Past exam of the physics 1 paper of the University of Oxford / 2021 / 9

Take $+y$ upwards in the source diagram. Write $w$ for $B$'s incoming speed, so the initial velocities areA contact-direction assumption is needed for a unique answer. The diagram specifies the incoming velocities but does not specify the contact normal or an outgoing angle. First adopt the usual intended model: the impulse is horizontal, so neither particle changes its $y$ component of velocity. This is the smooth oblique collision model with a horizontal line of centres at impact. Bringing $A$ to rest in the $x$ direction then brings it entirely to rest.

Let $b$ be $B$'s outgoing $x$ component of velocity. Conservation of momentum and conservation of kinetic energy givePut $q=(\sqrt{3}/2)w$. The second equation becomes $3u^2+q^2=(q-3u{)}^2$, so $q=u$. ThereforeThe outgoing kinetic energies areThe outgoing speed of $B$ is $u\sqrt{13/3}$, directed above the negative $x$ axis at angle ${\tan }^{-1}[1/(2\sqrt{3})]\simeq 16.1^{\circ }$. The initial kinetic energy is also $(3/2+2/3)m{u}^2=13m{u}^2/6$, as required for an elastic collision.

For the perfectly inelastic collision with these same incoming velocities, both particles share a common outgoing velocity. Conservation of momentum givesIts direction is again $16.1^{\circ }$ above the negative $x$ axis. The separate translational kinetic energies of the constituents and their total areThe loss of translational kinetic energy isThis becomes internal energy, for example deformation and heating; total energy is conserved. In the elastic collision no kinetic energy is lost and the particles separate with different velocities. In the perfectly inelastic collision relative motion disappears and the common velocity is the original centre of mass velocity.

The limitation of the printed data can be made explicit. Without the horizontal-impulse assumption, write the final velocity of $A$ as $(0,h)$. Then conservation of momentum determinesand conservation of kinetic energy gives onlyThere is a family of possibilities. For a concrete second solution, $h=u/\sqrt{3}$ givesThese velocities conserve both components of momentum and conserve kinetic energy, and they satisfy the specified vanishing $x$ component for $A$, while requiring a different incoming speed. They can arise from a smooth contact normal proportional to $(1,1/\sqrt{3})$. Thus the horizontal-impulse result is conditional, rather than uniquely implied by the printed statement. For any member of this family the corresponding sticking result remains well determined once $w$ is chosen:The last expression is the lost relative kinetic energy in a perfectly inelastic collision of these two point masses.