chap. 4: work and energy - bits pilani · 2012-09-25 · chap. 4: work and energy r i s h i k e s h...
TRANSCRIPT
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Chap. 4: Work and Energy
R i s h i k e s h V a i d y a
Theoretical Particle Physics
Office: 3265
Physics Group, B I T S Pilani
September 7, 2012
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Outline
1 Work Energy Theorem
2 Potential Energy
3 Non Conservative Force
4 Conservation Laws and Particle Collisions
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The Fundamental Problem of Mechanics
Given all forces, solve
~F = m~a
to find ~r(t).
Problem!
What we know is F(x) and not F(t).
Solve half the problem!
md~v
dt= ~F(~r)
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Equation of Motion in One Dimension
md2x
dt2= F(x) or m
dv
dt= F(x)
Integrate with respect to x
m
∫ xb
xa
dv
dtdx =
∫ xb
xa
F(x)dx
Change variable from x to t using
differentials:
dx =
(dx
dt
)
dt = vdt
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Equation of Motion in One Dimension
md2x
dt2= F(x) or m
dv
dt= F(x)
Integrate with respect to x
m
∫ xb
xa
dv
dtdx=m
∫ tb
ta
dv
dtvdt
=m
∫ tb
ta
d
dt
(1
2v2)
dt =1
2mv2
∣∣∣∣
ta
tb
=1
2mv2b −
1
2mv2a
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Equation of Motion in One Dimension
md2x
dt2= F(x) or m
dv
dt= F(x)
Work-Energy Theorem:
1
2mv2b −
1
2mv2a =
∫ xb
xa
F(x)dx
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Equation of Motion in One Dimension
md2x
dt2= F(x) or m
dv
dt= F(x)
Work-Energy Theorem: Motion in 3-D
1
2mv2b −
1
2mv2a =
∫ b
a
~F · d~r
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How Useful is Work-Energy Theorem
Evaluation of∫ b
a~F · d~r is tentamount to
knowing the trajectory beforehand !Two cases where Work-EnergyTheorem is useful.
◮ Conservative Forces: For many
interesting forces∫ b
a~F · d~r depends
only on end-points !◮ Constrained Motion: External
Constraints pre-determine the
trajectory.
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How Useful is Work-Energy Theorem
Regardless of whether forces are
conservative or not the Work-
Energy theorem is always true
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Prob.4.5 Mass m whirls on a frictionless
table, held to circular motion by a string
which passes through a hole in the table.
The string is slowly pulled through the
hole so that the radius of the circle
changes from l1 to l2. Show that the work
done in pulling the string equals the
increase in kinetic energy of the mass.
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Prob.4.5 Mass m whirls on a frictionless
table, held to circular motion by a string
which passes through a hole in the table.
The string is slowly pulled through the
hole so that the radius of the circle
changes from l1 to l2. Show that the work
done in pulling the string equals the
increase in kinetic energy of the mass.
The fundamental mystery here is ... how can
you pull radially and still end up changing the
angular velocity of the mass.
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Solution 4.5 For cicular motion:
Fr = mv21l1
= m l1ω21
If Fr is increased, m will move to smaller r:
Fr(r) = mv2(r)
r= mrω2(r)
Evaluate:
W = −∫
l2
l1
Fr(r)dr = −∫
l2
l1
mrω2(r)dr
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~a = (r − rθ2)r + (rθ + 2rθ)︸ ︷︷ ︸
=0
θ
rω + 2rω=0 [θ = ω]1
ω
dω
dt=−2
1
r
dr
dt∫ ω(r)
ω1
dω
ω=−2
∫ r
l1
dr
r⇒ ω(r) = l
21
ω1
r2
W = −∫
l2
l1
Fr(r)dr = −∫
l2
l1
mrω2(r)dr
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W=−ml41ω
21
∫l2
l1
dr
r3=
1
2ml
41ω
21
(
1
l22
− 1
l21
)
=1
2m
(
l41ω
21
l22
− l21ω
21
)
W=1
2m[(l2ω2)
2 − (l1ω1)2] [
ω2l22 = ω1l
21
]
=K2 − K1
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Prob. 4.7 A ring of mass M, hangs from a
thread, and two beads of mass m slide on
it without friction. The beads are released
simultaneously from the top of the ring
and slide down opposite side. Show that
the ring will start to rise if m > 3M/2, and
find the angle at which this occurs.
