Past exam of the physics 1 paper of the University of Oxford

Solution

TODO: source missing. Printed page 2, containing question 1, is absent from the supplied damaged PDF. Its statement, subparts and numerical data are unavailable, so no solution can be inferred responsibly.

2

Solution

TODO: source missing. Printed page 2, containing question 2, is absent from the supplied damaged PDF. Its statement, subparts and numerical data are unavailable, so no solution can be inferred responsibly.

3

Solution

TODO: source missing. Printed page 2, containing question 3, is absent from the supplied damaged PDF. Its statement, subparts and numerical data are unavailable, so no solution can be inferred responsibly.

4

a

Solution

For motion in a central potential, conservation of angular momentum gives

L = m r^{2} \dot{θ}

. In polar coordinates, the conserved mechanical energy is

E = \frac{m}{2} \overset{r}{˙}^{2} + \frac{m}{2} r^{2} \dot{θ}^{2} - G (r) = \frac{m}{2} \overset{r}{˙}^{2} + \frac{L ^{2}}{2 m r ^{2}} - G (r) .

Consequently, for

L \neq = 0

\overset{r}{˙} = \pm \frac{2}{m} [E + G (r)] - \frac{L ^{2}}{m ^{2} r ^{2}}, \frac{d r}{d θ} = \frac{m r ^{2}}{L} \overset{r}{˙} = \pm \frac{r ^{2}}{L} 2 m [E + G (r)] - \frac{L ^{2}}{r ^{2}} .

The sign in

\overset{r}{˙}

distinguishes inward and outward motion. Choosing the orientation so

L > 0

, it is also the sign of

d r / d θ

. The expression under the square root must be nonnegative. If

L = 0

, the motion is radial and

θ

is constant, so

r

cannot be parametrized by

θ

this way.

b

Solution

Take

β > 0

. At a radial turning point,

\overset{r}{˙} = 0

, and the effective potential gives

E = \frac{L ^{2}}{2 m r ^{2}} - \frac{β}{r}, E r_{m i n}^{2} + β r_{m i n} - \frac{L ^{2}}{2 m} = 0.

For

L \neq = 0

, the smaller positive radial turning point is conveniently written without division by

E

r_{m i n} = \frac{L ^{2}}{m β + m ^{2} β ^{2} + 2 m E L ^{2}} .

This includes

E = 0

, where

r_{m i n} = L^{2} / (2 m β)

. For

E < 0

there is also an outer radial turning point; the two coincide for the circular orbit at

E = - m β^{2} / (2 L^{2})

. Smaller energies are inaccessible at that

L

. For

E > 0

there is just one positive radial turning point. If

L = 0

, there is no centrifugal barrier and an inward trajectory can reach the singular centre; its limiting closest distance is zero.

The requested omission of $β$ is not possible. For example, at

E = 0

and fixed nonzero

L, m

, doubling

β

halves

r_{m i n}

. Thus

E, L, m

alone cannot determine closest approach.

5

a

Solution

Use upward generalized coordinates. Let

h

be the displacement of the lower pulley and

z

that of the

4 m

mass. The two inextensible mechanical strings impose

x + y - 2 h = 0, z + h = 0, h = \frac{x + y}{2}, z = - \frac{x + y}{2} .

For ideal massless pulleys and mechanical strings, only the three masses contribute kinetic energy. Hence

K = \frac{1}{2} (4 m) (\frac{x ˙ + y ˙}{2})^{2} + \frac{1}{2} (5 m) \overset{x}{˙}^{2} + \frac{1}{2} (3 m) \overset{y}{˙}^{2} = m (3 \overset{x}{˙}^{2} + \overset{x}{˙} \overset{y}{˙} + 2 \overset{y}{˙}^{2}) .

Up to a constant, their gravitational potential energy is

U = 4 m g z + 5 m gx + 3 m g y = m g (3 x + y) .

The Lagrangian of this compound Atwood machine is therefore

L = K - U = m (3 \overset{x}{˙}^{2} + \overset{x}{˙} \overset{y}{˙} + 2 \overset{y}{˙}^{2}) - m g (3 x + y) .

b

Solution

A constant displacement proportional to

(1, - 3)

leaves the velocities unchanged and changes

3 x + y

by zero. It is therefore a continuous symmetry of the Lagrangian. The generalized momentum components are

p_{x} = \frac{\partial L}{\partial x ˙} = m (6 \overset{x}{˙} + \overset{y}{˙}), p_{y} = \frac{\partial L}{\partial y ˙} = m (\overset{x}{˙} + 4 \overset{y}{˙}) .

Noether's theorem gives the conserved quantity

P = p_{x} - 3 p_{y} = m (3 \overset{x}{˙} - 11 \overset{y}{˙}) = constant .

A direct check using the Euler-Lagrange equations is

6 \overset{x}{¨} + \overset{y}{¨} = - 3 g, \overset{x}{¨} + 4 \overset{y}{¨} = - g ⟹ 3 \overset{x}{¨} - 11 \overset{y}{¨} = 0.

The normalization of a symmetry generator is arbitrary. One can take the transformation parameter to have length units and

(K_{x}, K_{y}) = (1, - 3)

dimensionless, giving

P

the units of momentum. Alternatively, for the paper's convention with dimensionless parameter and length-valued

K_{i}

, use

(K_{x}, K_{y}) = (ℓ, - 3 ℓ)

; the conserved quantity is then

ℓ P

. The conserved content is identical.

