# Nuclear collisions

1. Nov 4, 2007

### bigtymer8700

Collisions between atomic and subatomic particles are usually perfectly elastic. In one such collision, a proton traveling to the right at 274 km/s collides elastically with a stationary alpha particle (a helium nucleus, having mass 6.65×10−27 kg . I am trying to figure out the velocity of after collision for helium nucleus

Ma=?
Mb=6.65 x10-27kg
Vai= 274000m/s

i started off using the cons. of momentum so i got

Ma(274000m/s)+6.65^-27 kg(0 m/s) = MaVaf + Mb+Vbf

i know im missing a step here to continue working the problem i just cant figure out where

Last edited: Nov 4, 2007
2. Nov 4, 2007

### Astronuc

Staff Emeritus
should be MaVaf + MbVbf, but I presume that is a typo.

Well - one has two unknowns Vaf and Vbf, so one needs two equations.

One is conservation of momentum and the other is conservation of energy.

Now, is one assuming a head on collision without any lateral deflection?

See - http://hyperphysics.phy-astr.gsu.edu/hbase/colsta.html

But look at Rutherford scattering -

http://hyperphysics.phy-astr.gsu.edu/hbase/nuclear/rutsca2.html#c4

3. Nov 4, 2007

### bigtymer8700

yes its a head on collision no lateral deflection. so how would you use the conservation of energy to solve for the unknown of the two equations?

4. Nov 4, 2007

### Astronuc

Staff Emeritus
Kinetic energy before = kinetic energy after, similar to conservation of momentum.

274 km/s = 0.000913c, so non-relativistic mechanics is reasonable.

5. Nov 4, 2007

### bigtymer8700

yes so would you then set up 1/2MaVai^2 +1/2MbVbi^2=1/2MaVaf^2+ 1/2MbVbf^2 to get the unknown Mass or velocity?

Last edited: Nov 4, 2007
6. Nov 4, 2007

### Astronuc

Staff Emeritus
Yes. One knows the mass of a proton and helium atom (nucleus = alpha particle). Assume the helium atom is stationary, which it should be, otherwise calculate its velocity assuming room temperature.

7. Nov 4, 2007

### bigtymer8700

thats where im stuck on i have both equations set up but i dont see what i have that can be substituted in for each other. i know its somewhere simple

I have 1/2Ma(274000m/s)^2+ 0 = 1/2MaVaf^2 + 1/2 6.65^-27kg Vbf^2

thats for the conservation of KE

you already see what i have for the cons. of momemtum up there i just cant see what unknown i have to isolate first to then solve for it

8. Nov 4, 2007

### Astronuc

Staff Emeritus

Besides the mass of the He nucleus (6.65 E-27 kg), one knows the mass of a proton, which is 1.67 E-27 kg. The proton is moving at 2.74 E5 m/s.

One can compute the momentum (mv) and kinetic energy (1/2mv2) of the proton. The values are 0 for the He nucleus.

Momentum eq: mpvpi= mpvpf + mavaf, where mp is the mass of the proton and ma is the mass of the alpha particle, and i,f subscripts denote initial and final quantities.

Now one could divide the momentum equation by mp and obtain an equation of one unknown velocity in terms of the other, then substitute that into the energy equation.

Energy eq: mpvpi2= mpvpf2 + mavaf2 (the 1/2 factored out)

Similarly, one could divide by mp.

The quantities on the left-hand sides of the momentum and energy equations are known.