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Forces are different for cosθ > 2/3 and
cos θ < 2/3
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Forces are different for cosθ > 2/3 and
cos θ < 2/3
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cos θ, N and T vs. θ (m = 2M)
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N and T vs. θ: Magnified view
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2N cos θ vs. θ
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Accelaration of the ring vs. θ
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Prob. 4.10
A block of mass M on a horizontal
frictionless table is connected to a spring
(spring constant k). The block is set in
motion so that it oscillates about its
equilibrium position with a certain
amplitude A0. The period of motion is
T0 = 2π√
M/k.
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Prob. 4.10
(a) A lump of sticky putty of mass m is
dropped onto the block. the putty
sticks without bouncing. The putty hits
M at the instant when the velocity of
M is zero. Find1 The new period
2 The new amplitude3 The change in mechanical energy of thesystem
(b) Repeat part a, assuming that the sticky
putty hits M at the instant when M has
its maximum velocity.
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Potential Energy
For a conservative Force Field:
∫ ~rb
~ra
~F · d~r= func. of (~rb) − func. of (~ra)
Wba=−U(~rb) + U(~ra)
Wba=Kb − Ka [Work − energyTh.]
Wba=−Ub + Ua = Kb − Ka
Ua + Ka = Ub + Kb = E
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What Does Potential Energy Tell us About Force?
Ub − Ua = −∫ xb
xa
F(x)dx
U(x + ∆x) − U(x) = ∆U
∆U=−∫ x+∆x
x
F(x)dx
∆U≈−F(x)(x + ∆x − x) = −F(x)∆x
F(x)≈−∆U
∆xor F(x)
lim∆x→0= −dU
dx
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Potential Energy Determins Stability of a System
Harmonic Oscillator: U = kx2/2
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Potential Energy Determins Stability of a System
Simple Pendulum:
U(θ) = mgl(1 − cos θ) and dUdθ
= mgl sin θ
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Potential Energy Determins Stability of a System
Simple Pendulum:
U(θ) = mgl(1 − cos θ) and dUdθ
= mgl sin θ
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Potential Energy Determins Stability of a System
General Stability Criteria:
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Potential Energy Determins Stability of a System
General Stability Criteria:
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Potential Energy Determins Stability of a System
General Stability Criteria:
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Rock me Spin me yet I Stand Tall:
The Amazingly Stable Teeter-Toy
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Rock me Spin me yet I Stand Tall:
The Amazingly Stable Teeter-Toy
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Rock me Spin me yet I Stand Tall:
The Amazingly Stable Teeter-Toy
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Rock me Spin me yet I Stand Tall:
The Amazingly Stable Teeter-Toy
![Page 37: Chap. 4: Work and Energy - BITS Pilani · 2012-09-25 · Chap. 4: Work and Energy R i s h i k e s h V a i d y a Theoretical Particle Physics Office: 3265 rishidilip@gmail.com Physics](https://reader034.vdocument.in/reader034/viewer/2022042023/5e7bd18c6d400d25160674fd/html5/thumbnails/37.jpg)
A Sports Car
Stability requires low center of mass and
hence the peculiar design of a sports car.
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Non-Conservative Forces
~F = ~Fc + ~Fnc
Wtotalba =
∫ b
a
~F · d~r
Wtotalba =
∫ b
a
~Fc · d~r +∫ b
a
~Fnc · d~rKb − Ka=−Ub + Ua + Wnc
ba
Kb + Ub − (Ka + Ua) = Wncba
Eb − Ea = Wncba
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Power: Time rate of doing Work
P =dW
dt= ~F · d
~r
dt= ~F ·~v
Units:
[S.I.] 1 W=1J/s [CGS]1 erg/s = 10−7W
[English] 1 hp=550 lb ft/s ≈ 746W
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Power: Time rate of doing Work
Typical Power Consumption
Man running upstairs: 1/2 to 1 hp for
30 s
A husky man can do work over a period
of 8 hours only at a rate of
0.2 hp ≈ 1000 Kcal
Per person energy use: India (0.7 kW),
Germany (6 kW), USA (11.4 kW)
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Prob. 4.18 A 160 lb man leaps into the
air from a crouching position. His center
of gravity rises 1.5 ft before he leaves the
ground, and it then rises 3 ft to the top of
his leap. What power does he develop
assuming that he pushes the ground with
constant force?
