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akatz
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[SOLVED] Completely Elastic Proton Collision
In my physics class we tried to solve this problem but nobody in the class could figure it out and the teacher was having difficulty with it. It is possible that the book put in a problem that isn't possible to solve...
i am not very good with latex coding so feel free to ask about my sub/superscript system
A proton (atomic mass 1.01 u) with a speed of 518 m/s collides elastically with another proton at rest. The original proton is scattered 64.0 degrees from its initial direction.
(A) What is the direction of the velocity of the target proton after the collision?
(B) What are the speeds of the two protons after the collision?
A proton (atomic mass 1.01 u) with a speed of 518 m/s collides elastically with another proton at rest. The original proton is scattered 64.0 degrees from its initial direction.
Variables
m_{1}=1.01 u
v_{1}=518 m/s
m_{2}=1.01 u
v_{2}=0 m/s
v^{'}_{1}=?
v^{'}_{2}=?
\vartheta=64 degrees (b/n path of first proton before and after collision)
\varphi=? (b/n path of second proton after collision and path of first proton before collision)
Total momentum before equals total momentum after
P_{T}=P^{'}_{T}
momentum = mass X velocity
P=mv
we set up a force diagram so that the path of proton #1 is a horizontal line
vvvv total momentum in X direction vvvv
P_{T,X}= 518 u\timesm/s=V^{'}_{1}cos(64)+V^{'}_{2}cos(\varphi)
we have three variables in this equation and this makes a problem to solve for just one variable
Total momentum in Y direction is zero prior to collision and therefore is zero after collision
P_{T,Y}=0 u\timesm/s=V^{'}_{1}sin(64)+V^{'}_{2}sin(\varphi)
this is the point where I have absolutely no idea what to do and apparently my teacher and the rest of the class as well...
Any help/hint(s) would be very greatly appreciated
Thanks,
Akatz
In my physics class we tried to solve this problem but nobody in the class could figure it out and the teacher was having difficulty with it. It is possible that the book put in a problem that isn't possible to solve...
i am not very good with latex coding so feel free to ask about my sub/superscript system
A proton (atomic mass 1.01 u) with a speed of 518 m/s collides elastically with another proton at rest. The original proton is scattered 64.0 degrees from its initial direction.
(A) What is the direction of the velocity of the target proton after the collision?
(B) What are the speeds of the two protons after the collision?
Homework Statement
A proton (atomic mass 1.01 u) with a speed of 518 m/s collides elastically with another proton at rest. The original proton is scattered 64.0 degrees from its initial direction.
Variables
m_{1}=1.01 u
v_{1}=518 m/s
m_{2}=1.01 u
v_{2}=0 m/s
v^{'}_{1}=?
v^{'}_{2}=?
\vartheta=64 degrees (b/n path of first proton before and after collision)
\varphi=? (b/n path of second proton after collision and path of first proton before collision)
Homework Equations
Total momentum before equals total momentum after
P_{T}=P^{'}_{T}
momentum = mass X velocity
P=mv
The Attempt at a Solution
we set up a force diagram so that the path of proton #1 is a horizontal line
vvvv total momentum in X direction vvvv
P_{T,X}= 518 u\timesm/s=V^{'}_{1}cos(64)+V^{'}_{2}cos(\varphi)
we have three variables in this equation and this makes a problem to solve for just one variable
Total momentum in Y direction is zero prior to collision and therefore is zero after collision
P_{T,Y}=0 u\timesm/s=V^{'}_{1}sin(64)+V^{'}_{2}sin(\varphi)
this is the point where I have absolutely no idea what to do and apparently my teacher and the rest of the class as well...
Any help/hint(s) would be very greatly appreciated
Thanks,
Akatz
