Your Answer : (Not Answered)

Correct Answer : D

The transformation is canonical for \(\alpha = 2, \beta = 2\).

A transformation is canonical when the Poisson bracket \(\{Q, P\}_{q,p} = 1\). With \(Q = q^{1/\alpha}\cos\beta p\) and \(P = q^{1/\alpha}\sin\beta p\):

\[ \begin{aligned} \{Q,P\} &= \frac{\partial Q}{\partial q}\frac{\partial P}{\partial p} - \frac{\partial Q}{\partial p}\frac{\partial P}{\partial q} \\ &= \frac{\beta}{\alpha}q^{\frac{2}{\alpha} - 1}(\cos^{2}\beta p \\ &\qquad + \sin^{2}\beta p) \\ &= \frac{\beta}{\alpha}q^{\frac{2}{\alpha} - 1} \end{aligned} \]

For this to equal 1 for every \(q\), the power of \(q\) must vanish, so \(2/\alpha - 1 = 0\) and \(\alpha = 2\); then \(\beta/\alpha = 1\) gives \(\beta = 2\).

\(\alpha = \beta = 1\) leaves a factor \(q\); \(\alpha = 1/2\) leaves \(q^{3}\); and \(\alpha = 2, \beta = 1/2\) gives the bracket 1/4.

Your Answer : (Not Answered)

Correct Answer : A

The correct expression is the one with \(+2\bar{\omega}\times\frac{d\bar{A}}{dt}\big|_{S'}\) and \(+\bar{\omega}\times\bar{\omega}\times\bar{A}\).

For any vector, the rates of change in the two frames are related by \(\frac{d}{dt}\big|_{S} = \frac{d}{dt}\big|_{S'} + \bar{\omega}\times\). Applying this operator twice:

\[ \begin{aligned} \ddot{\bar{A}}_{S} &= \Big(\tfrac{d}{dt}\Big|_{S'} + \bar{\omega}\times\Big)(\dot{\bar{A}}' + \bar{\omega}\times\bar{A}) \\ &= \ddot{\bar{A}}' + \dot{\bar{\omega}}\times\bar{A} + 2\bar{\omega}\times\dot{\bar{A}}' \\ &\quad + \bar{\omega}\times(\bar{\omega}\times\bar{A}) \end{aligned} \]

where primes mean rates seen in \(S'\). The cross term appears twice, once from each application, which gives the factor 2 of the Coriolis term; the last term is the centripetal term. Both carry a plus sign here. The minus signs in the other options are what appear when the equation is solved for the acceleration in \(S'\), where the Coriolis and centrifugal terms move to the other side.

Your Answer : (Not Answered)

Correct Answer : B

The plane rotates 15° per hour at the pole.

The precession rate of a Foucault pendulum is \(\Omega\sin\phi\), where \(\Omega = 360^{\circ}\) per sidereal day, about 15 degrees per hour, and \(\phi\) is the latitude. At a pole, \(\sin 90^{\circ} = 1\), so the plane turns a full circle in a day, 15 degrees per hour.

At the equator \(\sin 0 = 0\), so there is no rotation. At latitude 60 degrees the rate is \(15 \times \sin 60^{\circ} \approx 13\) degrees per hour, not 7.5 (which would be \(15\sin 30^{\circ}\)) or 30.

Your Answer : (Not Answered)

Correct Answer : C

The Laplace-Runge-Lenz vector stays fixed in direction, so the answer is the figure in which every arrow points the same way: (C).

For a \(-k/r\) potential the vector \(\vec{A}\) is a constant of the motion. It lies along the major axis of the ellipse, pointing from the focus towards the perihelion, and neither its length nor its direction changes as the particle goes round the orbit. That is also why the orbit closes and does not precess.

In (C) all four arrows are parallel to the major axis and point the same way. In (A) the arrows turn through left, up, right and down as the particle moves around the ellipse, as a tangent or a velocity would. In (B) and (D) the arrows also change direction from point to point. A vector that changes direction around the orbit cannot be conserved.

Your Answer : (Not Answered)

Correct Answer : C

The moment of inertia about the x-axis is \(I_{xx} = \sum_{i} m_{i}(y_{i}^{2} + z_{i}^{2})\).

The moment of inertia about an axis uses the squared distance of each particle from that axis. The distance of \((x_{i}, y_{i}, z_{i})\) from the x-axis is \(\sqrt{y_{i}^{2} + z_{i}^{2}}\), which gives option C.

