Transcript No Slide Title

Chapter 7
Linear Momentum
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中微子)
http://www.hk-phy.org/articles/neutrino/neutrino.html
Linear Momentum
- Like energy, it is conserved

Define momentum p


p  mv
Consider the rate of change of momentum


0
dp
dv  dm   dm
dt
m
dt
v
dt
 F v
dt
Mass of the system does not change in general
Newton’s 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.
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
q
The final velocity is [-40, 75]
q
Example 7.1
(a) Calculate the momentum of a 110Kg football player running at 8 m/s.
(b) Compare the player’s momentum with that of a hard-thrown 0.41Kg
football that has a speed of 25 m/s.
m=110Kg, v=8 m/s, mfb=0.41 Kg, vfb=25 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 ball’s 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?
m=0.14 Kg, v0=45 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
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, I=Dp notice I // Dp
體操 : for a given Dp change, net F ↓ if Dt ↑  less impact
t2
I  Dp   F (t )dt  Feff (t 2  t1 )
t1
A piano hammer striking a string would
generate a force similar to Factual but its
impluse might be the same as that of Feff.
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.
M=110 Kg, vi=8 m/s, net F=17600 N, Dt=0.0550 s
 vf = ?
Use I = (net F) Dt = M(vf – vi)
-17600*0.0550 = 110(vf – 8)
vf = -0.800 m/s
The minus sign indicates the player bounces backward.
7.3 Conservation of momentum
Under what condition is momentum conserved ?
Net external force = 0  consider a larger system
Consider the impulse
Dp1 = F1 Dt
Dp2 = F2 Dt
Newton’s Third Law F2 = -F1
 Dp2 = (-F1)Dt = -Dp1
 Dp1 + Dp2 =0
 p1 + p2 = constant
Ptotal = p1 + p2 = p1’ + p2’
Conservation of Momentum
(isolated system, net F = 0)
Ptotal = constant
Ptotal = P’total
- 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
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
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 vB=0,  pA = pA’ + pB’
mA vA = mA vA’ + mB vB’
½ mAvA2 = ½ (mA vA’2 + mB vB’2)
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).
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, KE’int = 0  internal KE not conserved
The two masses come to rest after collision  momentum conserved
Perfectly inelastic, KE’int = 0
Example 7.5 Inelastic collision of puck and goalie
(a) 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.
(b) How much KE is lost in the collision? Assume friction between the ice and
the puck-goalie system is negligible.
m1=0.15 Kg, v1=35 m/s, m2=70.0 Kg, v2 = 0
(a) v1’=v2’=v’= ?
(b) How much KE is lost in collision ?
(a) m1 v1 + m2 v2 = (m1 + m2) v’  solve for v’
v’ = 0.0748 m/s
(b) KE’int – KEint = ½ (m1+m2)v’2 – ½ m1v12
= -91.7 J
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, v2’=3.7m/s
(b) KEint = ½ (m1v12 + m2 v22)= 0.763 J
KE’int = ½ (m1 v1’2 + m2 v2’2) = 6.22 J
KE’int – KEint = 5.46 J
(internal KE increased energy released by the spring
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 = m1v1’cosq1 + m2v2’cosq2
Y component : 0 = m1v1’sinq1 + m2v2’sinq2
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 m2=0.4, v2=0. 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 v1=2 m/s, and it is 1.5 m/s after the
collision. Calculate the magnitude and direction of the velocity (v’2 and h) of the 0.4Kg mass
after the collision.
X : m1v1 = m1v1’cos45 + m2v2’cosq2
Y : 0 = m1v1’sin45 + m2v2’sinq2
m1= 0.250Kg, m2=0.400Kg
v1= 2.00m/s, v2=0
v1’=1.5 m/s
solve for sinq2 and cosq2  tanq2
q2 = -48.5o
Given v2’  one can determine mass of the unseen mass m2
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 (v1’2 + v2’2)
X and Y components of momentum conserve
X component : m1v1 = m1v1’cosq1 + m2v2’cosq2
Y component : 0 = m1v1’sinq1 + m2v2’sinq2
 (X component)2 : v12= (v1’cosq1 + v2’cosq2)2
(Y component)2 : 0 = (v1’sinq1 + v2’sinq2)2
½ mv12 = ½ m (v1’2 + v2’2) + m v1’v2’ cos(q1 + q2)
 m v1’v2’ 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
Particle Accerlator
CERN - European Organization for Nuclear Research
27 Km in diameter
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/antimatter/hi
story/historypictures/LHC-drawing-half.jpg
• FermiLab
• http://www.cs.cmu.edu/~zollmann/pics/2004_10_chicago/s
lides/fermilab.html
• Stanford Linear Accelerator Center (SLAC)
• http://www.pbs.org/wgbh/nova/einstein/toda-ocon-01.html
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.
7.7 Rockect Propulsion