Lec9 Collisions

8/8/2019 Lec9 Collisions

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Lecture 7

COLLISIONS
http://find/http://goback/


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Lecture 7

COLLISIONS

1 2-d elastic collision in Lab Frame

2 Collision in CM reference frame

3 Advantages of CM Frame


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The Large Hadron Collider (LHC)

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The aim of the exercise:

To smash protons moving at 99.999999% of the speed oflight into each other and so recreate conditions a fraction of a

second after the big bang. The LHC experiments try and

work out what happened.

Collisions 2/16


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proton-proton collisions at 7 TeV each (14 TeV in CM frame!)

Collisions 2/16


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2808 bunches of 1.15 1011 protons per bunch

Collisions 2/16


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2808 bunches of 1.15 1011 protons per bunch

to find The Higgs Boson New particles

Collisions 2/16


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Collisions

Standard experimental method to discover the inner constituents

of matter

To find out the nature of sub-atomic forces

Collisions 4/16


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Collisions

Ernst Rutherford (1910): -particle-gold collisions

discovered NucleusCollisions 5/16
http://goback/http://find/http://find/http://goback/


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Collisions

James Chadwick (1932): -particle-Beryllium collisions

discovered Neutron

Collisions 6/16


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Laws of Collisions

Characterizing a Collision

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Laws of Collisions


1 A very short-duration interaction

Collisions 7/16

C


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Laws of Collisions


1 A very short-duration interaction2 External forces are ignorable

Collisions 7/16

L f C lli i


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Laws of Collisions



Conservations that help solve the problem:

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L f C lli i


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Laws of Collisions




1 Linear momentum is conserved:P = 0

Collisions 7/16

L f C lli i


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Laws of Collisions





2 Kinetic Energy:

Collisions 7/16

La s of Collisions


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Laws of Collisions





2 Kinetic Energy:

Elastic: KE conserved

Collisions 7/16

Laws of Collisions


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Laws of Collisions





2 Kinetic Energy:

Elastic: KE conservedInelastic: KE = Q

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Laws of Collisions


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Laws of Collisions





2 Kinetic Energy:


Q < 0: KE converted to energy released (Explosions)

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Laws of Collisions


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Laws of Collisions





2 Kinetic Energy:


Q < 0: KE converted to energy released (Explosions)Q > 0 KE is absorbed

Collisions 7/16

2 d elastic collision in Lab Frame


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2-d elastic collision in Lab Frame

Collisions 2-d elastic collision in Lab Frame 8/16



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Conservation Equations:




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px: m1v1 = m1v

1 cos 1 + m2v

2 cos 2




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px: m1v1 = m1v

1 cos 1 + m2v

2 cos 2

py: m1v

1 sin 1 = m2v

2 sin 2




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px: m1v1 = m1v

1 cos 1 + m2v

2 cos 2

py: m1v

1 sin 1 = m2v

2 sin 2

KE:1

2m1v

2

1 =1

2m1v

2

1 +1

2m2v

2

2




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px: m1v1 = m1v

1 cos 1 + m2v

2 cos 2

py: m1v

1 sin 1 = m2v

2 sin 2

KE:1

2m1v

2

1 =1

2m1v

2

1 +1

2m2v

2

2

4 eqns, 3 unknowns!




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px: m1v1 = m1v

1 cos 1 + m2v

2 cos 2

py: m1v

1 sin 1 = m2v

2 sin 2

KE:1

2m1v

2

1 =1

2m1v

2

1 +1

2m2v

2

2

4 eqns, 3 unknowns!

One final stage parameter has to be expt. measured! e.g 1.




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px: m1v1 = m1v1 cos 1 + m2v

2 cos 2

py: m1v

1 sin 1 = m2v

2 sin 2

KE:1

2m1v

2

1 =1

2m1v

2

1 +1

2m2v

2

2

4 eqns, 3 unknowns!

One final stage parameter has to be expt. measured! e.g 1.