\(\sum m_{i}(x_{i}^{2} + y_{i}^{2})\) is \(I_{zz}\) and \(\sum m_{i}(x_{i}^{2} + z_{i}^{2})\) is \(I_{yy}\). A negative sum like \(-\sum m_{i} x_{i}^{2}\) resembles a product of inertia such as \(I_{xy} = -\sum m_{i} x_{i}y_{i}\), which carries the minus sign in this convention, but a moment of inertia is never negative.

Your Answer : (Not Answered)

Correct Answer : C

On the key, a swimmer has 6 degrees of freedom.

The standard textbook treatment models the swimmer as a rigid body free to move in three-dimensional space (the water allows motion up and down as well as across the pool). A rigid body needs three coordinates to locate a reference point, its centre of mass, and three angles to fix its orientation, giving 3 + 3 = 6 degrees of freedom.

Three would describe the swimmer only as a point particle, ignoring orientation. Five is the count for a rigid linear body such as a diatomic molecule (three translations and two rotations), and nine has no basis for a single rigid body.

Your Answer : (Not Answered)

Correct Answer : B

With \(z\) measured downward along the line of fall, the Lagrangian is \(L = \frac{1}{2}m\dot{z}^{2} + mgz\).

The motion is one-dimensional, so a single coordinate \(z\) suffices, which rules out the two-coordinate forms in \(x\) and \(y\). For a falling particle the usual textbook choice measures \(z\) downward from the point of release. Gravity then does positive work as \(z\) increases, the potential energy is \(V = -mgz\), and

\[ \begin{aligned} L &= T - V = \frac{1}{2}m\dot{z}^{2} + mgz \\ m\ddot{z} &= mg \end{aligned} \]

which is free fall with acceleration \(g\) in the direction of increasing \(z\), as it should be.

The form \(\frac{1}{2}m\dot{z}^{2} - mgz\) is the same physics with \(z\) measured upward; the paper's key takes the downward convention, which is the natural one for a body described as falling.

Your Answer : (Not Answered)

Correct Answer : B

The correct statement is H is conserved.

The Lagrangian does not depend explicitly on time, so the Hamiltonian (the Jacobi integral) is a constant of the motion: \(dH/dt = -\partial L/\partial t = 0\).

H is not the total energy here, because the potential depends on the velocity. From the previous computation,

\[ \begin{aligned} H &= \frac{1}{2}m\dot{r}^{2} - \frac{\dot{r}^{2}}{rc^{2}} + \frac{1}{r} \\ E &= T + U \\ &= \frac{1}{2}m\dot{r}^{2} + \frac{\dot{r}^{2}}{rc^{2}} + \frac{1}{r} \end{aligned} \]

They differ by \(2\dot{r}^{2}/(rc^{2})\), which changes along the motion. So 'H represents the total energy' is false, E is not conserved, and 'both H and E are conserved' fails with it. \(H = T + U\) needs a potential independent of velocity and a time-independent constraint.

Your Answer : (Not Answered)

Correct Answer : B

The speed satisfies v² = u² − 2gr(1 − cos θ).

The tension in the string is perpendicular to the motion and does no work, so energy is conserved. After turning through angle \(\theta\) from the lowest point, the particle has risen

\[ h = r - r\cos\theta = r(1 - \cos\theta) \]

so \(\frac{1}{2}mv^{2} = \frac{1}{2}mu^{2} - mgr(1 - \cos\theta)\), which gives \(v^{2} = u^{2} - 2gr(1 - \cos\theta)\).

The options with a plus sign would make the particle speed up as it rises, and \(1 - \sin\theta\) does not give the height. \(v^{2} = (r\dot{\theta})^{2}\) is true for circular motion but does not answer the question in terms of u and \(\theta\).

Your Answer : (Not Answered)

Correct Answer : D

The equations of motion are \(m\ddot{x} + k\dot{x} = 0\) and \(m\ddot{y} + k\dot{y} + mg = 0\).

Take x horizontal and y vertically upward. The resistance is \(-k\vec{v}\), opposing the velocity, and gravity is \(-mg\) in the y direction. Newton's second law componentwise gives

\[ \begin{aligned} m\ddot{x} &= -k\dot{x} \\ m\ddot{y} &= -k\dot{y} - mg \end{aligned} \]

which is option D. \(\ddot{x} = 0,\ \ddot{y} + g = 0\) ignores the resistance. The option with \(-k\dot{x}\) moved to the other side as \(m\ddot{x} - k\dot{x} = 0\) makes the resistance accelerate the particle. The option putting mg in the horizontal equation applies gravity in the wrong direction.