6

a

Solution

The special relativity expression for momentum is

p = γm v

, with Lorentz factor

γ = (1 - v^{2} / c^{2})^{- 1/2}

. Equal speeds imply equal Lorentz factors, so the pion and proton momenta are in the ratio of their rest masses:

p_{π} = p_{p} \frac{m _{π}}{m _{p}} = 400 \frac{0.14}{0.94} ≃ 59.6 GeV / c .

The common dimensionless product is

γ v / c = p_{p} / (m_{p} c) = 400/0.94 ≃ 425.53

γ ≃ 425.53

and

v / c ≃ 0.99999724

b

Solution

By time dilation, the laboratory mean lifetime is

τ = γ τ_{0}

, where

τ_{0} = 2.6 \times 1 0^{- 8} s

. The exponential decay law evaluated at travel time

d / v

gives the surviving pion fraction

\frac{n}{n _{0}} = exp (- \frac{d}{γ v τ _{0}}) .

Use the exact product

γ v = (400/0.94) c

from part (a), rather than approximating the speed separately. With

c = 2.99792458 \times 1 0^{8} m/s

γ v τ_{0} ≃ 3317 m, \frac{n}{n _{0}} = exp [- \frac{400}{( 400/0.94 ) c ( 2.6 \times 1 0 ^{- 8} )}] ≃ 0.886.

Thus about

88.6%

emerge without decaying, and about

11.4%

decay inside the pipe. This assumes the given common longitudinal speed and neglects losses from other interactions.

7

a

Solution

Align the spatial axes and make the origins coincide at

t = t^{'} = 0

. For an inertial frame moving at velocity

+ v

along

x

, the Lorentz transformation is

x^{'} = γ (x - v t), t^{'} = γ (t - \frac{vx}{c ^{2}}), y^{'} = y, z^{'} = z, γ = \frac{1}{1 - v ^{2} / c ^{2}} .

The inverse Lorentz transformation follows by interchanging primed and unprimed coordinates and replacing

v

- v

b

Solution

Let the object be at rest in

S^{'}

, with proper length

L_{0} = x_{2}^{'} - x_{1}^{'}

. Measuring its length in

S

requires two endpoint events simultaneous in

S

, so

Δ t = 0

. The Lorentz transformation then gives

L_{0} = Δ x^{'} = γ (Δ x - v Δ t) = γ L, L = \frac{L _{0}}{γ} = L_{0} 1 - v^{2} / c^{2} .

This is length contraction. The simultaneous endpoint events in

S

are not simultaneous in

S^{'}

; that relativity of simultaneity is essential to the measurement.

For the two rockets, take the first rocket's Earth-frame velocity as

+ c /2

and the second's as

- c /2

. The velocity-addition formula gives the second rocket's velocity in the first rocket's rest frame:

u = \frac{- c /2 - c /2}{1 - ( - c /2 ) ( c /2 ) / c ^{2}} = - \frac{4}{5} c .

Its Lorentz factor is

γ_{rel} = 5/3

, and therefore its measured length is

L_{other} = \frac{3}{5} L_{0} .

Here “appear” is interpreted as a length measured using simultaneous endpoint positions in the observer's inertial frame. A photograph, involving unequal light travel times, is a different measurement.

c

Solution

Write the incident and scattered photon energies as

ε = h c / λ

and

ε^{'} = h c / λ^{'}

. Conservation of relativistic energy gives the recoiling electron energy

E_{e} = m c^{2} + ε - ε^{'}

. By momentum conservation, its momentum is the difference of the two photon momenta; hence

p_{e}^{2} c^{2} = ε^{2} + ε^{'2} - 2 ε ε^{'} cos θ .

Insert these into the relativistic energy-momentum relation for the electron:

(m c^{2} + ε - ε^{'})^{2} = m^{2} c^{4} + ε^{2} + ε^{'2} - 2 ε ε^{'} cos θ .

Expanding and cancelling terms gives

m c^{2} (ε - ε^{'}) = ε ε^{'} (1 - cos θ), \frac{1}{ε ^{'}} - \frac{1}{ε} = \frac{1 - c o s θ}{m c ^{2}} .

Since

1/ ε = λ / (h c)

, the Compton scattering formula follows:

λ^{'} = λ + \frac{h}{m c} (1 - cos θ) .

The scale

h / (m c)

is the electron Compton wavelength. At

θ = 0

λ^{'} = λ

: there is no energy or momentum transfer in this ideal forward limit, so the electron remains at rest. At

θ = π

, the wavelength shift is maximal:

Δ λ_{m a x} = \frac{2 h}{m c}, ε_{m i n}^{'} = \frac{ε}{1 + 2 ε / ( m c ^{2} )} .

The electron gains its largest kinetic energy,

ε - ε_{m i n}^{'}

, and recoils forward while the photon reverses direction.

8

a

Solution

Define

f (r) > 0

as the magnitude of the inward central force, so

F = - f (r) r

and

V^{'} (r) = f (r)

. The conserved angular momentum is

J = m r^{2} \dot{θ}

. Eliminating

\dot{θ}

from the mechanical energy yields

V_{eff} (r) = V (r) + \frac{J ^{2}}{2 m r ^{2}}, E = \frac{1}{2} m \overset{r}{˙}^{2} + V_{eff} (r) .

The added term in the effective potential is the centrifugal barrier. If instead

f

denotes the signed radial force component, the convention becomes

V^{'} = - f

b

Solution

A circular orbit has constant

r = r_{0} > 0

. Both its initial radial velocity and radial acceleration must vanish, giving

r (0) = r_{0}, \overset{r}{˙} (0) = 0, V_{eff}^{'} (r_{0}) = V^{'} (r_{0}) - \frac{J ^{2}}{m r _{0}^{3}} = 0.

