Energy and momentum: Collisions and conservation laws

Rest Energy is real: Nuclear fissionInitial mass of a neutron and a 235U nucleus.

Final products have less mass, but much more kinetic energy. Conversion of mass to kinetic energy.

Oh yes, and more neutrons, so the reaction can run wild (chain reaction!).

Protons: U=92, Ba = 56, Kr=36

Suppose that isolated object A has rest mass = 9m0 and speed vA=0.8c (A=5/3).

Object B has mass 12m0 and speed vB=−0.6c (B=5/4).

The objects collide and stick together (completely inelastic collision)

From Physics 1110 we know all collisions conserve momentum

A relativistic inelastic collision

How about energy conservation?

A)The above collision will conserve total kinetic energy

B)The above collision will conserve total rest energy Erest=mAc2 + mBc2

C)The above collision will conserve total energy Etotal = AmAc2 + BmBc2

D)It is an inelastic collision so heat will be generated. None of the above will hold.

Classically, what is the total initial momentum?

02.72.76.0128.09 0000classical

initial cmcmcmcmvmvmP BBAA

Let’s start with momentum conservation.

What is the total relativistic momentum?

cmcmcmcmcmvmvmP BBBAAAi 000045

035 39122.72.7

So it does not end at rest as predicted classically!

Suppose that object A has rest mass = 9m0 and speed vA=0.8c (A=5/3).

Object B has mass 12m0 and speed vB=−0.6c (B=5/4).

A relativistic inelastic collision

Momentum conservation gives us: fff umcm 03fi PP

Remember that mf may not be mA+mB as it would be classically.

Now let’s look at the total energy. The initial energy is

20

20

20

204

5203

522 301515129 cmcmcmcmcmcmcmE BBAAi

So conservation of energy gives us: 22030 cmcm fffi EE

Dividing these two equations: 22

0

0

303

cmum

cmcm

ff

fff

orcu f

101 so cu f 1.0

00 85.29/30 mmm ff

Furthermore: 005.1)1.01( 2/12 f

conservation of energy equation for mf:

so we can solve the

Object A has rest mass = 9m0 and speed vA=0.8c (A=5/3).

Object B has mass 12m0 and speed vB=−0.6c (B=5/4).

A relativistic inelastic collision

Classically, total momentum is 0 but in reality it is cm03

Classically, but in reality, so 8.85m0 of mass is gained!

021mmmm BAf 085.29 mm f

So the change in KE is 20

20

20 85.8915.0 cmcmcmKEKE if

20

204

1203

222 9129)1()1( cmcmcmcmcmKE BBAAi 2

02

02 15.085.29)1005.1()1( cmcmcmKE fff

The initial and final kinetic energies are:

The “lost” kinetic energy appears as gained mass in the total energy

A relativistic inelastic collision

Really, mass gets created!CERN in Geneva, Switzerland

Before the LHC (Large Hadron Collider) CERN operated LEP, the Large Electron-Positron collider in the same underground tunnel.

Electron and positrons have a mass of 9x10-31 kg. They were accelerated to very high energies so when they annihilate, they create a Z0 particle with a mass of 1.6x10-25 kg.

8

A proton has a mass of 938 MeV/c2. What is this in kg?

An important unit of energy is the electron-volt (eV). It’s the energy obtained by an electron moving through 1 V.

It is not an SI unit but is very common.

ΔE = qΔV = 1 eV = 1.6•10-19 C • 1 V = 1.6•10-19 J

kg 1067.1m/s 103.00

J/eV 1060.1eV 10938/MeV 938 2728

1962

c

Also use eV/c or MeV/c units for momentum

Since mc2 is a unit of energy, dividing energy by c2 gives a unit of mass. Also, dividing energy by c gives a unit of momentum.

At what speed is the total energy of a particle equal to twice its rest mass energy?

A. 0

B. 0.7c

C. 0.87c

D. 0.94c

E. c

To have total energy equal to twice the rest mass energy, need =2

2rest mcE 2mcE u 2)1( mcKE u

2/12)1( Solve for .

22

11 2

2 11 22 11

87.075.025.0121111 22

so you need to be moving pretty fast to get your kinetic energy close to your rest mass energy!