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 Incoming links: Relative speed retention in an oblique collision

When a particle of mass

M_{1}

and initial speed

u_{1}

strikes a stationary mass

M_{2}

V = M_{1} u_{1} x / (M_{1} + M_{2})

q

, its possible outgoing velocity vectors lie on the displayed circle in a scattering plane. Intersect a ray of fixed laboratory direction with that circle to obtain the quadratic formula for outgoing speed. A negative root is inadmissible, and the origin has no defined scattering angle. Tangency gives the maximum laboratory angle when the circle excludes the origin.

In the laboratory frame, the initial target is at rest and the outgoing directions are directly related to detector positions. The centre-of-momentum frame has zero total momentum, so the two outgoing momenta are equal and opposite; it cleanly separates bulk motion from relative kinetic energy. Transforming back is needed to predict laboratory angles and speeds. Neither reference frame removes the need for an appropriate collision law.

Take the incident direction as

x

and put

S = M_{1} + M_{2}

V = \frac{M _{1} u _{1}}{S} x, w_{1} = \frac{M _{2} u _{1}}{S} x, w_{2} = - \frac{M _{1} u _{1}}{S} x .

After the collision,

M_{1} w_{1}^{'} + M_{2} w_{2}^{'} = 0

. Hence both particles have the same final-to-initial speed ratio,

q = \frac{∣ w _{1}^{'} ∣}{∣ w _{1} ∣} = \frac{∣ w _{2}^{'} ∣}{∣ w _{2} ∣} = \frac{∣ v _{1}^{'} - v _{2}^{'} ∣}{u _{1}} .

The relative kinetic energy is multiplied by

q^{2}

. This common ratio exists independently of how the outgoing direction is selected.

The meaning of restitution matters here. The usual coefficient of restitution

e_{n}

is a ratio of normal separation and approach speeds, not total speeds. In a smooth oblique collision, the tangential relative velocity is unchanged. If the incoming relative velocity makes angle