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[SOLVED] Elastic collision
A proton of mass 'm' collides with a particle of unknown mass at rest. After the collision, the proton and the unknown particle are seen moving at an angle of 90 degrees with respect to each other. What is the mass of the unknown particle.
Known quantities:
mass of proton = m
velocity of proton = v
mass of unknown particle = M
Angle between M and m after collision is 90 degrees. [itex]\alpha + \beta = {\pi \over 2}[/itex]
Conservation of Momentum. Momentum must be conserved in x and ydirection:
Xdirection:
[tex]mv = mv'\cos \alpha + Mv'\cos \beta[/tex]
Ydirection:
[tex] 0 = mv'\sin \alpha + Mv'\sin \beta [/tex]
Since the angle between the particles is known i.e. 90 degrees,
[itex] \beta = {\pi \over 2}  \alpha[/itex] & [itex]\cos ({\pi \over 2}  \alpha) = \sin \alpha[/itex].
The momentum conservation equations modify as:
Xdirection:
[tex]mv = mv'\cos \alpha + Mv'\sin \alpha[/tex]
Ydirection:
[tex] 0 = mv'\sin \alpha + Mv'\cos \alpha [/tex]
I am not able to proceed beyond this, I have two unknown quantities on the RightSide viz. [itex]\alpha[/itex] and the final velocity v'. I have to obtain the unknown mass in terms of m. Am I missing out something here?
Homework Statement
A proton of mass 'm' collides with a particle of unknown mass at rest. After the collision, the proton and the unknown particle are seen moving at an angle of 90 degrees with respect to each other. What is the mass of the unknown particle.
Known quantities:
mass of proton = m
velocity of proton = v
mass of unknown particle = M
Angle between M and m after collision is 90 degrees. [itex]\alpha + \beta = {\pi \over 2}[/itex]
Homework Equations
Conservation of Momentum. Momentum must be conserved in x and ydirection:
Xdirection:
[tex]mv = mv'\cos \alpha + Mv'\cos \beta[/tex]
Ydirection:
[tex] 0 = mv'\sin \alpha + Mv'\sin \beta [/tex]
The Attempt at a Solution
Since the angle between the particles is known i.e. 90 degrees,
[itex] \beta = {\pi \over 2}  \alpha[/itex] & [itex]\cos ({\pi \over 2}  \alpha) = \sin \alpha[/itex].
The momentum conservation equations modify as:
Xdirection:
[tex]mv = mv'\cos \alpha + Mv'\sin \alpha[/tex]
Ydirection:
[tex] 0 = mv'\sin \alpha + Mv'\cos \alpha [/tex]
I am not able to proceed beyond this, I have two unknown quantities on the RightSide viz. [itex]\alpha[/itex] and the final velocity v'. I have to obtain the unknown mass in terms of m. Am I missing out something here?
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