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## Homework Statement

An atomic nucleous of mass

*m*travelling with speed

*v*collides elastically with a target particle of mass

*3.0m*(initially at rest) and is scattered at 45

^{o}

(a). What are the final speeds of the two particles?

Advice: eliminate the target particle's recoil angle by manipulating the equation of momentum conservation so that can use the identity

*sin*

^{2}

*Φ + cos*

^{2}

*Φ = 1*

__Variables (that I used):__

*v*= speed of the atom before the collision (along the x-axis, i.e. parallel to the x-axis)

*u*final velocity of the atom after the collision (goes above the x-axis by 45

_{1}=^{o})

*u*= final velocity of the target particle after the collision (goes below the x-axis by

_{2}*Φ)*

## Homework Equations

[/B]

__Conservation of Momentum:__

*m*

_{1}v_{1}+ m_{2}v_{2}= m_{1}u_{1}m_{2}u_{2}## The Attempt at a Solution

[/B]

I used the conservation of momentum formula and simplified it as follows:

*m*

mv = mu

mv = m(u

v = u

_{1}v_{1}+ m_{2}v_{2}= m_{1}u_{1}+ m_{2}u_{2}mv = mu

_{1}+ 3.0mu_{2}mv = m(u

_{1}+ 3.0u_{2})v = u

_{1}+ 3.0u_{2}*u*and

_{1}*u*into component forms:

_{2}**

v = u

0 = u

__x-direction:__v = u

_{1}cos(45^{o}) + 3.0u_{2}cosΦ__y-direction:__0 = u

_{1}sin(45^{o}) - 3.0u_{2}sin*Φ*

*v*

^{2}= u_{1}^{2}+ 3.0u_{2}^{2}*sin*

^{2}

*Φ + cos*

^{2}

*Φ = 1*as the Advice recommended me to do. I asked my Prof. and he said that I had to break the final velocities into its component forms to get that trig. identity before rushing to a meeting.

__Note:__Please correct me if I'm wrong but, I think that the angle between these two final velocities

*(45*, since the conservation of kinetic energy equation:

^{o}+ Φ) = 90^{o}*v*looks like Pythagoras Theorem, where

^{2}= u_{1}^{2}+ 3.0u_{2}^{2},*v*is the hypotenuse, and all these 3 velocities form a right triangle.