Equivalently, the initial tangential speed and angular velocity obey

v_{θ, 0}^{2} = \frac{r _{0} f ( r _{0} )}{m}, \dot{θ} (0) = \frac{J}{m r _{0}^{2}}, J^{2} = m r_{0}^{3} f (r_{0}) .

The initial polar angle is arbitrary, and the energy is fixed at

E = V_{eff} (r_{0})

. Either orientation is possible. These conditions give an exact circular orbit even if it is unstable; stability requires an additional positive second derivative of the effective potential.

c

Solution

For the power law central potential

V = β r^{k}

, differentiate the effective potential:

V_{eff}^{'} (r) = β k r^{k - 1} - \frac{J ^{2}}{m r ^{3}} .

The circular orbit condition is

m β k r_{0}^{k + 2} = J^{2}

. Thus, when

J \neq = 0

β k > 0

and

k \neq = - 2

r_{0} = (\frac{J ^{2}}{m β k})^{1/ (k + 2)} .

The sign condition

β k > 0

expresses attraction. At

k = 0

the force vanishes and there is no nontrivial circular orbit. At

k = - 2

, attraction requires

β < 0

and the condition reduces to

J^{2} = - 2 m β

. In that exceptional case the effective potential is identically zero and every

r_{0} > 0

is a possible circular radius for the tuned

J

; there is no isolated preferred radius. A small radial velocity then gives

\overset{r}{¨} = 0

, rather than an oscillation.

d

Solution

Put

r = r_{0} + η

, where

∣ η ∣ ≪ r_{0}

. The Taylor expansion at a circular orbit is

V_{eff} (r_{0} + η) = V_{eff} (r_{0}) + \frac{1}{2} V_{eff}^{''} (r_{0}) η^{2} + O (η^{3}) .

The constant

V_{eff} (r_{0})

has no effect on the force. The paper's displayed harmonic potential should be read as this expansion after subtracting that constant and neglecting higher orders; it is not an equality for the absolute effective potential in general. For example, when

m = J = β = 1

and

k = 2

r_{0} = 2^{- 1/4}

and

V_{eff} (r_{0}) = 2

, whereas the printed quadratic expression is zero at

r_{0}

Differentiating again and then using

β k r_{0}^{k - 2} = J^{2} / (m r_{0}^{4})

gives

V_{eff}^{''} (r_{0}) = β k (k - 1) r_{0}^{k - 2} + \frac{3 J ^{2}}{m r _{0}^{4}} = (k + 2) \frac{J ^{2}}{m r _{0}^{4}} .

The linearized radial equation of motion is

m \overset{η}{¨} = - V_{eff}^{''} (r_{0}) η

, so the radial frequency in a power-law central potential is

ω_{r} = k + 2 \frac{∣ J ∣}{m r _{0}^{2}}, k > - 2.

For

k < - 2

the circular orbit is unstable and radial disturbances grow exponentially, with growth rate

- (k + 2) ∣ J ∣/ (m r_{0}^{2})

. At

k = - 2

the linear restoring force vanishes, as described in part (c). Thus an attractive central force alone does not guarantee small oscillations.

e

Solution

Choose the orientation

J > 0

. To leading order in the radial disturbance, the azimuthal angular frequency is

ω_{θ} = J / (m r_{0}^{2})

, so

\frac{ω _{r}}{ω _{θ}} = k + 2 .

For either orientation the positive frequency ratio is

ω_{r} /∣ ω_{θ} ∣

. At

k = - 1

, the ratio is

1

: there is one radial oscillation per revolution. The bound Kepler orbit is an ellipse with the force centre at a focus, with one closest and one furthest approach per revolution. Its exact polar coordinates equation is

r (θ) = \frac{p}{1 + e c o s θ}, p = \frac{J ^{2}}{m ∣ β ∣}, 0 < e ≪ 1.

k = 2

, the ratio is

2

: there are two radial oscillations per revolution. The isotropic harmonic oscillator trajectory is an ellipse centred on the force centre. An exact parametrization is

x (t) = a cos Ω t, y (t) = b sin Ω t, Ω = \frac{2 β}{m}, a ≃ b .

Here

r^{2} = a^{2} cos^{2} Ω t + b^{2} sin^{2} Ω t

repeats twice during one full circuit. The two different placements of the force centre distinguish these ellipses. These are the two all-bound-orbit cases of Bertrand's theorem.

Figure 1.
Nearly circular ellipses: one radial cycle about a focus for the inverse-distance potential, and two about the centre for the quadratic potential
. **Details**.

9

a

Solution

The diagram makes

x

the remaining distance to the wall:

x

decreases from

L

0

. With the endpoint condition

m (0) = 0

, integrating the supplied mass gradient gives

m (x) = \frac{M x}{L} .

Assume escaping sand has the same horizontal velocity as the bucket at release. There is then no horizontal recoil: the momentum flux of the outgoing sand cancels the

\overset{m}{˙} \overset{x}{˙}

contribution to the bucket's momentum derivative. The variable-mass system equation is therefore

m \overset{x}{¨} = - T .

Set

u = - \overset{x}{˙} \geq 0

, the speed towards the wall. Since

\overset{x}{¨} = u d u / d x

, the initial condition

u (L) = 0

gives

u \frac{d u}{d x} = - \frac{T L}{M x}, u^{2} (x) = \frac{2 T L}{M} ln \frac{L}{x} .

Consequently the remaining sand has kinetic energy

K (x) = \frac{1}{2} m (x) u^{2} (x) = T x ln \frac{L}{x} .

