Past exam of the physics 1 paper of the University of Oxford / 2017 resit / 9

For two particles, let $M=m_1+m_2$ and ${\mathbf{V}}_{\mathrm{cm}}=(m_1{\mathbf{v}}_1+m_2{\mathbf{v}}_2)/M$. The centre-of-mass frame velocities are ${\mathbf{v}}_i^{\ast }={\mathbf{v}}_i-{\mathbf{V}}_{\mathrm{cm}}$, so $m_1{\mathbf{v}}_1^{\ast }+m_2{\mathbf{v}}_2^{\ast }=0$. Thus before and after an isolated collision the two momentum vectors are equal and opposite. For an elastic collision their common magnitude is also unchanged: only their direction rotates.

The kinetic energy decomposition about the center of mass separates unchanged bulk translation from relative motion:The reduced mass $\mu$ reduces the calculation to one relative velocity. After solving the geometry in the centre-of-mass frame, add ${\mathbf{V}}_{\mathrm{cm}}$ to each velocity to return to the laboratory. The four panels below illustrate this for the subsequent example.

For equal masses $m$, the reduced mass is $m/2$. The kinetic energy of their centre of mass cannot be removed by an isolated collision, since total momentum is fixed. Therefore the maximum energy available for conversion to internal forms is the initial relative kinetic energy,It is attained by a perfectly inelastic collision in which the final relative velocity is zero. This concerns the convertible initial kinetic energy, not an arbitrary pre-existing internal energy reservoir.

For masses $m$ and $2m$, the constant centre-of-mass velocity is $(V/3){\mathbf{e}}_x$. Initially the laboratory velocities are $V{\mathbf{e}}_x$ and zero, while the centre-of-mass frame velocities are $(2V/3){\mathbf{e}}_x$ and $-(V/3){\mathbf{e}}_x$. After the elastic collision they areAdding $(V/3,0)$ produces the lower-left panel's laboratory velocity vectors. The given $30^{\circ }$ laboratory angle impliesThe centre-of-mass frame scattering angle is therefore different from the laboratory angle. Reflection across the $x$-axis gives the equally possible negative-angle configuration.

Let $u>0$ be the projectile's final speed, taking its deflection above the $x$-axis. Momentum conservation givesSubstituting these into kinetic energy conservation, $V^2=u^2+2|{\mathbf{v}}_2{|}^2$, gives $3u^2-2Vu\cos 30^{\circ }-V^2=0$. Its positive root is $u=V(\sqrt{3}+\sqrt{15})/6$. HenceThese vectors satisfy ${\mathbf{v}}_1+2{\mathbf{v}}_2=(V,0)$ and $|{\mathbf{v}}_1{|}^2+2|{\mathbf{v}}_2{|}^2=V^2$. For the reflected deflection, reverse both $y$ components.