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Sol.4.18
P=W/T (W = work done by N)
W=N · 1.5 (c.g. rises by 1.5ft)
N=mg + ma or N = 160 +160
32a
a=v2
2s= 64
[
v =√
2gs′ =√2 · 32 · 3 = 8
√3]
N=480 lb W = 720 lb.ft T = v/a =√3/8
P = W/T = 3325lb.ft/s ≈ 6hp
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Prob. 4.19
In the preceeding problem take
F = F0 cosωt where F0 is the peak force,
and the contact with ground ends at
ωt = π/2. Find the peak power that the
man develops during the jump.
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Sol.4.19
P(t)=N(t)v(t) [N(t) = −F(t)]
N(t) − mg=ma(t)
m
∫ v(t)
0
dv=
∫ t
0
(F0 cosωt − mg) dt
v(t)=F0
mωsinωt − gt [F0, ω?]
x(t)=F0
mω2(1 − cosωt) − 1
2gt2
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Sol.4.19
Boundary conditions:
v(t = π/2ω)=8√3 =
F0
mω− gπ
2ω
x(t = π/2ω)=1.5 ft =F0
mω2− gπ2
8ω2
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Sol.4.19
ω = 9.96s−1 F0 = 832 lb t =π
2ω= 0.16s
P(t)=F(t)v(t)
F0
2mω
F0 sin 2ωt︸ ︷︷ ︸
1
− 2mgωt cosωt︸ ︷︷ ︸
2
A reasonable approximation: F0 >> mg
then 1st >> 2nd
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Sol.4.19
P(t)≈ F20
2mωsin 2ωt
For Pmax. :dP
dt=0
dP
dt=
F20
mcos 2ωt = 0
Pmax. = P|t= π4ω
=F20
2mωsin
(
2ωπ
4ω
)
=F20
2mω
![Page 49: Chap. 4: Work and Energy - BITS Pilani · 2012-09-25 · Chap. 4: Work and Energy R i s h i k e s h V a i d y a Theoretical Particle Physics Office: 3265 rishidilip@gmail.com Physics](https://reader034.vdocument.in/reader034/viewer/2022042023/5e7bd18c6d400d25160674fd/html5/thumbnails/49.jpg)
Sol.4.19
Check:
d2P
dt2
∣∣∣∣∣t= π
4ω
=−F2
0
m2ω sin 2ωt
∣∣∣∣∣t= π
4ω
< 0
![Page 50: Chap. 4: Work and Energy - BITS Pilani · 2012-09-25 · Chap. 4: Work and Energy R i s h i k e s h V a i d y a Theoretical Particle Physics Office: 3265 rishidilip@gmail.com Physics](https://reader034.vdocument.in/reader034/viewer/2022042023/5e7bd18c6d400d25160674fd/html5/thumbnails/50.jpg)
Sol.4.19
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Sol.4.19
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What do collisions teach us?