Your Answer : (Not Answered)

Correct Answer : A

The rest mass is 12 GeV/c\(^{2}\).

The relativistic energy-momentum relation gives

\[ \begin{aligned} (mc^{2})^{2} &= E^{2} - (pc)^{2} \\ &= 13^{2} - 5^{2} = 144\ \text{GeV}^{2} \\ mc^{2} &= 12\ \text{GeV} \end{aligned} \]

The numbers form the 5-12-13 right triangle. The MeV options are a thousand times too small, and 1.2 GeV/c\(^{2}\) misplaces the decimal point; the rest energy must be of the same order as the 13 GeV total energy because the momentum is comparatively small.

Your Answer : (Not Answered)

Correct Answer : A

The rocket's velocity is 0.75 c.

Velocities must be combined with the relativistic addition formula, not simply added:

\[ \begin{aligned} u &= \frac{v + u'}{1 + vu'/c^{2}} = \frac{0.4c + 0.5c}{1 + 0.2} \\ &= \frac{0.9c}{1.2} = 0.75c \end{aligned} \]

The option 0.90 c is the Galilean sum, which ignores the denominator. The other options do not follow from either rule.

Your Answer : (Not Answered)

Correct Answer : B

The period is \(2\pi\sqrt{R/g}\).

For a satellite close to the earth (\(H \ll R\)), gravity supplies the centripetal force, and \(GM = gR^{2}\):

\[ \begin{aligned} \frac{mv^{2}}{R} &= mg \Rightarrow v = \sqrt{gR} \\ T &= \frac{2\pi R}{v} = 2\pi\sqrt{\frac{R}{g}} \end{aligned} \]

In such an orbit the man and the satellite fall freely together, so he is weightless. This is about 84 minutes. The option \(2\pi\sqrt{g/R}\) has the ratio inverted and is not even a time, and the \(1/2\pi\) options are frequencies rather than periods.

Your Answer : (Not Answered)

Correct Answer : C

The Lagrangian is \(L = \dfrac{\dot{q}_{1}^{2}}{4} + \dfrac{\dot{q}_{2}^{2}}{4} + kq_{1}^{2}\).

Hamilton's equation \(\dot{q}_{i} = \partial H / \partial p_{i} = 2p_{i}\) gives \(p_{i} = \dot{q}_{i}/2\). Then the Legendre transform \(L = \sum p_{i}\dot{q}_{i} - H\) gives

\[ \begin{aligned} L &= \frac{\dot{q}_{1}^{2} + \dot{q}_{2}^{2}}{2} - \frac{\dot{q}_{1}^{2} + \dot{q}_{2}^{2}}{4} + kq_{1}^{2} \\ &= \frac{\dot{q}_{1}^{2}}{4} + \frac{\dot{q}_{2}^{2}}{4} + kq_{1}^{2} \end{aligned} \]

The sign of \(kq_{1}^{2}\) flips because it enters H as part of the potential (\(H = T + V\), \(L = T - V\) with \(V = -kq_{1}^{2}\)). Options with \(-kq_{1}^{2}\) miss that, and those with denominators 2 or 1 use the wrong relation between \(p\) and \(\dot{q}\).

Your Answer : (Not Answered)

Correct Answer : A

As \(t \to \infty\), \(q \to \infty\) and \(p \to 0\).

Hamilton's equations for \(H = pq\):

\[ \begin{aligned} \dot{q} &= \frac{\partial H}{\partial p} = q \\ \dot{p} &= -\frac{\partial H}{\partial q} = -p \end{aligned} \]

So \(q = q_{0}e^{t}\) and \(p = p_{0}e^{-t}\). For a non-zero starting \(q_{0}\) (positive, as the option assumes), \(q\) grows without bound, while \(p\) decays to 0. The product \(pq = H\) stays constant, as energy conservation requires.

Both tending to 0 or both to \(\infty\) would change \(pq\), and the last option reverses the two equations.

Your Answer : (Not Answered)

Correct Answer : B

The correct statement is that \(n\) particles with \(k\) holonomic constraints have \(3n - k\) degrees of freedom.

