What are the Final Velocities After a Proton Collides with a Carbon Atom?

[Xerxes1986]

A proton is traveling to the right at 2.0E7 m/s. It has a head-on perfectly elastic collision with a carbon atom. The mass of the carbon atom is 12 times the mass of the proton. What are the speeds of each after the collision?

i was trying to tinker with the fact that the KE before and after the collision will be equal and so will be the momentum. but when i solve the equation i get 5773502 m/s but it doesn't work...i don't know how to even start this problem...

 

[dextercioby]

There's a trick here.v/c~1/15,sot the errors in chosing Newton's dynamics to Einstein's one is less than 1%.If you're not interested in accuracy,you might chose the simpler framework offered by nonrelativistic physics.

What's the intial velocity of the C atom...?

Daniel.

 

[Xerxes1986]

There's a trick here.v/c~1/15,sot the errors in chosing Newton's dynamics to Einstein's one is less than 1%.If you're not interested in accuracy,you might chose the simpler framework offered by nonrelativistic physics.

What's the intial velocity of the C atom...?

Daniel.

ok that just flew right over my head...einstein? umm how bout a simple solution? lol this is supposed to be simple...

the C atom isn't moving when the proton hits it

 

[dextercioby]

Perfect.Then apply the laws of conservation of KE and momentum...Write the latter in vector form and choose the axis of movement as the axis of projection with the positive sense being the sense of the initial proton's velocity.

Daniel.

 

[Xerxes1986]

Perfect.Then apply the laws of conservation of KE and momentum...Write the latter in vector form and choose the axis of movement as the axis of projection with the positive sense being the sense of the initial proton's velocity.

Daniel.

i tried but i getg like some really REALLY weird equation that doesn't work...what equations should i try to manipulate around

 

[dextercioby]

Post the 2 equations.

Daniel.

 

[Xerxes1986]

Post the 2 equations.

Daniel.

lol ill try

.5*m_p*2E7^2=.5*m_p*v_fp^2 + .5*12m_p*v_fc^2

and

m_p*2E7=m_p*v_fp + 12m_p*v_fc

and then substitute right? well i got a equation with v_fc^4 in it and it has 4 roots...none of which are the right anser

 

[dextercioby]

The system:

$$12(v_{C}^{fin})^{2}+(v_{p}^{fin})^{2}=(v_{p}^{init})^{2}$$

$$12v_{C}^{fin}+v_{p}^{fin}=v_{p}^{init}$$


has the physically acceptable sollutions

$$v_{C}^{fin}\approx 3\cdot 10^6 \ m \ s^{-1}$$

$$v_{p}^{fin}\approx -1.7\cdot 10^{7} \ m \ s^{-1}$$

Can u convince yourself of the validity of these answers?

Daniel.

 

[Xerxes1986]

The system:

$$12(v_{C}^{fin})^{2}+(v_{p}^{fin})^{2}=(v_{p}^{init})^{2}$$

$$12v_{C}^{fin}+v_{p}^{fin}=v_{p}^{init}$$


has the physically acceptable sollutions

$$v_{C}^{fin}\approx 3\cdot 10^6 \ m \ s^{-1}$$

$$v_{p}^{fin}\approx -1.7\cdot 10^{7} \ m \ s^{-1}$$

Can u convince yourself of the validity of these answers?

Daniel.

i don't know howyou got them but the v_c one isn't right because when i enter it it says "your close but your answer is off...rounding error or something"

 

[dextercioby]

Rounding error...?That's bull****.The next figure after 3 is 0...Anyway,to get the full #,solve the system.It's basically a quadratic that u must solve,the only problem being that it has quite large coefficients.

Daniel.

 

[Doc Al]

lol ill try

.5*m_p*2E7^2=.5*m_p*v_fp^2 + .5*12m_p*v_fc^2

and

m_p*2E7=m_p*v_fp + 12m_p*v_fc

So far, so good.

and then substitute right? well i got a equation with v_fc^4 in it and it has 4 roots...none of which are the right anser

You must be making an error here. If you eliminate v_fp, then you'll get a quadratic equation with v_fc. Do it over.