Two particles with masses 2m and 5m are moving toward each other along the x axi

Question

Two particles with masses 2m and 5m are moving toward each other along the x axis with the same initial speeds vi. Particle 2m is traveling to the left, while particle 5m is traveling to the right. They undergo an elastic, glancing collision such that particle 2m is moving in the negative y direction after the collision at a right angle from its initial direction. (a) Find the final speeds of the two particles in terms of vi. particle 2m_____x Vi particle 5m_____x Vi (b) What is the angle ? at which the particle 5m is scattered? _________

Explanation / Answer

K = 1/2 m1 v1^2 + 1/2 m2 v2^2 = 3.5m vi^2
px = m1vx1 + m2vx2 = -2mvi + 5mvi = 3mvi
py = 0
With momentum decomposed in its x and y components, and posing that the 7m particle is the one heading in the positive direction.

After the collision, the only information we're given is that the particle 2m's speed in in the direction. We must still have the same values for the sum of the two masses' kinetic energy and momentum.
K' = m*(v1')^2 + 2.5m (v2')^2 = 3.5m vi^2
px' = 0 + 5m*v2' *cos? = 3m vi
py' = -2m v1' + 5m*v2' *sin? = 0
Where ? is the scattering angle for the 5m particle

That gives use a system of three equations, with three unkowns (v1, v2, and ?).

Simplifying a little, we get
(v1')^2 + 2.5 (v2')^2 = 3.5 vi^2
5*v2' *cos? = 3 vi
2v1' = 5v2' *sin?

This gives us, going up from the third equation
v2' = 2v1' / (5*sin?)
5*[2v1' / (5*sin?)] *cos? = 3 vi
0.667v1' *cos? /sin? = vi

(v1')^2 + 2.5 [2v1' / (5*sin?)]^2 = 3.5 [0.667v1' *cos? /sin?]^2
v1'^2 + 2/5 v1'^2 / (sin?)^2 = 1.5557 v1'^2 * (cos? /sin?) ^2

Dividing everything by v1^2, then multiplying by (sin?)^2, we have
(sin?)^2 + 2/5 = 1.5557 (cos?)^2
(sin?)^2 + 2/5 = 1.5557 (cos?)^2 + (sin?)^2 + (cos?)^2 -1
(Since (sin

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