Linear Momentum

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Chapter 7

• Linear Momentum

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• Why study momentum ?
• In an isolated system, net force is zero,
• ?momentum is a conserved quantity
• Applications
• collision problems ? mass or velocity
• determination
• discovery of missing objects or
• sub-atomic particles (neutrino???)

http//www.hk-phy.org/articles/neutrino/neutrino.h
tml
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Linear Momentum - Like energy, it is conserved
Define momentum
Consider the rate of change of momentum
0
Mass of the system does not change in general
Newtons Second Law
Newton actually stated his Second Law of motion
as ? Net External force equals the change in
momentum of a system divided by the time over
which it changes.
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• Collision with a stationary object
• Axis-aligning bounding box ? means that the sides
  of the box are horizontal and vertical
• If the stationary boundary is vertical, vf
  -vix, viy
• If the stationary boundary is horizontal, vf
  vix, -viy
• For incoming velocity vi vix, viy
• The angle that the ball comes in at must equal
  the angle at which it leaves, that is the angle
  of incidence (incoming) must equal the angle of
  reflection (outgoing).
• Example
• Suppose you are coding a simple Pong game, and
  you want to model the collision of the ball with
  the paddle. If the ball is approaching the paddle
  with an incoming velocity of 40, 75 when it
  hits, what should be its resulting velocity be?
• Answer
• The final velocity is -40, 75

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• Example 7.1
• Calculate the momentum of a 110Kg football player
  running at 8 m/s.
• Compare the players momentum with that of a
  hard-thrown 0.41Kg football that has a speed of
  25 m/s.
• m110Kg, v8 m/s, mfb0.41 Kg, vfb25 m/s
• p/pfb ? 85.9
• Example 7.2
• What is the average force does exert on a 0.14Kg
  baseball by a bat, given that the balls initial
  velocity is 45 m/s and that its final velocity,
  after a 1.3 ms impact, is 65 m/s in exactly the
  opposite direction?
• m0.14 Kg, v045 m/s, vf-65 m/s, in 1.3 ms
• F ? ? Dp/Dt (pf pi)/Dt
  m(vf-v0)/0.0013 11800 N ? 2600 lb

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7.2 Impluse
Effect of a force on an object depends on how
long it acts. Dp
(net F) Dt The change in momentum equals the
average net external force multiplied by the
time this force acts.
(net F) Dt is called the impulse, I Impulse is
the same as change in momentum, IDp notice I //
Dp ?? for a given Dp change, net F ? if Dt ? ?
less impact
A piano hammer striking a string would generate a
force similar to Factual but its impluse might
be the same as that of Feff.
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Example 7.3 Calculate the final speed of a 110Kg
football player running at 8 m/s who collides
head on with a padded goalpost and experiences a
backward force of 17600 N for 0.055 s. M110 Kg,
vi8 m/s, net F17600 N, Dt0.0550 s ? vf
? Use I (net F) Dt M(vf vi)
-176000.0550 110(vf 8)
vf -0.800 m/s The minus sign
indicates the player bounces backward.
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7.3 Conservation of momentum
Under what condition is momentum conserved ? Net
external force 0 ? consider a larger system

• p1 p2 constant
• Ptotal p1 p2 p1 p2
• Conservation of Momentum (isolated system, net F
  0)
• Ptotal constant
• Ptotal Ptotal

• Consider the impulse
• Dp1 F1 Dt
• Dp2 F2 Dt
• Newtons Third Law F2 -F1
• Dp2 (-F1)Dt -Dp1
• Dp1 Dp2 0

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- The three dimension in nature is independent -
Momentum can be conserved along one direction and
not another. - Momentum is conserved along the
X-direction, but not in Y-direction
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7.4 Elastic collisions in one dimension
The two collided bodies moving the along the same
direction Elastic collision both momentum and
internal kinetic energy conserved
Very nearly elastic collision because some KE ?
heat, sound
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• Example 7.4 An elastic collision
• Calculate the velocities of two masses following
  an elastic collision,
• given that mA 0.500 Kg, mB 3.5 Kg, vA 4.00
  m/s, vB 0.
• What are the final velocities of mA and mB ? ?
  vA ? , vB ?
• Since vB0, ? pA pA pB
• mA vA mA vA mB vB
• ½ mAvA2 ½ (mA vA2 mB vB2)
• Solve the conservation of momentum for vB first,
  and substitute into
• conservation of internal KE to eliminate vB,
  leaving only vA unknown.
• There are two solutions for vA, vA 4.00 m/s or
  -3.00 m/s
• First solution ? same as initial condition ?
  discarded 4.00 m/s
• vB 1.00 m/s
• A small mass m, collide with a larger mass M, the
  larger mass
• is knocked forward with a lower speed (here
  mB 7mA).