~F = m~a is a double-edge sword
Collisions studies gave most profound
knowledge about fundamental Physics
Constraints from energy and
momentum conservation severe enough
to extract vital information about
scattering
![Page 53: Chap. 4: Work and Energy - BITS Pilani · 2012-09-25 · Chap. 4: Work and Energy R i s h i k e s h V a i d y a Theoretical Particle Physics Office: 3265 rishidilip@gmail.com Physics](https://reader034.vdocument.in/reader034/viewer/2022042023/5e7bd18c6d400d25160674fd/html5/thumbnails/53.jpg)
LHC@CERN Geneva
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Fermilab@Illinois USA
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Classical Collisions
A + B → C + D
Mass is conserved:
mA + mB → mC + mD
Momentum is conserved:
pA + pB = pC + pD
K.E. may or may not be conserved
![Page 56: Chap. 4: Work and Energy - BITS Pilani · 2012-09-25 · Chap. 4: Work and Energy R i s h i k e s h V a i d y a Theoretical Particle Physics Office: 3265 rishidilip@gmail.com Physics](https://reader034.vdocument.in/reader034/viewer/2022042023/5e7bd18c6d400d25160674fd/html5/thumbnails/56.jpg)
Classical Collisions
A + B → C + D
Types of Collisionssticky: K.E. decreases
KA + KB > KC + KD
Explosive: K.E. increases
KA + KB < KC + KD
Elastic: K.E. is conserved
KA + KB = KC + KD
![Page 57: Chap. 4: Work and Energy - BITS Pilani · 2012-09-25 · Chap. 4: Work and Energy R i s h i k e s h V a i d y a Theoretical Particle Physics Office: 3265 rishidilip@gmail.com Physics](https://reader034.vdocument.in/reader034/viewer/2022042023/5e7bd18c6d400d25160674fd/html5/thumbnails/57.jpg)
Prob. 4.23 A small ball of
mass m is placed on top of the
a “superball” of mass M, and
the two balls are dropped to
the floor from height h.How
high does the small ball rise
after the collision? Assume
that the collision is perfecly
elastic, and that m << M.
![Page 58: Chap. 4: Work and Energy - BITS Pilani · 2012-09-25 · Chap. 4: Work and Energy R i s h i k e s h V a i d y a Theoretical Particle Physics Office: 3265 rishidilip@gmail.com Physics](https://reader034.vdocument.in/reader034/viewer/2022042023/5e7bd18c6d400d25160674fd/html5/thumbnails/58.jpg)
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Sol. 4.23 Let v1 and v2 be the initial
velocities of m and M before collision and
v′1 and v′2, after collision. In a two body
one dimensional collision:
v′1=(m − M)v1 + 2Mv2
m + M
v′2=(M − m)v2 + 2mv1
m + M
Here: v1 = −√2gh, v2 =
√2gh, and
M >> m
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Sol. 4.23
v′1 =3M · √2gh
M= 3
√
2gh
If m rises to height h′ after collision:
1
2m1v
′21 = m1gh
′
h′ =v′21
2g= 9h
![Page 61: Chap. 4: Work and Energy - BITS Pilani · 2012-09-25 · Chap. 4: Work and Energy R i s h i k e s h V a i d y a Theoretical Particle Physics Office: 3265 rishidilip@gmail.com Physics](https://reader034.vdocument.in/reader034/viewer/2022042023/5e7bd18c6d400d25160674fd/html5/thumbnails/61.jpg)
Problem 4.9 A “superball” of mass m bounces
back and forth between two surfaces with speed v0.Gravity is neglected and collisions are perfectly
elastic.
(a) Find the average force F on each wall.
(b) If one surface is moving uniformly toward the
other with speed V << v0, the bounce rate willincrease due to shorter distance between
collisions, and because the ball’s speed increaseswhen it bounces from the moving surface. Find
the F in terms of separation of surfaces, x.
(c) Show that work needed to push the surface
from l to x equals gain in kinetic energy of theball.
![Page 62: Chap. 4: Work and Energy - BITS Pilani · 2012-09-25 · Chap. 4: Work and Energy R i s h i k e s h V a i d y a Theoretical Particle Physics Office: 3265 rishidilip@gmail.com Physics](https://reader034.vdocument.in/reader034/viewer/2022042023/5e7bd18c6d400d25160674fd/html5/thumbnails/62.jpg)
Problem 4.29
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Sol. 4.29
(a) Momentum transfer to wall in 1
bounce: m(2v0)
No. of bounce per unit time: v0/2l
F = 2mv0v0
2l=
mv20l
(b) F(x) = mv(x)2
x
Integrate a(x) = dv(x)dt
to find v(x).