In terms of

z = x / L

K / (T L) = - z ln z

. Its derivative is

- ln z - 1

and its second derivative is

- 1/ z < 0

x_{K} = \frac{L}{e}, K_{m a x} = \frac{T L}{e} .

The magnitude of the remaining sand's momentum is

p (x) = m (x) u (x) = x \frac{2 MT}{L} ln \frac{L}{x} .

It is easier to maximize its square,

p^{2} / (2 MT L) = - z^{2} ln z

. Its derivative is

- z (2 ln z + 1)

, giving the unique interior maximum

x_{p} = \frac{L}{e}, p_{m a x} = \frac{MT L}{e} .

Both kinetic energy and momentum tend to zero at

x \to 0^{+}

, even though the idealized speed diverges. There is no finite-mass bucket left at the endpoint, so that divergence signals the limit of the idealization.

The leaking bucket under constant tension has

x_{K} = L / e < L / e = x_{p}

: the kinetic energy maximum is closer to the wall. As the bucket loses mass, its increasing speed continues to boost

K = p u /2

after

p

has begun to fall. Equivalently,

p \propto z - ln z

and

K \propto z (- ln z)

, so kinetic energy has the stronger dependence on the growing speed. The work

T (L - x)

is shared between the remaining sand and the sand already lost; it is not all the bucket's instantaneous kinetic energy.

b

Solution

Take upward velocity as positive, and let

w > 0

be the rearward rocket exhaust speed. During

d t

, the ejected parcel has mass

- d m > 0

and laboratory velocity

v - w

to first order. The momentum change of the rocket and this parcel equals the gravitational impulse:

(m + d m) (v + d v) + (- d m) (v - w) - m v = - m g d t .

Discarding the second-order product

d m d v

gives the vertical rocket equation

m \frac{d v}{d t} = - w \frac{d m}{d t} - m g .

With

\overset{m}{˙} = - α

and

m = m_{0} - α t

, integration for free flight starting from rest at

t = 0

gives

v (t) = w ln \frac{m _{0}}{m _{0} - α t} - g t, 0 \leq t \leq T .

The logarithmic term is the ideal Tsiolkovsky rocket equation contribution;

g t

is the gravity loss. Immediate upward departure requires positive initial acceleration, or

α w > m_{0} g, w_{threshold} = \frac{m _{0} g}{α} .

At exact equality the initial acceleration is zero, but further burning reduces the weight and permits departure immediately thereafter. If departure is allowed to be delayed while fuel burns on the pad, the condition is instead

α w > m (t) g

at some time before burnout. With final rocket dry mass

m_{f}

, any powered departure before burnout requires

w > g m_{f} / α

; the free-flight formula then starts at the actual departure time

t_{ℓ}

with

m (t_{ℓ})

v = w ln [m (t_{ℓ}) / m (t)] - g (t - t_{ℓ})

For the specified data, the initial mass includes the payload. The rocket fuel fraction gives

m_{f} = (1 - f) m_{0} = 2500 kg, T = \frac{f m _{0}}{α} = 15 s .

The

100 kg

payload is already part of the

2500 kg

final mass; it is not the entire rocket dry mass and must not be added a second time. Here

m_{0} g / α = 196 m/s

, well below the stated rocket exhaust speed, so departure is immediate. The burnout speed is

v (T) = 2500 ln 4 - 9.8 (15) ≃ 3319 m/s .

This is the speed at burnout. Subsequent motion under the assumed uniform gravity has

v (t) = v (T) - g (t - T)

10

a

Solution

For an equal-speed elastic impact on a free rod, let

h

be the signed distance of the impact point from the rod's center of mass,

I = m l^{2} /12

, and

J

the horizontal impulse delivered to the rod. For a smooth impact the impulse is perpendicular to the rod. If the ball's signed final velocity is

u

, the rod's center of mass velocity is

V

and its angular velocity is

ω

, momentum balance and angular impulse give

m (v_{0} - u) = J = mV, I ω = J h .

Both objects have positive final translational velocities for a nonzero impact away from the centre. More explicitly, restitution in an elastic collision gives

V + ωh - u = v_{0}

, and hence

V = \frac{2 v _{0}}{2 + m h ^{2} / I}, u = \frac{( m h ^{2} / I ) v _{0}}{2 + m h ^{2} / I} \geq 0.

Equal final speeds thus require

u = V = v_{0} /2

. One can also see that the alternative

u = - V

would violate

u + V = v_{0}

By conservation of kinetic energy in the elastic collision,

\frac{1}{2} m v_{0}^{2} = \frac{1}{2} m (\frac{v _{0}}{2})^{2} + \frac{1}{2} m (\frac{v _{0}}{2})^{2} + \frac{1}{2} I ω^{2} .

Thus

I ω^{2} = m v_{0}^{2} /2

. Since

I ω = mVh = m v_{0} h /2

, elimination of

ω

gives

h^{2} = \frac{2 I}{m} = \frac{l ^{2}}{6}, ∣ h ∣ = \frac{l}{6} .

Either side of the center of mass works; the nearer end is a distance

l (1/2 - 1/ 6)

away. The final angular speed is

∣ ω ∣ = 6 v_{0} / l

. These values also satisfy the restitution condition because the struck point's velocity is

V + ωh = 3 v_{0} /2

, reversing the initial relative contact velocity.

b

Solution

The free rod has the same mass

m

and length

l

as in part (a). There is no pivot. Put the rod's initial center of mass at

y = 0

and the impact end at

y = l /2

. At impact the combined center of mass is

y_{C} = \frac{m ( 0 ) + m ( l /2 )}{2 m} = \frac{l}{4} .