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7.5 Inelastic collision in one dimension
Inelastic collision ? internal kinetic energy is
not conserved, some internal KE may be converted
into heat or sound energy
Two equal masses head toward one another at equal
speeds and stick together. KEint mv2, KEint
0 ? internal KE not conserved The two masses
come to rest after collision ? momentum conserved
Perfectly inelastic, KEint 0
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• Example 7.5 Inelastic collision of puck and
  goalie
• Find the recoil velocity of a 70 Kg hockey
  goalie, originally at rest, who catches a 0.15 Kg
  hockey puck slapped at him at a velocity of 35
  m/s.
• How much KE is lost in the collision? Assume
  friction between the ice and the puck-goalie
  system is negligible.
• m10.15 Kg, v135 m/s, m270.0 Kg, v2 0
• v1v2v ?
• How much KE is lost in collision ?

• (a) m1 v1 m2 v2 (m1 m2) v ? solve for v
• v 0.0748 m/s
• KEint KEint ½ (m1m2)v2 ½ m1v12
• -91.7 J

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Example 7.6 A collision that releases stored
energy is inelastic Two carts collide
inelastically, and a spring releases its PE and
converts it into internal KE. Mass of the cart on
the left is m1 0.35Kg, v1 2.00m/s. Cart on the
right has m2 0.5Kg, v2 -0.5m/s. After
collision, the first cart has a recoil velocity
v1 -4.00m/s. (a) What is the final vel. of the
first cart, v2? (b) how much energy is
released by the spring (assume all internal PE is
converted into internal KE)
(a) m1v1 m2v2 m1v1 m2v2 Solve the
conservation of momentum for v2 first,
v23.7m/s (b) KEint ½ (m1v12 m2 v22) 0.763
J KEint ½ (m1 v12 m2 v22) 6.22 J
KEint KEint 5.46 J (internal KE
increased ?energy released by the spring
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7.6 Collision of point masses in two dimensions

• Complication, if
• object rotation
• Point masses
• Neglect rotation
• Many scattering
• experiments have a
• target mass that is
• stationary in laboratory

X component m1v1 m1v1cosq1 m2v2cosq2 Y
component 0 m1v1sinq1 m2v2sinq2
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Example 7.7 Determine the final velocity of an
unseen mass from the scattering of another mass A
0.25Kg mass is slid on a frictionless surface
into a dark room, where it strikes an object with
m20.4, v20. The 0.25Kg mass emerges from the
room at an angle of 45 with its incoming
direction, The speed of the 0.25Kg mass was v12
m/s, and it is 1.5 m/s after the collision.
Calculate the magnitude and direction of the
velocity (v2 and h) of the 0.4Kg mass after the
collision.
X m1v1 m1v1cos45 m2v2cosq2 Y 0
m1v1sin45 m2v2sinq2 m1 0.250Kg,
m20.400Kg v1 2.00m/s, v20 v11.5 m/s solve
for sinq2 and cosq2 ? tanq2 q2 -48.5o
Given v2 ? one can determine mass of the unseen
mass m2
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Elastic Collisions of Two Equal Masses

• See p.185
• a special case m1 m2 like the case with
  billiard balls, or the case
• with some subatomic collisions
• assume v2 0, and elastic collisions ? internal
  KE conserve
• ½ mv12 ½ m (v12 v22)
• X and Y components of momentum conserve
• X component m1v1 m1v1cosq1 m2v2cosq2
• Y component 0 m1v1sinq1 m2v2sinq2
• ? (X component)2 v12 (v1cosq1 v2cosq2)2
• (Y component)2 0 (v1sinq1 v2sinq2)2
• ½ mv12 ½ m (v12 v22) m
  v1v2 cos(q1 q2)
• ? m v1v2 cos(q1 q2)0
• (1) v1 0 ? head on collision, incoming mass
  stop
• (2) v2 0 ? no collision, incoming mass
  unaffected
• (3) cos(q1 q2) 0 ? angle of separation is 90o
  after collision

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Particle Accerlator CERN - European Organization
for Nuclear Research
27 Km in diameter
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World Class Particle accelerator

• CERN
• http//cernenviro.web.cern.ch/CERNenviro/web/main/
  main.php
• Large Hadron Collider (LHC)
• http//livefromcern.web.cern.ch/livefromcern/antim
  atter/history/historypictures/LHC-drawing-half.jpg
  
• FermiLab
• http//www.cs.cmu.edu/zollmann/pics/2004_10_chica
  go/slides/fermilab.html
• Stanford Linear Accelerator Center (SLAC)
• http//www.pbs.org/wgbh/nova/einstein/toda-ocon-01
  .html

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A submicroscopic particle scatters straight
backward from a target particle. In experiments
seeking evidence for quarks, electrons were
observed t occasionally scatter straight backward
from a proton.
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7.7 Rockect Propulsion