![Page 64: Chap. 4: Work and Energy - BITS Pilani · 2012-09-25 · Chap. 4: Work and Energy R i s h i k e s h V a i d y a Theoretical Particle Physics Office: 3265 rishidilip@gmail.com Physics](https://reader034.vdocument.in/reader034/viewer/2022042023/5e7bd18c6d400d25160674fd/html5/thumbnails/64.jpg)
Sol. 4.29
(b) ∆v after each bounce ∆v = 2V
(Sling-shot effect)
This change happens in time ∆t = 2x/v
dv
dt=
Vv
x∫ v
v0
dv′
v′=
∫ t
0
Vdt
x= −
∫ x
l
dx′
x′
v(x) =v0l
x⇒ F =
mv(x)2
x=
ml2v20x3
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Sol. 4.29
(c) Work done in moving the surface:
W=−∫ x
l
F(x′) dx′ = −ml2v20
∫ x
l
dx′
x′3
=ml
2v202
[1
x2− 1
l2
]
=ml
2v202x2
− 1
2mv20
=1
2mv(x)2 − 1
2mv20
=Change in K.E. of ball
![Page 66: Chap. 4: Work and Energy - BITS Pilani · 2012-09-25 · Chap. 4: Work and Energy R i s h i k e s h V a i d y a Theoretical Particle Physics Office: 3265 rishidilip@gmail.com Physics](https://reader034.vdocument.in/reader034/viewer/2022042023/5e7bd18c6d400d25160674fd/html5/thumbnails/66.jpg)
Sling-Shot (Gravity-Assist) and NASA’s Cassini
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Collisions and Center of Mass Coordinates
How does it help?
Total momentum in C-system is zero
Initial and final velocities lie in the
same plane
Each particle is scattered through the
same angle θ in the plane of scattering.
For elastic collisions, the speed of each
particle is same before & after the
collision
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Collisions and Center of Mass Coordinates
m1 & m2 having velocity
v1 & v2.
V =m1v1 + m2v2
m1 + m2
![Page 69: Chap. 4: Work and Energy - BITS Pilani · 2012-09-25 · Chap. 4: Work and Energy R i s h i k e s h V a i d y a Theoretical Particle Physics Office: 3265 rishidilip@gmail.com Physics](https://reader034.vdocument.in/reader034/viewer/2022042023/5e7bd18c6d400d25160674fd/html5/thumbnails/69.jpg)
Collisions and Center of Mass Coordinates
v1c=v1 − V
=m2
m1 + m2
(v1 − v2)
v2c=v2 − V
=− m1
m1 + m2
(v1 − v2)
![Page 70: Chap. 4: Work and Energy - BITS Pilani · 2012-09-25 · Chap. 4: Work and Energy R i s h i k e s h V a i d y a Theoretical Particle Physics Office: 3265 rishidilip@gmail.com Physics](https://reader034.vdocument.in/reader034/viewer/2022042023/5e7bd18c6d400d25160674fd/html5/thumbnails/70.jpg)
Collisions and Center of Mass Coordinates
The momenta in
C-system:
p1c=m1v1c
=m1m2
m1 + m2
(v1 − v2)
=µv
p2c=−µv
0=p1c + p2c
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C-System Plane of Scattering
m1 and m2 striking with
v1 & v2.
Initial velocities in L &
C-system.
![Page 72: Chap. 4: Work and Energy - BITS Pilani · 2012-09-25 · Chap. 4: Work and Energy R i s h i k e s h V a i d y a Theoretical Particle Physics Office: 3265 rishidilip@gmail.com Physics](https://reader034.vdocument.in/reader034/viewer/2022042023/5e7bd18c6d400d25160674fd/html5/thumbnails/72.jpg)
C-System Plane of Scattering
m1 and m2 striking with
v1 & v2. Final velocities in L &
C-system (The plane in
general is different)
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C-System Plane of Scattering
C-system plane of
scattering.
For elastic scattering,
v1c = v′1c and v2c = v′2c.
Velocity vectors simple
rotate.
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If m2 is at rest
V=m1
m1 + m2
v1
v1c=v1 − V
=m2
m1 + m2
v1
v2c=−V
=− m1
m1 + m2
v1
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If m2 is at rest
V=m1
m1 + m2
v1
v1c=v1 − V
=m2
m1 + m2
v1
v2c=−V
=− m1
m1 + m2
v1