Past exam of the physics course of the year 1 of the natural sciences course of the University of Cambridge / 2012 / Paper 1 / Section B

Let $s$ increase down the ramp. On a plane, static friction $F$ acts uphill. Translation and rotation giveWith $I=2m{a}^2/5$, elimination yields $\ddot{s}=(5/7)g\sin \theta$. Static friction does no work at the stationary contact point.

For a sphere rolling in a symmetric groove, the distance from the centre to the instantaneous contact axis AB is $h=a\sin (\varphi /2)$, so $v=h\omega$. The kinetic energy and work-energy equation areHenceFor the stated angles this is $5g/18$. Using $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and a one-second descent,On the level floor the rolling radius becomes $a$, so an energy-preserving change of rolling radius gives $K_g=\frac{7}{10}m{v}_f^2$. ThereforeThe floor speed exceeds the groove speed because less of the same total kinetic energy is rotational. Holding the translational speed fixed at the transition would violate the stipulated energy balance with the new no-slip constraint.

For $m>0$, $E={\gamma }_{u}m{c}^2$ and $\mathbf{p}={\gamma }_{u}m\mathbf{u}$, with ${\gamma }_u=(1-u^2/c^2{)}^{-1/2}$. ThusA photon obeys $E=pc$; $E=hf$ and $p=h/\lambda$ agree because $f\lambda =c$ in vacuum.

Take the incident photon along $+x$, and write $\beta =v/c$ for the boost. In the boosted frame the initial momenta areTheir sum vanishes when $1-2\beta =0$, giving $v_{\mathrm{ZM}}=c/2$ and ${\gamma }_v=2/\sqrt{3}$. The initial photon energy there is $m{c}^2/\sqrt{3}$, and total energy is $\sqrt{3}m{c}^2$.

For right-angle Compton scattering in the centre-of-momentum frame, choose the outgoing ZM photon along $+y$. The electron has the opposite momentum $-p^{\prime }\hat{\mathbf{y}}$. Conservation of energy determines its magnitude:Thus ${\mathbf{u}}_e^{\prime }=-(c/2)\hat{\mathbf{y}}$. Inverting the velocity transformation givesConsequentlyThe electron recoils below the incident direction for our choice of photon scattering above it; the reflected configuration reverses both transverse signs. Finally,As a check, $E_{e,\mathrm{lab}}=4m{c}^2/3$, so the final energies sum to the initial $2m{c}^2$. The outgoing laboratory photon makes $60^{\circ }$ with $+x$.

For a fixed set of bodies in an inertial frame, total linear momentum is constant if the resultant external force vanishes. Summing Newton's second law cancels internal force pairs by Newton's third law. For an open subsystem, momentum carried across its boundary must also be included.

For an equal-intake-fuel mass-loading rocket, enclose the original craft and the stationary material collected during $dt$. Its initial momentum is $mu$. At the end, the craft has mass $m-\mu dt$ and the exhaust has mass $2\mu dt$. To first order,The first term is rocket thrust; the second is the momentum cost of collecting stationary matter. It vanishes in the conventional rocket equation, which therefore has no analogous fixed exhaust-speed ceiling in Newtonian mechanics. Here the acceleration vanishes at $u_{\max }=2u_0$; from rest it stays positive below that value while fuel remains. Relativistic limits are outside this model.

For constant $\mu$, mass conservation gives $m=m_i-\mu t$. Separating variables,ThereforeThe formal limiting speed would require $t=m_i/\mu$ and zero remaining craft mass. A positive rocket dry mass and finite carried fuel terminate the burn earlier. Maintaining nonzero collection rate even at rest is an idealization explicitly prescribed by the problem.