momentum consider systems consisting of more than a single particle and how these multiple particles...

Momentum• Consider systems consisting of more than a single

particle and how these multiple particles interact with each other– “Many–body” problems more complicated, so we need

effective ways of describing “systems” of particles

• Start with Newton’s 2nd Law:• Quantity is called the momentum of a particle:• For a system of particles:

t

vm

t

vmamF

net

vm

vmp

m1

m2

m3m1v1 m2v2

m3v3

Total momentum of system: 321321tot vmvmvmpppp

Impulse• Thus Newton’s 2nd Law can be written as:

– This is (essentially) the original way Isaac Newton wrote it!– Assumes constant force producing constant acceleration– The shorter the time period over which a given momentum

change occurs, the larger the external force

• Product of the net force and the time interval over which it acts is called the impulse of the net force:– Assumes constant force– If net force is not constant,

impulse is determined by 1 of 2 ways:

t

pF

net

J

12netnet ttFtFJ

Area under F vs. t curve

Fave t

Impulse – Momentum Theorem– So and thus

– Same result holds when net force is not constant

• Now consider a system consisting of the Sun and the Earth:

– Sun and Earth exert equal magnitude (but opposite direction) gravitational forces on each other

– No other external forces are acting on them

• Apply Newton’s 2nd Law to each mass:

12

12net tt

ppF

12 ppJ

(Impulse – Momentum Theorem)

Sun (m1)Earth (m2)F

t

pF

1

t

pF

2

F

Conservation of Momentum• Thus

• Where = total momentum of the system• This means that the total momentum of the system

is conserved (or constant) if the net external force acting on the system is zero

• Note that this also follows from Newton’s 3rd Law• Objects exchange momentum through mutual

interactions (gravity, collisions, etc.)• Momentum conservation can hold even when

mechanical energy isn’t conserved

021

FFt

p

t

p 021

t

pp

0tot

t

p

21tot ppp

CQ 1: Two 1-kg carts with spring bumpers undergo a collision on a frictionless track as shown in the before and after pictures below. The total momentum of the

system is equal to:

A) 0 kg-m/s before the collision and 0 kg-m/s after the collision.B) –4 kg-m/s before the collision and 4 kg-m/s after the collision.C) –8 kg-m/s before the collision and 8 kg-m/s after the collision.D) 8 kg-m/s before the collision and 0 kg-m/s after the collision.

CQ2: Interactive Conceptual Example: Basketball vs. Flower Pot Drop

How does the change in momentum of the basketball compare with the change in momentum of the flowerpot after each object hits the table?

PHYSLET Exercise 8.5.1, Prentice Hall (2001)

A) The change in momentum of the basketball is the same as the change in momentum of the flowerpot.

B) The change in momentum of the basketball is twice the change in momentum of the flowerpot.

C) The change in momentum of the basketball is half of the change in momentum of the flowerpot.

D) The change in momentum of the flowerpot is twice the change in momentum of the basketball.

Example Problem #6.24

Solution (details given in class):

62 s

A 730-N man stands in the middle of a frozen pond of radius 5.0 m. He is unable to get to the other side because of a lack of friction between his shoes and the ice. To overcome this difficulty, he throws his 1.2-kg physics textbook horizontally toward the north shore at a speed of 5.0 m/s. How long does it take him to reach the south shore?

CQ 3: Interactive Example Problem:Saving an Astronaut

(ActivPhysics Online Problem #6.6, Pearson/Addison Wesley)

How much time elapses from when the astronaut removes her oxygen tank until she arrives back at the ship?

A) 5 sB) 10 sC) 15 sD) 20 sE) 60 s

CQ 4: A boy is sliding down a long icy hill on his sled. In order to decrease his mass and

increase his velocity, he drops his heavy winter coat and heavy boots from the sled while he is

moving. Will his strategy work?

A) No, because he loses the potential energy of the objects that he leaves behind.

B) No, because although his kinetic energy increases, his momentum decreases.

C) Yes, because although his kinetic energy decreases, his momentum increases.

D) Yes, because although his momentum decreases, his kinetic energy decreases.

Collisions• Inelastic collisions

– Example: cars stick together after a collisionBEFORE collision:

Me (m2) v2i = 0You (m1)

v1i

AFTER collision (cars locked together):vf

Energy used to do work (destroy fenders)What is vf?

