionless wagon of mass 15.0 kg at point B. This wagon is attached to a spring of negligible mass which has a spring constant of 250 N/m. The other end of the spring is fixed to the wall. Upon impact, the two wagons stick together and compress the spring.

B

C. Determine the speed of the two wagons immediately after they stick together. (3 points max)

1 point for a correct application of conservation of momentum

pwagon1+pwagon2=pwagon1+wagon2 20kg(3.75m/s) + 0 = (20kg+15kg)vwagon1&2

2.14

1 point for using the speed found in part b 1 point for the correct answer including correct units and reasonable number of significant digits

m = vwagon1&2 s

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Linear Momentum & Impulse

D. Determine the kinetic energy of the two wagons immediately after they stick together.

(3 points max)

1 point for a correct expression for Kinetic energy

K = ½ mv2 K= ½(35)(2.14)

1 point for correct substitution using answer in part c

2

1 point for the correct answer including correct units and reasonable number of significant digits

K=80 J

E. Determine the distance the wagons travel until the spring is fully compressed.

1 point for a correct expression for conversation of energy

(4 points max) K=U

1 point for the correct expression for the potential energy of a spring

U = ½ kx2 80 J = ½ (250 N/m)(x2)

1 point for correct substitution using answer in part d

x = 0.80 m

1 point for the correct answer including correct units and reasonable number of significant digits

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Linear Momentum & Impulse

Question 2 (15 pts) An arrow with a mass of 0.20 kg attached to a bowstring experiences a variable force for 0.85 s as it is shot from a bow as shown in the diagram below.

50

Force (N)

0

Time (s)

.85

A. Determine the Impulse applied to the arrow by the bowstring during the 0.85 s that they are in contact. (3 points max) J = F Δt = mΔ v 1 J = F Δt = bh 2 1 J = ( 0.85s )( 50N ) 2 J = 21.25 N is

1 point for any statement indicating that impulse equals change in momentum 1 point for using the area under the curve of the F vs. t graph to determine the impulse or change in momentum of the arrow

1 point for the correct answer including correct units and reasonable number of significant digits

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Linear Momentum & Impulse

B. Determine the speed of the arrow at the instant it leaves the bow. (2 points max) J = F Δ t = mΔ v

1 point for the correct application of the impulse momentum theorem

kgm s 0.2 kg m Δv = 106.25 s

1 point for the correct answer or an answer consistent with part (A), including correct units and reasonable number of significant digits

J Δv = = m

21.25

The arrow then travels in a parabolic path until it hits an apple atop a man’s head. When the arrow hits the apple, it is at the same height as it was when it left the bow. The arrow and apple stick together and fly off his head landing a certain distance behind the man. The mass of the apple is 0.4 kg. C. Determine the speed of the apple/arrow as they leave the man’s head.

1 point for any statement that momentum is conserved

(4 points max) P1 + P2 = P1'+ 2 m1v1 + 0 = ( m1 + m2 ) v ' m⎞ ⎛ 0.2 kg ) ⎜ 106.25 ⎟ ( m1v1 s⎠ ⎝ = v' = m1 + m2 ( 0.2 kg + 0.4 kg ) v ' = 35.4

m s

1 point for a correct expression of momentum p=mv 1 point for adding the masses of the arrow and the apple 1 point for the correct answer or an answer consistent with part (B), including correct units and reasonable number of significant digits

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Linear Momentum & Impulse

D. Calculate the kinetic energy of the two-object system before and after the collision.

1 point for a correct expression of kinetic energy

(4 points max) 1 KE = mv 2 2 Before 2

1 m⎞ ⎛ KE= ( 0.2 kg ) ⎜ 106.25 ⎟ = 1130 J 2 s⎠ ⎝ After 2

1 m⎞ ⎛ KE= ( 0.2 kg + 0.4 kg ) ⎜ 35.4 ⎟ = 376 J 2 s⎠ ⎝

1 point for a correct quantity or answer consistent with part B 1 point for a correct quantity or answer consistent with part C 1 point for a correct identification of before and after

E. Is the collision elastic? Justify your answer. (2 points max) No

1 point for indicating the collision is NOT elastic

KEBefore ≠ KE After

1 point for a statement that kinetic energy is not conserved in the collision

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Linear Momentum & Impulse

Multiple Choice

1. Two objects of mass 3m and 4m are initially at rest on a frictionless horizontal surface. An ideal spring is compressed between the objects which are held together by a thread of negligible mass. After the thread is cut and the spring is released, the object with a mass of 3m has a speed v. What is the speed of the object with the mass of 4m? A)

1/7 v

B)

3/4 v

C)

v

D)

4/3 v

E)

7v Conservation of p (explosion)

p1 + p 2 = p1' + p'2

B

0 + 0 = 3mv + 4mV −3mv = 4mV −3/ 4 = V

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Linear Momentum & Impulse

question 2 4 m/s 10 kg

2 m/s 5 kg

2. A 10 kg block moving to the right with a speed of 4 m/s on a frictionless surface collides with a 5 kg block also moving to the right with a speed of 2 m/s. If the blocks stick together upon impact, what is their new speed? A) 3

3 m/s 4

B) 3

1 m/s 2

C) 3

1 m/s 3

D) 3 m/s E) 2

1 m/s 2

Conservation of p (objects in same direction stick together)

p1 + p 2 = p1' + p '2

C

40 + 10 = 15v 50 /15 = v 3 13 = v

3. In a demonstration a physics professor throws an egg against a blanket and it does not break. When the same egg is thrown against a brick wall, it breaks. Which of the following best describes the egg’s momentum change? A) B) C) D)

The brick wall provides a greater momentum change than the blanket. The blanket provides a greater momentum change than the brick wall. The momentum changes in both cases are the same. The momentum change between the egg and the brick wall occurs over more time with more force. E) The momentum change between the egg and the blanket occurs over less time with less force.

