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m 1. m 3. m 1 v 1. m 3 v 3. m 2. m 2 v 2. Total momentum of system:. Momentum. Consider systems consisting of more than a single particle and how these multiple particles interact with each other

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momentum

m1

m3

m1v1

m3v3

m2

m2v2

Total momentum of system:

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:
impulse
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:

Area underF vs. tcurve

FaveDt

impulse momentum theorem
Impulse – Momentum Theorem

(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:

Sun (m1)

Earth (m2)

conservation of momentum
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
slide5

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:

  • 0 kg-m/s before the collision and 0 kg-m/s after the collision.
  • –4 kg-m/s before the collision and 4 kg-m/s after the collision.
  • –8 kg-m/s before the collision and 8 kg-m/s after the collision.
  • 8 kg-m/s before the collision and 0 kg-m/s after the collision.
cq2 interactive conceptual example basketball vs flower pot drop
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?

  • The change in momentum of the basketball is the same as the change in momentum of the flowerpot.
  • The change in momentum of the basketball is twice the change in momentum of the flowerpot.
  • The change in momentum of the basketball is half of the change in momentum of the flowerpot.
  • The change in momentum of the flowerpot is twice the change in momentum of the basketball.

PHYSLET Exercise 8.5.1, Prentice Hall (2001)

example problem 6 24
Example Problem #6.24

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?

Solution (details given in class):

62 s

cq 3 interactive example problem saving an astronaut
CQ 3: Interactive Example Problem:Saving an Astronaut

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

  • 5 s
  • 10 s
  • 15 s
  • 20 s
  • 60 s

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

slide9

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?

  • No, because he loses the potential energy of the objects that he leaves behind.
  • No, because although his kinetic energy increases, his momentum decreases.
  • Yes, because although his kinetic energy decreases, his momentum increases.
  • Yes, because although his momentum decreases, his kinetic energy decreases.
collisions

v1i

You (m1)

vf

Energy used to do work (destroy fenders)

Conservation of momentum:

Collisions

BEFORE collision:

  • Inelastic collisions
    • Example: cars stick together after a collision

Me (m2)

v2i = 0

AFTER collision (cars locked together):

What is vf?

collisions11
Collisions

  • Note that ifm1 = m2thenvf = 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 m1v1i2
    • K.E. after collision = K2 = ½ (m1 + m2)vf2 = ½ (m1 + m2)[m1 / (m1 + m2)]v1i2
    • SoK2 / 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
elastic collisions

m2, v2i = 0

m1, v1f

m2, v2f

m1, v1i

Elastic Collisions
  • Head-on collision where one object is at rest before the collision:
  • Conservation of kinetic energy gives:

½ m1v1i2 = ½ m1v1f2 + ½ m2v2f2

  • Conservation of momentum gives:

m1v1i = m1v1f + m2v2f

  • Combining these 2 equations:
elastic collisions13
Elastic Collisions
  • Interesting cases: (1) m1 = m2
    • v1f = 0 andv2f = 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)
slide14

CQ 5: A block of mass m1slides across a frictionless surface with speed v1and 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?

  • I only
  • II only
  • I and II only
  • I, II and III
example problem 6 73
Example Problem #6.73

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.

  • Find the magnitude of the downward velocity with which the basketball reaches the ground.
  • 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?

Solution (details given in class):

  • 4.85 m/s
  • 8.41 m
2 d collisions

m2, v2i = 0

m1, v1f

q

j

m1, v1i

m2, v2f

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

2-D Collisions Interactive

example problem 6 49

vfy

vfx

Example Problem #6.49

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.

BEFORE collision:

y (N)

AFTER collision:

y (N)

v1 = 10 m/s

vf = 5.22 m/s

x (E)

40°

m1 = 2000 kg

v2

x (E)

m1+m2

vfx = vfcos40° = 4.00 m/s

vfy = vf sin40° = 3.36 m/s

m2 = 3000 kg

Solution (details given in class):

5.59 m/s

slide18

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)

  • 3 m/s
  • 6 m/s
  • 9 m/s
  • 12 m/s
example problem 6 41
Example Problem #6.41

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?

Solution (details given in class):

273 m/s

example problem 6 56
Example Problem #6.56

Solution (details given in class):

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?

rocket propulsion
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!