The initial total momentum is zero, so the outgoing neutron recoils with momentum magnitude $E_{\gamma }/c$. Put $W=(m_p+m_{\pi })c^2$. Conservation of relativistic energy givesFor example the rest energies $938.2721$, $139.5704$ and $939.5654\mathrm{MeV}$ for the proton, charged pion and neutron give this value.^[ref] The result is smaller than the initial-to-final rest-energy difference of about $138.3\mathrm{MeV}$, because some released energy becomes neutron recoil kinetic energy. Atomic binding of a captured pion is neglected in the stated stationary-particle model.

Choose the emitted beam direction as positive $x$. In the original laboratory frame the sensor moves with velocity $-0.2c$; let the laser recede from it with velocity $-\beta c$. The collinear relativistic Doppler shift from source to laboratory and then to sensor givesAt the detection threshold this ratio is $633/1000$. HenceThe minimum collinear speed is therefore about $0.578c$, away from the approaching sensor, while the laser is still aimed towards it. The source must recede fast enough to overcome the sensor's approach-induced blueshift. A larger recession speed gives a longer received wavelength, if the detector has no additional upper cutoff.

As a check, the required relative recession speed is $(1-0.633^2)c/(1+0.633^2)$; relativistic velocity addition gives this same value from $(\beta -0.2)c/(1-0.2\beta )$.

Centre the square on the $xy$ plane, with uniform surface mass density $m/a^2$. For an in-plane axis at angle $\alpha$ to the $x$ direction, the perpendicular distance of $(x,y)$ from the axis is $-x\sin \alpha +y\cos \alpha$. Therefore its moment of inertia isThe mixed integral vanishes by symmetry, while the $x^2$ and $y^2$ integrals are equal. ThusThis is the in-plane isotropy of the square's inertia tensor: its two central in-plane principal moments coincide. It does not assert isotropy about arbitrary three-dimensional axes, nor about parallel axes that miss the centre.

Let the motor's constant power be $P>0$ and put $\omega =\dot{\theta }$. The braking coefficient in the proposed formula must describe a retarding torque $k\omega$. This constant-power rotational acceleration obeys the energy balance:Consequently the steady angular velocity is ${\omega }_0=\sqrt{P/k}$, and the ordinary rotational frequency is ${\omega }_0/(2\pi )$. Solving the linear equation for ${\omega }^2$ from rest givesThe printed time formula is missing a minus sign. Its logarithm is negative for $0<\omega <{\omega }_0$, whereas elapsed time must be positive. The terminal speed is approached only as $t\to \infty$.

A constant-power motor starting exactly from rest is an idealization: its driving torque $P/\omega$ would diverge. The energy equation has a regular solution for ${\omega }^2$, although the initial angular acceleration is singular. If $k\omega$ were literally a tangential force, a lever arm would also be needed; the dimensions of the requested expression identify $k$ as a rotational damping coefficient.

In spherical coordinates, the Lagrangian isA central force exerts zero torque, so the angular momentum vector is constant and the orbit lies in a fixed plane. Choose that plane as $\theta =\pi /2$; then $\ell =m{r}^2\dot{\varphi }$ is conserved. Substituting this relation into the energy givesThe effective potential reduces radial motion to a one-dimensional problem. Allowed radii have $E\ge V_{\mathrm{eff}}$, radial turning points satisfy equality, and a circular orbit occurs at $V_{\mathrm{eff}}^{\prime }=0$. A strict local minimum gives stable radial oscillations. Eliminating $\dot{\varphi }$ directly inside the original Lagrangian would require a Routhian; the energy derivation avoids giving the centrifugal term the wrong sign.

The familiar right-angle result requires an elastic collision with conserved horizontal momentum and the appropriate treatment of rotational kinetic energy. In the ideal smooth-sphere model the spins do not change during the collision, so the translational kinetic energy is separately conserved. For equal masses, with initial velocity $\mathbf{u}$ and outgoing velocities ${\mathbf{v}}_1,{\mathbf{v}}_2$,Squaring the first equality and using the second givesThus the two nonzero outgoing directions are perpendicular. In the head-on limiting case one sphere stops, so an angle for its zero velocity is undefined. An ideal elastic model that enforces rolling without slipping at both ends gives the same energy equation: each solid sphere has total kinetic energy $7m{v}^2/10$, and the common factor cancels.

Elasticity is not stated explicitly in the question. It cannot be omitted from the argument: for a smooth equal-mass collision with normal incoming component $u_n$ and coefficient of restitution $e<1$, the outgoing scalar product is $(1-e^2)u_n^2/4>0$. Moreover, after a smooth impact the spheres need not still be rolling; subsequent frictional relaxation can change their directions. The right-angle statement therefore concerns the specified ideal elastic collision, not arbitrary later rolling trajectories.

Use the test-particle approximation, neglecting the planet's recoil. The conserved specific angular momentum is $j=bv$ and the specific energy at infinity is $v^2/2$. At closest approach the radial velocity vanishes, so the effective potential givesThe positive root and the tangential speed areThe second form for $r_0$ avoids subtracting two nearly equal numbers for a small impact parameter. These formulas describe a nonradial orbit; $b=0$ gives a collision rather than a finite nonzero pericentre in the point-planet model.

Take upward velocity as positive. The linear drag force is $-mkv$, so it opposes the motion both during ascent and descent. Newton's second law gives $\dot{v}+kv=-g$. With $v(0)=u$, the integrating factor $e^{kt}$ yieldsThe terminal velocity is $-g/k$. The limit $k\to 0$ at fixed $t$ recovers $u-gt$; expanding the exponential is necessary to see the cancellation of the apparently singular terms.

Assume the cylinders are released from rest. In each case, “there is friction” is interpreted as sufficient static friction for rolling without slipping at the rough contacts. Without a coefficient or a no-slip assumption, sliding cases would not have a unique specified acceleration. On release the two lower cylinders separate; their mutual contact must not be retained as a rigid constraint.

Let the top cylinder's initial downward acceleration be $w$, and each lower cylinder's outward horizontal acceleration be $u$. The top cylinder initially has no horizontal acceleration. Differentiating the fixed upper-to-lower centre separation twice at the initial zero velocity givesBy reflection symmetry the top cylinder has zero angular acceleration, while the lower cylinders rotate in opposite senses.