Summary Lecture 8

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Summary Lecture 8
Systems of Particles 9.2 Centre of
mass 9.3 Newton 2 for system of
particles 9.4-7 Conservation of
momentum 9.8-11 Collisions 9.12 Rocket
propulsion
Week March 20 24 20-minute test on material in
lectures 1-7
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ProblemsChap. 9 1, 6, 10 , 15, 20, 27,
40, 71, 73, 78
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System of Particles
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So far we have considered the motion of POINT
PARTICLES
FINITE OBJECTS can move as a whole
(translational motion) and also rotate about the
Centre of Mass
The Centre of Mass is that point where if we
apply a force, the object will not rotate.
What happens depends on where we apply the force
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The Centre of Mass
The motion of the Centre of Mass is a simple
parabola. (just like a point particle)
The motion of the entire object is complicated.

• This motion resolves to
• motion of the CM

• motion of points around the CM

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The Centre of Mass
M m1 m2
M
m1g x d1 m2g x d2
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Centre of Mass (1D)
M m1 m2
M xcm m1 x1 m2 x2
moment of M moment of individual masses
In general
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Centre of Mass (3D)
For a collection of masses in 3D
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Centre of Mass (3D)
For a collection of masses in 3D
rcm ixcm jycm
So in a solid body we can find the CM by finding
xcm and ycm
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Xcm 16/15 1.07 m
ycm 20/15 1.33 m
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Finding the Centre of Mass
For odd shaped objects this probably needs to be
determined experimentally
For symmetric objects this can often be calculated

1. Look for a symmetry axis
2. Then carry out the integral to find the position
   of xcm along the axis.

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Solid cone
dm ??r2 dx but r (R/h)x dm ?? (R/h)2x2 dx
Mass of cone M 1/3 ??R2 h
xcm ¾ h
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For a system of particles, the dynamics of the
Centre of Mass obeys Newton 2.
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This also applies to a solid body, where the
individual particles are rigidly connected. The
dynamics of the Centre of Mass obeys Newton 2
For a system of particles, the dynamics of the
Centre of Mass obeys Newton 2.
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Linear Momentum of system of particles
For a system of particles P ?p
Mvcm
This also applies to extended objects
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Conservation of Linear Momentum
NO EXTERNAL forces act on the system
If Fext 0
That is Px, Py and Pz remain constant if
Fext-x, Fext-y and Fext-z are zero
In an isolated system, momentum is conserved.
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Exploding rocket
Why? No external horizontal forces so horiz
momentum unchanged
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m 3.8 g, n 12 v 1100 m s-1

Initial momentum Pi n mv M Vi n mv 0
Pi
Final momentum pf (M nm) V
n mv
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m 3.8 g, n 12 v 1100 m s-1

KE initial ½ n mv2 ½ x 12 x 0.0038 x
(1100)2 27588 J
Initial momentum Pi n mv M Vi n mv 0
KE final ½ (M 12m)V2 ½ x (12.0456) x 4.22 163 J
Final momentum pf (M nm) V Pi n mv
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coll
isions
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Collisions
What is a collision?
An isolated event involving 2 or more objects
Usually interact (often strongly) for short time
Equal and opposite impulses are exerted on each
other
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Collisions
Elastic collisions Energy and momentum are
conserved
Inelastic collisions Only momentum is conserved
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Elastic collision
In 1 dimension
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Elastic Collision 1D
We want to find V1f and V2f
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Mom. Cons. m1v1i m1v1f m2v2f(1)
? m2v2f m1(v1i- v1f)(2)
Energy Cons ½ m1v1f2 ½ m2v2f2 ½ m1v1i2
? ½ m2v2f2 ½ m1(v1i2 - v1f2)
Mult. by 2 and factorise
? m2v2f2 m1(v1i- v1f) (v1i v1f) (3)
Divide equ. (3) by (2)
? v2f v1i v1f .(4)
V1i is usually given, so to find v2f we need to
find an expression for v1f. Get this from equ.
(1).
m1v1f m1v1i - m2v2f ?
Substitute this form of v1f into equ 4
? v2f v1i v1i m2/m1 v2f
? v2f(1 m2/m1) 2v1i
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If m1gtgt m2
v2f ? 2v1i
v1f ? v1i
If m2gtgtm1
v2f ? 0
v1f ? -v1i
If m1 m2
v1f ? 0
v2f ? v1i
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That's all folks
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Motion of the C of M
m1 v1i
m2 v2i 0
vcm
What is Vcm? Mom of CM mom of m1 mom of
m2 (m1 m2 ) Vcm m1v1i m2v2i
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Motion of the C of M
m1 v1i
m2 v2i 0
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Lets observe the elastic collision from the view
point of the centre of mass
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inelastic collision
In 1 dimension
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m1 v1i
m2 v2i 0
What is Vcm? Mom of CM mom of m1 mom of
m2 (m1 m2 ) Vcm m1v1i m2v2i
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m1 v1i
m2 v2i
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Lets observe the elastic collision from the view
point of the centre of mass
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Collisions in 2 dimensions
Elastic billiard balls comets a-particle
scattering
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Elastic collisions in 2-D
Momentum is conserved
Consider x-components
m1v1i m1v1f cos ?1 m2v2f cos ?2
Consider y-components
0 -m1v1f sin ?1 m2v2f sin ?2
Since elastic collision energy is conserved
7 variables!
3 equations
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Collisions in 2 dimensions
Inelastic Almost any real collision! an
example Automobile collision
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Cons. Momentum gt pA pB pf X component PA
Pf cosq mAvA (mA mB) vf cosq.(1) Y
component PB Pf sinq mBvB (mA mB) vf
sinq.(2)
____________________ mAvA (mA mB) vf cosq
Divide equ (2) by (1)
Gives q 39.80
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Gives Vf 48.6 kph
Use equ 2 to find Vf
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Can the investigators determine who was speeding?
http//www.physics.ubc.ca/outreach/phys420/p420_9
6/danny/danweb.htm
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momentum conservation
and
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Dm
IN THE EARTH REF. FRAME Vel of gas rel me vel
of gas rel. rocket - vel of rocket rel me U
- v
Mom. of gas ?m(U - v) -change in mom. of
rocket (impulse) i.e. F dt ?m(v - U)
v dm - U dm
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F dt ?m(v - U) v dm - U dm
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If I want to find out the TOTAL effect of
throwing out gas, from when the mass was mi and
velocity was vi, to the time when the mass is mf
and the velocity vf, I must integrate.
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An example
Mi 850 kg mf 180 kg U 2800 m s-1 dm/dt
2.3 kg s-1
Thrust dp/dt of gas dm/dt U
2.3 x 2800 6400 N
Initial acceleration F ma gt a F/m
6400/850 7.6 m s-2
Final vel.