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Linear Momentum & Impulse

Momentum change

The momentum change depends only on mass as well as initial and final velocities.

C

4. If the momentum of an object changes and its mass remains the same, what else must be happening to the object? I. it is accelerating II. its velocity is changing III. there is a net force acting on it A) B) C) D) E)

I only II only III only I and II only I , II, and III

What is momentum

If p is changing and m stays the same, then v must be changing. Therefore it must be accelerating, and to accelerate and net force must be acting on it

E

5. A 2 kg ball traveling eastward at 3.0 m/s is kicked in such a way that it is now traveling westward with a speed of 8.0 m/s. The impulse applied to the ball is A) B) C) D) E)

- 6 Ns - 10 Ns - 18 Ns - 22 Ns - 48 Ns Impulse = Δp

Δp = p 2 − p1

D

Δp = (2)( −8) − (2)(3) Δp = −16 − 6 Δp = −22

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Linear Momentum & Impulse

Question 6

M

2v

v

m

6. An object with mass M and speed v collides with a second object of mass m traveling to the left with a speed 2v. They are both traveling on a frictionless surface. If they collide and stick together upon impact, they will move off together with a new speed of (M ≠ m) A) B)

1 2

v M +m ( M + 2m ) v

C)

M +m ( M − 2m )v

D)

( M + 2m ) v M +m

E)

( M − 2m )v M +m

Mv + m( −2v ) = ( M + m )V ( M − 2m )v = ( M + m )V

Conservation of p (objects in opposite directions stick together)

( M −2 m ) v ( M +m )

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E

=V

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Linear Momentum & Impulse

Questions 7-9 A net force that varies as a function of time is applied to a mass of 7.0 kg according to the graph as shown below:

Force (N) 4

0

10

25

Time (s) 7. Calculate the impulse applied to the mass during the first 25 seconds A) 70 Ns B) 40 Ns C) 30 Ns D) - 2/25 Ns E) - 4/25 Ns Impulse = area under F vs. t curve

Area = 4(10)+1/2(15)(4) Area = 40 + 30 Area = 70

A

8. If the object had an initial velocity of 3.0 m/s at time t = 0, it’s velocity after 25 s would be A) B) C) D) E)

7 m/s 10 m/s 13 m/s 15 m/s 25 m/s

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Linear Momentum & Impulse

Impulse = Δp Using data from graph

Δp = m( v 2 − v1 )

C

70 = 7( v 2 − 3) 10 = ( v 2 − 3) 13 = v 2

9. What is the acceleration of the object during the first 10 seconds? 0 m/s2 4/7 m/s2 7/4 m/s2 28 m/s2 40 m/s2

A) B) C) D) E)

a=F/m using data from graph

a=F/m a=4/7

B

10. In an elastic collision I. II. III. A) B) C) D) E)

Linear momentum is conserved Kinetic Energy is conserved Elastic Potential Energy is conserved

I only II only III only I & II only II & III only

Definition of an elastic collision

In an elastic collision, both momentum and Kinetic Energy are conserved

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D

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Linear Momentum & Impulse

11. Two objects, one of which is at rest, collide. The arrows below represent the momenta of the two objects after the collision.

Which of the following may represent the initial momentum of the moving object before the collision? A)

B)

C)

D)

E)

zero

Two dimensional momentum Adding the two vectors tip-to-tail gives a resultant which points in the direction of the vector arrow D.

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D

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Linear Momentum & Impulse

12. A circus performer is shot out of a cannon and flies through the air until landing harmlessly in a net. Which of the following best describes why the performer is not injured after this “death defying” event? A) The net decreases the amount of time the force is acting on the person and increases the change in momentum. B) The net increases the amount of time the force is acting on the person and decreases the change in momentum. C) The net decreases the amount of time the momentum is changing and increases the average force acting on the person. D) The net increases the amount of time the momentum is changing and decreases the average force acting on the person. E) The net increases the amount of time the momentum is changing and decreases the impulse acting on the person. The nature of impulse

Ft = Δp by increasing the time of the momentum change, the force applied to the circus performer is reduced

D

13. A 0.1 kg ball traveling horizontally westward at 20 m/s strikes a vertical wall and bounces off the wall eastward with a speed of 20 m/s. What is the impulse acting on the ball? A) B) C) D) E)

zero 2 Ns westward 2 Ns eastward 4 Ns westward 4 Ns eastward Impulse = Δp

Assume east is positive I = Δp Δp = p 2 − p1

E

Δp = (.1)(20) − (.1)( −20) Δp = 4 The positive sign indicates eastward

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Linear Momentum & Impulse

14. Object A has a mass m and a velocity v. Object B has the same mass but a velocity V. If the magnitude of object A’s momentum equals the magnitude of object B’s kinetic energy, what is V in terms of v? A) v B) 2v C) 2 v D) 1 2 v 2 E) v 2

Relating momentum to KE

mv = 1/ 2mV 2

D

2v = V 2 V2 v= 2 Their magnitudes are equal (obviously p does not equal KE)

15. Which of the following have equivalent SI units? (I) (II) (III) A) B) C) D) E)

Force Momentum Impulse

I only II only III only I and II only II and III only

Units of Momentum and Impulse

Momentum’s unit Kg m/s is equivalent to Impulse’s unit Ns

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E

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