Past exam of the physics 1 paper of the University of Oxford / 2019 resit / 8

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 $\hat{\mathbf{x}}$ and put $S=M_1+M_2$. Momentum conservation fixes the centre-of-momentum frame velocity, and a Galilean transformation gives the initial velocities in that frame:After the collision, $M_1{\mathbf{w}}_1^{\prime }+M_2{\mathbf{w}}_2^{\prime }=0$. Hence both particles have the same final-to-initial speed ratio,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 $\chi$ to the contact normal, thenFor example, $e_n=1/2$ and $\chi =\pi /4$ give $q=\sqrt{5/8}\ne 1/2$. Therefore the requested identification is not generally true with the standard normal definition. The printed equation is valid if its $e$ denotes the relative speed retention in an oblique collision, namely $e=q$, or if an additional collision model supplies that equality. The elastic cases have $q=e_n=1$ and are unaffected by this qualification.

To derive the scattering formula under that convention, write ${\mathbf{v}}_1^{\prime }=\mathbf{V}+{\mathbf{w}}_1^{\prime }$. Its components give the laboratory scattering circleSolving the resulting quadratic for the nonnegative speed yieldsWith $q=e$, this is the requested formula. A real square root and $v_1\ge 0$ are both necessary: a negative algebraic root is not a second physical speed, and $v_1=0$ does not define a scattering angle. The subsequent limits are taken using this laboratory scattering circle.

For equal masses and an elastic collision, $q=1$ andThus the nonzero outgoing branch isand the other nonnegative algebraic root is $v_1=0$. The latter is the head-on transfer of the incoming momentum and kinetic energy to particle 2: $v_2=u_1$. A stationary particle has no defined outgoing direction, so this root is not a separate physical outcome for every chosen ${\theta }_1$.

On the nonzero branch, momentum conservation givesThe two nonzero outgoing velocities are perpendicular: squaring momentum conservation and comparing with kinetic energy conservation gives ${\mathbf{v}}_1^{\prime }\cdot {\mathbf{v}}_2^{\prime }=0$. For $\cos {\theta }_1<0$, the nonzero root is negative and must be rejected; backward scattering of a moving incident particle is impossible in this equal-mass, initially stationary-target case.

For an elastic collision with $M_2>M_1$, the positive root in the laboratory scattering circle formula is the only nonnegative one. With $\epsilon =M_1/M_2\ll 1$,ConsequentlyThe heavy target can absorb a finite momentum with negligible velocity and negligible kinetic energy. Thus the projectile may change direction substantially while retaining its speed; a backward reflection also has $v_1\simeq u_1$.

A totally inelastic collision in the sticking sense has zero final relative velocity, so $q=0$. Both velocities are the conserved centre-of-momentum frame velocity:Thus the bodies move together along the original incident direction; the moving projectile has ${\theta }_1=0$. The common outgoing state is stationary in the centre-of-momentum frame. The lost translational kinetic energy iswhich can become internal energy. As explained in part 8, setting only the normal coefficient of restitution to zero in a smooth oblique collision does not force sticking: tangential relative motion can remain.

Set $q=1$ in the laboratory scattering circle formula. For $M_1>M_2$, a real outgoing speed requiresAlso $v_1\ge 0$ excludes the backward-angle interval: the square root is smaller than $|\cos {\theta }_1|$ when $M_2/M_1<1$, so both roots would be negative for $\cos {\theta }_1<0$. ThereforeAt this boundary the two possible speeds coincide, with $v_1/u_1=\sqrt{(M_1-M_2)/(M_1+M_2)}$. Geometrically the outgoing ray is tangent to the laboratory scattering circle. As $M_2/M_1\to 0$ the maximum tends to zero, while it approaches $\pi /2$ as the masses approach equality from $M_1>M_2$.