Solving,

v1 =m1v1 cos 1

m1 + m2 v1

m2

1cos2 1

(m1 + m2)2

m1 m2m1 + m2




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Analysis:

1 Ifm

1

> m2,

1 has a max. limiting value:




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Analysis:

1 If m1 > m2, 1 has a max. limiting value: cos2

max= 1

m22

m21




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Analysis:


max= 1

m22

m21 For 1 < max two possible values of v

1




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Analysis:


max= 1

m22


1

m1 m2, max 0




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Analysis:


max= 1

m22


1

m1 m2, max 0

2

Head-on collisions: 1

= 0

= v1 = v1m1 m2m1 + m2

,




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Analysis:

1 If m1 > m2, 1 has a max. limiting value: cos2 max = 1

m22


1

m1 m2, max 0

2


= 0

= v1 = v1m1 m2m1 + m2

, v2 = v12m1

m1 + m2



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Analysis:


m22


1

m1 m2, max 0

2


= 0

= v1 = v1m1 m2m1 + m2

, v2 = v12m1

m1 + m2

Final KE of m2 =4 m1/m2

(1 + m1/m2 )2intial KE of m1.

i.e. measuring initial & final KE gives ratio of masses




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Analysis:


m22

m2

1

For 1 < max two possible values of v

1

m1 m2, max 0

2 Head-on collisions:1

= 0

= v1 = v1m1 m2m1 + m2

, v2 = v12m1

m1 + m2

Final KE of m2 =4 m1/m2

(1 + m1/m2 )2intial KE of m1.

i.e. measuring initial & final KE gives ratio of masses

Chadwicks discovery of neutron


Collision in CM reference frame


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CM frame is inertial:

Collisions Collision in CM reference frame 10/16



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RCM =m1r1 + m2r2

m1 + m2




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RCM =m1r1 + m2r2

m1 + m2

Here, VCM =m1v1

m1 + m2, constant.




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RCM =m1r1 + m2r2

m1 + m2

Here, VCM =m1v1

m1 + m2, constant.

Velocity transformations:

v1c = v1 VCM =m2

m1 + m2v1




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RCM =m1r1 + m2r2

m1 + m2

Here, VCM =m1v1

m1 + m2, constant.


v1c = v1 VCM =m2

m1 + m2v1

v2c = v2 VCM = m1

m1 + m2v1




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RCM =m1r1 + m2r2

m1 + m2

Here, VCM =m1v1

m1 + m2, constant.


v1c = v1 VCM =m2

m1 + m2v1

v2c = v2 VCM = m1

m1 + m2v1

v

1c = v

1

VCM




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RCM =m1r1 + m2r2

m1 + m2

Here, VCM =m1v1

m1 + m2, constant.


v1c = v1 VCM =m2

m1 + m2v1

v2c = v2 VCM = m1

m1 + m2v1

v

1c = v

1

VCM




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RCM =m1r1 + m2r2

m1 + m2

Here, VCM =m1v1

m1 + m2, constant.


v1c = v1 VCM =m2

m1 + m2v1

v2c = v2 VCM = m1

m1 + m2v1

v

1c =v

1 V

CM

v

2c = v

2 VCM




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This is a zero-momentum frame




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Momentum Conservation:

m1v1c = m2v2c

m1v

1c = m2v

2c

Trajectories turn through anangle .



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Momentum Conservation:

m1v1c = m2v2c

m1v

1c = m2v

2c

Trajectories turn through anangle .

Magnitudes of velocities: v2c =m1m2

v1c; v

2c =m1m2

v1c

KE Conservation: v1c

= v2c

i.e speeds of the particles are the same in CM frame for an elastic

collision.


Advantages of CM frame


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1 Momentum is automatically conserved

2

Only one angle to worry about


Relationship between scattering angles and 1


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tan 1 =




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tan 1 =

v1c sin



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tan 1 =

v1c sin

VCM + v

1c cos

=sin

cos + VCMv1c




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Since v1c = v1c =m2m1

VCM,

tan 1 =

v1c sin

VCM + v

1c cos

=sin

cos + VCMv1c




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VCM,

tan 1 =

v1c sin

VCM + v

1c cos

=sin

cos + VCMv1c

tan 1 =sin

cos + m1m2

Cases:




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VCM,

tan 1 =

v1c sin

VCM + v

1c cos

=sin

cos + VCMv1c

tan 1 =sin

cos + m1m2

Cases:

1 m1 < m2: No restriction on 1




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Since v1c = v1c = m2

m1VCM,

tan 1 =

v1c sin

VCM + v

1c cos

=sin

cos + VCMv1c

tan 1 =sin

cos + m1m2

Cases:


2 m1 = m2: 1 [0,

2]




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Since v1c = v1c = m2

m1VCM,

tan 1 =

v1c sin

VCM + v

1c cos

=sin

cos + VCMv1c

tan 1 = sincos + m1

m2

Cases:


2 m1 = m2: 1 [0,

2]

3 m1 > m2: 1 has a maximum limiting value


Advantages of CM frame (contd)


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Energy expressions:

Collisions Advantages of CM Frame 14/16



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Energy expressions:

KE in lab frame expressed in terms of VCM, v1c and v2c




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Energy expressions:


KE =1

2m1(v1c + VCM)

2 +1

2m2(v2c + VCM)

2

= 12

m1v2

1c +1

2m2v

2

2c

+ (m1v1c + m2v2c) VCM

+ 12

(m1 + m2)V2

CM




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Energy expressions:


KE =1

2m1(v1c + VCM)

2 +1

2m2(v2c + VCM)

2

= 12

m1v2

1c +1

2m2v

2

2c KE1c + KE2c

+ (m1v1c + m2v2c) VCM

+ 12

(m1 + m2)V2

CM



E i


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Energy expressions:


KE =1

2m1(v1c + VCM)

2 +1

2m2(v2c + VCM)

2

= 12

m1v2

1c +1

2m2v

2

2c KE1c + KE2c

+ (m1v1c + m2v2c) VCM 0

+ 12

(m1 + m2)V2

CM



E i


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Energy expressions:

KE in lab frame expressed in terms of VCM, v1c and v

2c

KE =1

2m1(v1c + VCM)

2 +1

2m2(v2c + VCM)

2

= 12

m1v2

1c +1

2m2v

2

2c KE1c + KE2c

+ (m1v1c + m2v2c) VCM 0

+ 12

(m1 + m2)V2

CM KECM



E i


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Energy expressions:


2c

KE =1

2m1(v1c + VCM)

2 +1

2m2(v2c + VCM)

2

= 12

m1v2

1c +1

2m2v

2

2c KE1c + KE2c

+ (m1v1c + m2v2c) VCM 0

+ 12

(m1 + m2)V2

CM KECM

KE = KEI + KECM



Energ e pressions


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Energy expressions:


2c

KE =1

2m1(v1c + VCM)

2 +1

2m2(v2c + VCM)

2

= 12

m1v2

1c +1

2m2v

2

2c KE1c + KE2c

+ (m1v1c + m2v2c) VCM 0

+ 12

(m1 + m2)V2

CM KECM

KE = KEI + KECM

Theorem

KE of a system =



Energy expressions:


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Energy expressions:


2c

KE =1

2m1(v1c + VCM)

2 +1

2m2(v2c + VCM)

2

= 12

m1v2

1c +1

2m2v

2

2c KE1c + KE2c

+ (m1v1c + m2v2c) VCM 0

+ 12

(m1 + m2)V2

CM KECM

KE = KEI + KECM

Theorem

KE of a system =internal KE (relative to CM)



Energy expressions:


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Energy expressions:


2c

KE =1

2m1(v1c + VCM)

2 +1

2m2(v2c + VCM)

2

= 12

m1v2

1c +1

2m2v

2

2c KE1c + KE2c

+ (m1v1c + m2v2c) VCM 0

+ 12

(m1 + m2)V2

CM KECM

KE = KEI + KECM

Theorem

KE of a system = internal KE (relative to CM) +KE of CM



Of the total initial KE of system only part is available for use:


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Of the total initial KE of system, only part is available for use:

KEc = KEinitial KECM

=1

2

m1m2m1 + m2

(v2 v1)2

=1

2

v2rel



Of the total initial KE of system only part is available for use:


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Of the total initial KE of system, only part is available for use:

KEc = KEinitial KECM

=1

2

m1m2m1 + m2

(v2 v1)2

=1

2

v2rel

For an inelastic collision, KE = Q we require the masses to

have a minimum available energy for the collision to be effective:

KEc|min = Q


References

LHC Homepage:


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LHC Homepage:

http://lhc-machine-outreach.web.cern.ch/lhc-machine-outreach/

Images:

http://www.boston.com/bigpicture/2008/08/the large hadron collider.


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