A proton is traveling to the right at 2.5×10^7m/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?
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To determine the speeds of the proton and carbon atom after the collision, we can make use of the principle of conservation of momentum and kinetic energy.
The principle of conservation of momentum states that the total momentum of an isolated system remains constant before and after a collision. In this case, the system consists of the proton and the carbon atom.
Let's denote the mass of the proton as mp and its velocity before the collision as vp. The mass of the carbon atom will be mc, and its velocity before the collision will be vc.
Using the principle of conservation of momentum, we can write the equation:
Initial momentum = Final momentum
(mp * vp) + (mc * vc) = (mp * vp') + (mc * vc')
where vp' and vc' are the velocities of the proton and carbon atom after the collision, respectively.
Since the collision is perfectly elastic, kinetic energy is also conserved. This means that:
Initial kinetic energy = Final kinetic energy
(1/2 * mp * vp^2) + (1/2 * mc * vc^2) = (1/2 * mp * vp'^2) + (1/2 * mc * vc'^2)
Now, we can use the above two equations to solve for vp' and vc'. However, we need one more equation to solve the system of equations. We can use the fact that the mass of the carbon atom is 12 times the mass of the proton (mc = 12 * mp).
Substituting mc = 12 * mp into the equations, we get:
(mp * vp) + (12 * mp * vc) = (mp * vp') + (12 * mp * vc')
(1/2 * mp * vp^2) + (1/2 * 12 * mp * vc^2) = (1/2 * mp * vp'^2) + (1/2 * 12 * mp * vc'^2)
From the first equation, we can simplify by dividing both sides by mp:
vp + 12 * vc = vp' + 12 * vc'
From the second equation, we can simplify by dividing both sides by (1/2 * mp):
vp^2 + 12 * vc^2 = vp'^2 + 12 * vc'^2
We now have a system of two equations with two unknowns. You can solve this system of equations using algebra or a numerical method such as substitution or elimination.
Solve the system of equations to find the values of vp' and vc', which represent the speeds of the proton and carbon atom after the collision, respectively.
Ignoring relativistic effects...
mp*vp=mp*vp' + 12mpv'
vp=vp' + 12v' all from conservation of momentum.
Energy:
1/2 mp vp^2=1/2mp*vp'^2+ 1/2 12mp v'^2
vp^2=vp'^2 + 12 v'^2
solve for vp' in the first equation
vp'=vp-12v' put that in the second equation
vp^2=(vp-12v')^2+12v'^2
expand it, and solve the quadratic equation.