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Protium_H1
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Homework Statement
An atomic nucleous of mass m traveling with speed v collides elastically with a target particle of mass 3.0m (initially at rest) and is scattered at 45o
(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 sin2Φ + cos2Φ = 1
Variables (that I used):
v = speed of the atom before the collision (along the x-axis, i.e. parallel to the x-axis)
u1 = final velocity of the atom after the collision (goes above the x-axis by 45o)
u2 = final velocity of the target particle after the collision (goes below the x-axis by Φ)
Homework Equations
[/B]
Conservation of Momentum:
m1v1 + m2v2 = m1u1 m2u2
The Attempt at a Solution
[/B]
I used the conservation of momentum formula and simplified it as follows:
m1v1 + m2v2 = m1u1 + m2u2
mv = mu1 + 3.0mu2
mv = m(u1 + 3.0u2)
v = u1 + 3.0u2
I broke u1 and u2 into component forms:mv = mu1 + 3.0mu2
mv = m(u1 + 3.0u2)
v = u1 + 3.0u2
x-direction:
v = u1cos(45o) + 3.0u2cosΦ
y-direction:
0 = u1sin(45o) - 3.0u2sinΦ
This is where I get stuck, I even tried to use the conservation of kinetic energy, v = u1cos(45o) + 3.0u2cosΦ
y-direction:
0 = u1sin(45o) - 3.0u2sinΦ
v2 = u12 + 3.0u22
To find for the unknowns, but for some reason, the equations just all cancel out or become impossible to solve when I tried to substitute and find the final velocities, and for some reason, I can't get the expression sin2Φ + cos2Φ = 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 (45o + Φ) = 90o, since the conservation of kinetic energy equation: v2 = u12 + 3.0u22, looks like Pythagoras Theorem, where v is the hypotenuse, and all these 3 velocities form a right triangle.