Each free particle in space needs three coordinates, so \(n\) particles need \(3n\). A holonomic constraint is an equation \(f(\mathbf{r}_{1}, \ldots, \mathbf{r}_{n}, t) = 0\) among the coordinates, and each independent one lets us eliminate one coordinate. So \(3n - k\) independent generalised coordinates remain.

A finite system of particles cannot have infinitely many degrees of freedom (that needs a continuum). \(n - k\) forgets the three dimensions of space, and 6 is the number for a single rigid body, not for \(n\) particles in general.

Your Answer : (Not Answered)

Correct Answer : A

The equation is \(m\ddot{y} + k\dot{y} + mg = 0\), with y measured upward.

With a dissipation function G, Lagrange's equation becomes

\[ \begin{aligned} &\frac{d}{dt}\frac{\partial L}{\partial \dot{y}} \\ &\quad - \frac{\partial L}{\partial y} + \frac{\partial G}{\partial \dot{y}} = 0 \end{aligned} \]

Taking \(L = \frac{1}{2}m\dot{y}^{2} - mgy\):

\[ \begin{aligned} \frac{d}{dt}(m\dot{y}) + mg + k\dot{y} &= 0 \\ m\ddot{y} + k\dot{y} + mg &= 0 \end{aligned} \]

The friction term \(k\dot{y}\) always opposes the velocity, so it enters with the same sign as \(m\ddot{y}\). Option D omits friction altogether, and options B and C give the friction term the wrong sign, which would make the resisting force speed the particle up.

Your Answer : (Not Answered)

Correct Answer : C

The frequency is unchanged: \(\omega = \omega_{0}\).

For small oscillations a simple pendulum has

\[ \omega_{0} = \sqrt{\frac{g}{l}} \]

which depends only on the length and the acceleration due to gravity. The mass cancels, because both the restoring force \(mg\sin\theta\) and the inertia \(ml\) are proportional to m. Halving the mass, with the length the same, therefore leaves the frequency as it was.

Doubling, halving or multiplying by \(\sqrt{2}\) would all require the frequency to depend on m, which it does not; such mass dependence belongs to a spring-mass oscillator, where \(\omega = \sqrt{k/m}\).

Your Answer : (Not Answered)

Correct Answer : A

The correct statement is \(v_{0}b = v_{m}r_{m}\).

The Coulomb force is central, so angular momentum about the nucleus is conserved. Far away it is \(mv_{0}b\); at closest approach the velocity is perpendicular to the radius, so it is \(mv_{m}r_{m}\). Hence \(v_{0}b = v_{m}r_{m}\).

Energy conservation, with the alpha particle's charge 2e, gives \(v_{0}^{2} = v_{m}^{2} + 4r_{0}v_{0}^{2}/r_{m}\). Using \(v_{m} = v_{0}b/r_{m}\):

\[ \begin{aligned} r_{m}^{2} - 4r_{0}r_{m} - b^{2} &= 0 \\ r_{m} &= 2r_{0} + \sqrt{4r_{0}^{2} + b^{2}} \\ &\approx 4r_{0} + \frac{b^{2}}{4r_{0}} \end{aligned} \]

So the correction is \(b^{2}/4r_{0}\), not \(b^{2}/2r_{0}\) or \(b^{2}/8r_{0}\). And \(v_{0}b = 2v_{m}r_{0}\) would need \(r_{m} = 2r_{0}\), less than the head-on value \(4r_{0}\). Only the angular momentum statement holds.

Your Answer : (Not Answered)

Correct Answer : B

The statement that does NOT apply is the normal coordinates are orthogonal to one another.

Normal coordinates are the combinations of the original coordinates that diagonalise both the kinetic and potential energies at once. In terms of them,

\[ \begin{aligned} T &= \tfrac{1}{2}\sum_{k}\dot{\eta}_{k}^{2} \\ V &= \tfrac{1}{2}\sum_{k}\omega_{k}^{2}\eta_{k}^{2} \end{aligned} \]

so the potential energy is a sum of squares with constant coefficients, each coordinate oscillates independently, and each has its own eigen-frequency \(\omega_{k}\). Those three statements are the defining properties.

Orthogonality is not a property of the coordinates themselves. The mode (eigen)vectors from which they are built are orthogonal only with respect to the kinetic-energy (mass) matrix, not in the ordinary sense, so the key treats this as the statement that does not apply.