Conservation of momentum gives the combined center of mass velocity

V_{C} = v_{0} /2

. Evaluate angular momentum about this combined center of mass; the incident ball's perpendicular lever arm is

l /4

, so its initial magnitude is

m v_{0} l /4

After the perfectly inelastic collision, the parallel axis theorem gives the combined moment of inertia

I_{C} = (\frac{m l ^{2}}{12} + m \frac{l ^{2}}{16}) + m \frac{l ^{2}}{16} = \frac{5 m l ^{2}}{24} .

With negligible external angular impulse, conservation of angular momentum therefore gives

∣ ω ∣ = \frac{m v _{0} l /4}{5 m l ^{2} /24} = \frac{6 v _{0}}{5 l} .

Its sense is that of the incident ball's angular momentum, clockwise in the source diagram. As a consistency check, the final kinetic energy is

K_{f} = \frac{1}{2} (2 m) (\frac{v _{0}}{2})^{2} + \frac{1}{2} \frac{5 m l ^{2}}{24} (\frac{6 v _{0}}{5 l})^{2} = \frac{2}{5} m v_{0}^{2} .

Thus

m v_{0}^{2} /10

, or one fifth of the initial kinetic energy, is dissipated in sticking. Treating the rod's original centre as a fixed pivot would give an incorrect angular velocity.

c

Solution

Let

a

be the center of mass acceleration downhill and

F

the uphill static friction. Translation and rotation about the center of mass obey

ma = m g sin θ - F, I α = F r .

The rolling without slipping constraint is

a = r α

. For the coin's moment of inertia

I = m r^{2} /2

, this gives

F = ma /2

, so

a = \frac{2}{3} g sin θ, F = \frac{1}{3} m g sin θ .

The normal force is

N = m g cos θ

. Static friction can enforce rolling without slipping only if

F \leq μ N

\frac{1}{3} m g sin θ \leq μ m g cos θ, θ_{m a x} = arctan (3 μ) .

The static friction acts uphill to provide the torque; it is not automatically equal to

μ N

, attaining that limiting value only at the threshold. For finite

μ \geq 0

the result lies between

0

and

π /2

2021 resit (A10287W1)

Source paper: www.whatdotheyknow.com/request/physics_preliminary_past_papers/response/2680788/attach/3/2018%202021%20Physics%20prelims.zip?cookie_passthrough=1

The cover dates this resit to 6 September 2021; its archive member is misleadingly labelled 2020/L/a10287w1.pdf.

1

Solution

Interpret

h

as the vertical drop of the cylinder's centre of mass, and initially take the suspended rod to be vertical. The moment of inertia of a uniform solid cylinder is

I = m r^{2} /2

. Before impact, rolling without slipping gives

ω = u / r

, so conservation of mechanical energy gives

m g h = \frac{1}{2} m u^{2} + \frac{1}{2} I \frac{u ^{2}}{r ^{2}} = \frac{3}{4} m u^{2}, u = \frac{4 g h}{3} .

Use the usual instantaneous smooth elastic collision model: the horizontal impact impulse passes through the cylinder's centre of mass, and the floor supplies negligible tangential impulse during impact. The cylinder's angular velocity is then unchanged during impact; its rotational kinetic energy does not transfer to the bob. The rod supplies no horizontal impulse when vertical. If

v

and

V

are the signed cylinder and bob velocities immediately afterwards, conservation of momentum and the unit coefficient of restitution give

m u = m v + 2 mV, V - v = u, v = - \frac{u}{3}, V = \frac{2 u}{3} .

During the subsequent simple pendulum swing, conservation of mechanical energy gives

\frac{1}{2} (2 m) V^{2} = 2 m g h (1 - cos θ_{m a x}), cos θ_{m a x} = \frac{19}{27}, θ_{m a x} = cos^{- 1} \frac{19}{27} ≃ 45.2 8^{\circ} .

The cylinder generally slips immediately after this impact, which is consistent with imposing rolling without slipping only on the descent.

Impact convention matters. The floor-impulse dependence of a rolling-cylinder collision gives a different answer if a different constraint is imposed. If instead the floor enforces rolling without slipping throughout the impulsive collision, it supplies a nonzero horizontal impulse. The cylinder's effective translational mass is then

m + I / r^{2} = 3 m /2

. Writing

J

for the impulse delivered to the bob gives

(3 m /2) (v - u) = - J

2 mV = J

and

V - v = u

. Hence

V = 6 u /7

and

cos θ_{m a x} = 25/49

59.3 2^{\circ}

. The original wording specifies the descent constraint but does not specify an impulsive floor-friction law; the first boxed value uses the smooth-impact convention.

2

Solution

Choose the heavier particle's outgoing velocity above the initial direction. Let its speed be

V

, and let the lighter particle have components

(v cos θ, - v sin θ)

. Thus

θ

and

ϕ

are nonnegative deflections on opposite sides. Conservation of momentum gives

m u = m v cos θ + M V cos ϕ, m v sin θ = M V sin ϕ .

The elastic collision conserves kinetic energy:

m u^{2} = m v^{2} + M V^{2} .

Taking the squared magnitude of the vector momentum equation yields

m^{2} v^{2} = m^{2} u^{2} + M^{2} V^{2} - 2 m M u V cos ϕ .

Eliminate

v^{2}

with the kinetic energy equation. For a nontrivial collision,

V \neq = 0

, so

(M + m) V = 2 m u cos ϕ .

Substituting back into the two momentum components gives

v cos θ = \frac{u ( m - M c o s 2 ϕ )}{m + M}, v sin θ = \frac{u M s i n 2 ϕ}{m + M} .