Conservation of momentum: afterbefore pp

ii vmvmp 1111before 0

fvmmp

21after

Collisions

• Note that if m1 = m2 then vf = ½ v1i

• In inelastic collisions, the kinetic energy of system after collision < kinetic energy of system before collision (since mechanical energy is typically expended to do work)– K.E. before collision = K1 = ½ m1v1i

2

– K.E. after collision = K2 = ½ (m1 + m2)vf2 =

½ (m1 + m2)[m1 / (m1 + m2)]v1i2

– So K2 / K1 = m1 / (m1 + m2) < 1 so K2 < K1

• Elastic collisions– Forces between colliding bodies are conservative– Kinetic energy is conserved (may be temporarily

converted to elastic potential energy)– Momentum is conserved

fi vmmvm

2111 if vmm

mv 1

21

1

Elastic Collisions• Head-on collision where one object is at rest before

the collision:

• Conservation of kinetic energy gives:

½ m1v1i2 = ½ m1v1f

2 + ½ m2v2f2

• Conservation of momentum gives:

m1v1i = m1v1f + m2v2f

• Combining these 2 equations:

m2, v2i = 0

m1, v1im1, v1f m2, v2f

if vmm

mmv 1

21

211

if v

mm

mv 1

21

12

2

Elastic Collisions• Interesting cases: (1) m1 = m2

– v1f = 0 and v2f = v1i

• (2) m1 << m2

– v1f – v1i

– v2f << v1i

– Similar to result of ping-pong ball (m1) striking stationary bowling ball (m2)

• (3) m1 >> m2

– v1f v1i

– v2f 2v1i

– Similar to result of bowling ball (m1) striking stationary ping-pong ball (m2)

• If v2i ≠ 0, then in general: v1i – v2i = –(v1f – v2f)

CQ 5: A block of mass m1 slides across a frictionless surface with speed v1 and collides with a stationary

block of mass m2. The blocks stick together after the collision and move away with speed v2. Which of the following statements is (are) true about the blocks?

A) I onlyB) II onlyC) I and II onlyD) I, II and III

21

2221

211

22111

.III2

1

2

1.II

.I

vv

vmmvm

vmmvm



(a) 4.85 m/s

(b) 8.41 m

A tennis ball of mass 57.0 g is held just above a basketball of mass 590 g. With their centers vertically aligned, both balls are released from rest at the same time, to fall through a distance of 1.20 m, as shown at right.

(a) Find the magnitude of the downward velocity with which the basketball reaches the ground.

(b) Assume that an elastic collision with the ground instantaneously reverses the velocity of the basketball while the tennis ball is still moving down. Next, the two balls meet in an elastic collision. To what height does the tennis ball rebound?

2 – D Collisions• Can also have a “glancing” collision:

• Now the final velocities (and hence the momenta) have horizontal and a vertical components

• Use the component form of conservation of momentum for the system of both balls:

px,initial = px,final

py,initial = py,final

• True for elastic or inelastic collisions

x

m2, v2i = 0

m1, v1i

m1, v1f

m2, v2f

2-D Collisions Interactive


BEFORE collision:

v1 = 10 m/s

m1 = 2000 kg

m2 = 3000 kg

v2

x (E)

y (N)AFTER collision:

40°

vf = 5.22 m/svfy

vfxm1 + m2

vfx = vf cos40° = 4.00 m/s

vfy = vf sin40° = 3.36 m/s


5.59 m/s

y (N)

x (E)

A 2000-kg car moving east at 10.0 m/s collides with a 3000-kg car moving north. The cars stick together and move as a unit after the collision, at an angle of 40.0° north of east and a speed of 5.22 m/s. Find the speed of the 3000-kg car before the collision.

CQ 6: Ball A moving at 12 m/s collides elastically with ball B as shown. If both balls have the same mass,

what is the final velocity of ball A? (Note: sin60° = 0.87; cos60° = 0.5)

A) 3 m/sB) 6 m/sC) 9 m/sD) 12 m/s



273 m/s

A 12.0-g bullet is fired horizontally into a 100-g wooden block that is initially at rest on a frictionless horizontal surface and connected to a spring having spring constant 150 N/m. The bullet becomes embedded in the block. If the bullet–block system compresses the spring by a maximum of 80.0 cm, what was the speed of the bullet at impact with the block?



A bullet of mass m and speed v passes completely through a pendulum bob of mass M as shown above. The bullet emerges with a speed of v/2. The pendulum bob is suspended by a stiff rod of length l and negligible mass. What is the minimum value of v such that the bob will barely swing through a complete vertical circle?

glm

Mv

4

Rocket Propulsion

• Rocket propulsion can be understood in two ways:

1. Conservation of momentum: Spent fuel (gasses) have momentum directed away from rocket; rocket has momentum in opposite direction

2. Newton’s Third Law: Rocket exerts force on spent fuel gasses, expelling them outward; gasses in response exert force on rocket in opposite direction

• Rocket needs nothing to “push against” in order to move, otherwise it would not work in the vacuum of space!

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