Therefore

tan θ = \frac{sin 2 ϕ}{m / M - cos 2 ϕ} .

The ratio alone does not determine the quadrant: for

M > m

, the lighter particle can move backwards. The two component equations select the correct

0 \leq θ \leq π

. At

m / M = cos 2 ϕ

, the outgoing direction is perpendicular to the initial direction; the displayed tangent has a pole there.

3

Solution

Let

t = 0

be the start of deposition and

I_{0} = M R^{2} /2

be the original moment of inertia of a uniform disc. Assume that all deposited sand sticks and remains at radius

r

, with negligible thickness and no losses. The sand's incoming angular momentum about the vertical axis is zero. Neither the bearing nor gravity supplies an axial torque, so conservation of angular momentum gives

I (t) = I_{0} + α r^{2} t, I (t) ω (t) = I_{0} ω_{0} .

Consequently the angular velocity and angular acceleration are

ω (t) = \frac{I _{0} ω _{0}}{I _{0} + α r ^{2} t} = \frac{M R ^{2} ω _{0}}{M R ^{2} + 2 α r ^{2} t}, \overset{ω}{˙} (t) = - \frac{α r ^{2} I _{0} ω _{0}}{( I _{0} + α r ^{2} t ) ^{2}} .

The rotational kinetic energy is

K (t) = \frac{1}{2} I (t) ω (t)^{2} = \frac{I _{0}^{2} ω _{0}^{2}}{2 ( I _{0} + α r ^{2} t )} = K (0) \frac{I _{0}}{I _{0} + α r ^{2} t} .

Thus, for

α r^{2} > 0

t_{1/2} = \frac{I _{0}}{α r ^{2}} = \frac{M R ^{2}}{2 α r ^{2}} .

This is mass-loaded rotation: the sticking process dissipates mechanical energy even though angular momentum is conserved. Indeed

\dot{K} = - α r^{2} ω^{2} /2

. If

r = 0

α = 0

, no additional axial moment of inertia is deposited and the nonzero initial rotational kinetic energy never halves.

4

Solution

In the inertial centre of mass frame, the Lagrangian mechanics description starts from kinetic energy minus potential energy:

L = \frac{1}{2} m_{1} ∣ \dot{r}_{1} ∣^{2} + \frac{1}{2} m_{2} ∣ \dot{r}_{2} ∣^{2} - V (∣ r_{1} - r_{2} ∣) .

Define the two-particle relative coordinate

r = r_{1} - r_{2}

. The centre of mass condition gives

m_{1} r_{1} + m_{2} r_{2} = 0, r_{1} = \frac{m _{2}}{m _{1} + m _{2}} r, r_{2} = - \frac{m _{1}}{m _{1} + m _{2}} r .

Substitution therefore gives

L = \frac{1}{2} μ ∣ \dot{r} ∣^{2} - V (r), μ = \frac{m _{1} m _{2}}{m _{1} + m _{2}}, r = ∣ r ∣.

The Euler-Lagrange equation is simply

μ \ddot{r} = - \nabla V (r)

. Thus center-of-mass reduction turns a two-body problem into a one-particle central force problem with reduced mass

μ

, reducing six position components to three relative components. In an arbitrary inertial frame one adds the independent centre of mass term

\frac{1}{2} (m_{1} + m_{2}) ∣ \dot{R} ∣^{2}

; that free motion does not affect the relative solution. Since

V

depends only on separation, conservation of angular momentum further confines a nonradial relative trajectory to a plane.

5

Solution

Interpret the percentage relative to the classical kinetic energy, so that

K_{rel} = 1.02 K_{cl}

. Set

β = v / c

and use the Lorentz factor

γ = (1 - β^{2})^{- 1/2}

. The relativistic kinetic energy and classical kinetic energy give

(γ - 1) m c^{2} = 1.02 (\frac{1}{2} m v^{2}), γ - 1 = 0.51 β^{2} .

For the positive nonzero speed, insert

β^{2} = 1 - γ^{- 2}

and cancel

γ - 1

\frac{γ ^{2}}{γ + 1} = 0.51, γ^{2} - 0.51 γ - 0.51 = 0, γ = \frac{0.51 + 2.3001}{2} ≃ 1.01330403.

It follows that

v = c 1 - γ^{- 2} ≃ 0.161513 c ≃ 4.8420 \times 1 0^{7} m s^{- 1}, E = γm c^{2} ≃ 950.479 MeV .

Here

m c^{2} = 938 MeV

is the proton's rest energy; the resulting relativistic kinetic energy is

12.479 MeV

. The zero-speed equality before cancellation has no meaningful percentage excess because both kinetic energies then vanish.

6

Solution

Take positive velocity towards

B

, and interpret

100 MHz

as the emitted frequency in the probe's rest frame. The relativistic velocity-addition formula gives the probe's station-frame velocity:

u_{P} = \frac{0.5 c + 0.5 c}{1 + ( 0.5 ) ( 0.5 )} = 0.8 c .

For reception at the station, photons propagate backwards while the probe recedes. The longitudinal relativistic Doppler effect gives

f_{station} = 100 \frac{1 - 0.8}{1 + 0.8} MHz = \frac{100}{3} MHz ≃ 33.33 MHz .

Before the probe overtakes

B

, their separation decreases. The probe's velocity in

B

's inertial frame is

u_{P ∣ B} = \frac{0.8 c - 0.4 c}{1 - ( 0.8 ) ( 0.4 )} = \frac{10}{17} c .

The approaching-source relativistic Doppler effect therefore gives

f_{B} = 100 \frac{1 + 10/17}{1 - 10/17} MHz = 100 \frac{27}{7} MHz ≃ 196.40 MHz .

An independent frame calculation gives the same result: forward photons have station-frame frequency

300 MHz

, and a receiver moving at

0.4 c

measures

300 0.6/1.4 MHz

. If reception occurs after the probe passes

B

, backwards photons instead give

f_{B} = 100 7/27 MHz ≃ 50.92 MHz

. The direction of propagation and the reception geometry are essential to choosing the Doppler factor.

7

Solution

Powered ascent. Use upward signed velocity, constant gravitational acceleration

g = 9.81 m s^{- 2}

, constant relative rocket exhaust speed

u = 2500 m s^{- 1}

, vertical flight from rest, no drag force, and no impulse at stage separation. The constant-

g

approximation is an explicit modelling assumption; a varying Newtonian gravitational field would require integrating a different ascent equation. Put

m_{0} = 540000 kg

and

q = 2500 kg s^{- 1}

. During ascent

m (t) = m_{0} - qt

, and momentum balance gives the rocket equation with gravity:

m \overset{v}{˙} = u q - m g, v (t) = u ln \frac{m _{0}}{m _{0} - qt} - g t .

The launch rocket thrust

u q = 6.25 \times 1 0^{6} N

exceeds

m_{0} g

, so immediate lift-off is consistent. At ascent burnout the entire attached vehicle still has mass

m_{b} = 140000 kg

, not

40000 kg

; the latter is the separated first stage. The burn duration and endpoint speed are

t_{b} = \frac{400000}{2500} = 160 s, v_{b} = u ln \frac{m _{0}}{m _{b}} - g t_{b} = 1805.22 m s^{- 1} .

Coasting height. Integrating the ascent velocity from zero height gives

z (t) = u [t - \frac{m _{0} - qt}{q} ln \frac{m _{0}}{m _{0} - qt}] - \frac{1}{2} g t^{2} .

For example this follows by substituting

m = m_{0} - qt

\int_{0}^{t} ln [m_{0} / (m_{0} - q s)] d s

. At separation,

z_{b} = u [t_{b} - \frac{m _{b}}{q} ln \frac{m _{0}}{m _{b}}] - \frac{1}{2} g t_{b}^{2} = 85442.26 m .

The separated stage then undergoes free fall in a constant gravitational field with initial upward speed

v_{b}

. Conservation of mechanical energy during this coast gives

H = z_{b} + \frac{v _{b}^{2}}{2 g} = 251538.47 m ≃ 251.54 km .

Its smaller mass after separation does not change that coasting height.

Reserve burn. On descending to

z_{i} = 100000 m

, conservation of mechanical energy gives downward speed

w_{i} = 2 g (H - z_{i}) = 1724.29255 m s^{- 1} .

Reset time to

s = 0

at reserve ignition. Now

m_{i} = 40000 kg

m_{f} = 14000 kg

and

q_{r} = 26000/92 kg s^{- 1}

. Downward ejection supplies upward rocket thrust. The upward signed velocity is

v_{r} (s) = - w_{i} + u ln \frac{m _{i}}{m _{i} - q _{r} s} - g s .

Thus the formal free-flight burnout value is

v_{r} (92) = - w_{i} + u ln \frac{40000}{14000} - 92 g = - 2.25724 m s^{- 1},

corresponding to a formal downward speed of $2.26 m s^{- 1}$ . Keeping

g = 9.8 m s^{- 2}

throughout instead gives

3.82 m s^{- 1}

downward; premature rounding is especially damaging because three large velocity terms almost cancel.

The free-flight endpoint is below ground. Integrating the same reserve-burn velocity gives

z_{r} (s) = z_{i} - w_{i} s + u [s - \frac{m _{i} - q _{r} s}{q _{r}} ln \frac{m _{i}}{m _{i} - q _{r} s}] - \frac{1}{2} g s^{2} .

At the stated endpoint this gives

z_{r} (92) = - 167.27 m

for

g = 9.81 m s^{- 2}

. The surface is crossed at approximately

s = 89.15 s

, before reserve burnout. The stage has already hit the surface, so a physical post-impact burnout speed cannot be inferred without a ground-contact model. The boxed velocity is the intended ideal continuation, not an achievable free-flight landing result under these exact assumptions. In particular, the given numbers do not describe an exact soft landing. The constant-rate retro-rocket landing must satisfy both its velocity and height conditions.

Figure 1.
Constant-gravity reserve descent: the formal trajectory crosses ground before its nominal burnout
. **Details**.

8

Solution

Axis theorems. The parallel axis theorem states

I_{O} = I_{G} + m d^{2},

where

I_{G}

is the moment of inertia about an axis through the centre of mass,

I_{O}

is about a parallel axis,

m

is the body's total mass, and

d

is the perpendicular separation of the axes. Expanding

\int [(x - d)^{2} + y^{2}] d m

proves the result because

\int x d m = 0

at the centre of mass. The perpendicular axis theorem applies to a planar lamina in the

x y

plane:

I_{z} = I_{x} + I_{y} .

All three axes meet at the same point,

x, y

lie in the lamina and

z

is normal to it; the identity follows from

x^{2} + y^{2}

being the squared distance to the

z

axis. It is not a theorem for an arbitrary three-dimensional body.

Cylinder integrals. Put the cylinder's midpoint at

z = 0

and its symmetry axis along

z

. Its uniform mass density is

ρ = m / (π r^{2} l)

. The transverse moment of inertia of a uniform solid cylinder about the

x

axis is

I_{⊥, G} = ρ \int_{- l /2}^{l /2} d z \int_{0}^{r} s d s \int_{0}^{2 π} d φ (z^{2} + s^{2} sin^{2} φ) = \frac{m l ^{2}}{12} + \frac{m r ^{2}}{4} .

Equivalently each thin circular slice has diametral moment of inertia

r^{2} d m /4

, by the perpendicular axis theorem and symmetry, and shifting it by

z

adds

z^{2} d m

. The parallel axis theorem then gives the transverse end-axis moment of inertia:

I_{⊥, G} = \frac{m}{12} (l^{2} + 3 r^{2}), I_{⊥, end} = \frac{m}{12} (4 l^{2} + 3 r^{2}) .

The axial moment of inertia of a uniform solid cylinder is

I_{∥} = m r^{2} /2

. Equating it to

I_{⊥, G}

gives

l^{2} = 3 r^{2}, l = 3 r .

Here the source's principal axis means the cylinder's symmetry axis; its transverse centre axes are also principal axes. This equality describes a short cylinder rather than the thin-rod limit.

Hammer dynamics and forces. Treat the compact head as a point mass and neglect its intrinsic moment of inertia. Let

θ

be the displacement from downward vertical, and let

e_{r}

point along the rod from pivot to head, with

e_{θ}

in the direction of increasing

θ

. Define

m_{T} = M + m, A = Ml + \frac{m l}{2}, d = \frac{A}{m _{T}}, I = M l^{2} + m (\frac{l ^{2}}{3} + \frac{r ^{2}}{4}) .

The two external weights

M g

and

m g

act at distances

l

and

l /2

from the pivot. The frictionless pivot supplies a reaction

R

but no torque about its axis. The total restoring torque and rotational equation of motion are

τ = - g A sin θ, I \ddot{θ} + g A sin θ = 0.

For completeness, Newton's second law for the centre of mass gives both components of the pivot force:

R_{r} = - m_{T} d \dot{θ}^{2} - m_{T} g cos θ, R_{θ} = m_{T} d \ddot{θ} + m_{T} g sin θ .

The negative radial component points towards the pivot. These follow from the centre of mass acceleration

- d \dot{θ}^{2} e_{r} + d \ddot{θ} e_{θ}

and the total weight

m_{T} g (cos θ e_{r} - sin θ e_{θ})

. They specify the external forces even when the oscillation is not small.

With the small-angle approximation,

sin θ ≃ θ

, this is simple harmonic motion. The physical pendulum has small-amplitude oscillation period

T_{0} = 2 π \frac{I}{g A} = 2 π \frac{M l ^{2} + m ( l ^{2} /3 + r ^{2} /4 )}{g l ( M + m /2 )} .

For release from a finite amplitude

0 < θ_{0} < π

, conservation of mechanical energy instead gives

\frac{1}{2} I \dot{θ}^{2} = g A (cos θ - cos θ_{0})

. Integrating a quarter-cycle gives the exact oscillation period

T (θ_{0}) = 4 \frac{I}{g A} K (sin \frac{θ _{0}}{2}), K (k) = \int_{0}^{π /2} \frac{d ψ}{1 - k ^{2} s i n ^{2} ψ},

(10)

where

K

is the complete elliptic integral of the first kind;

K (0) = π /2

recovers

T_{0}

For

m = M /10

and

r = l /20

, compare with the simple pendulum value

T_{s} = 2 π l / g

\frac{T _{0}}{T _{s}} = \frac{1 + \frac{1}{10} ( \frac{1}{3} + \frac{1}{1600} )}{1 + \frac{1}{20}} = \frac{49603}{50400} ≃ 0.992062.

(11)

For small oscillations, the hammer's oscillation period is about

0.794%

shorter. The extra rod weight adds restoring torque proportionately faster than it adds pivot moment of inertia.

9

a

Solution

Assume the station's mass remains fixed through an instantaneous kick, with subsequent motion governed only by the Earth's Newtonian gravitational field. Put

κ = G M_{E}

, and work per unit station mass. The initial circular orbit has specific angular momentum

j

and specific orbital energy

ε_{0}

given by

v_{0}^{2} = \frac{κ}{r _{o}}, j^{2} = κ r_{o}, ε_{0} = - \frac{κ}{2 r _{o}} .

Let

δ

denote the fractional increase of kinetic energy. The position and potential energy are unchanged during the kick, so

ε = ε_{0} + δ \frac{κ}{2 r _{o}} = - \frac{κ ( 1 - δ )}{2 r _{o}} .

Unchanged angular momentum at unchanged position fixes the tangential velocity; the added kinetic energy must be radial. Thus this is a radial kick, not a purely tangential speed increase. The new orbital eccentricity follows from

e^{2} = 1 + \frac{2 ε j ^{2}}{κ ^{2}} = δ .

At either radial turning point, the Kepler radial energy equation gives

ε = j^{2} / (2 r^{2}) - κ / r

. Its two roots, equivalently the periapsis and apoapsis of the elliptic orbit, are

r_{p} = \frac{r _{o}}{1 + δ}, r_{a} = \frac{r _{o}}{1 - δ} .

For

δ = 0.01

e = 0.1

and

r_{p} = \frac{10}{11} r_{o}, r_{a} = \frac{10}{9} r_{o} .

An inward or outward kick gives the same pair of radial turning points, although it places the station at a different phase of the orbit. These two finite positive radii require

0 \leq δ < 1

b

Solution

Retain the unchanged-angular momentum assumption of the preceding part. The initial orbit has radius

r_{o} = 6400 + 500 = 6900 km

, and the desired pericentre distance is

r_{p} = 6400 + 100 = 6500 km

. Using

r_{p} = r_{o} / (1 + e)

gives

e = \frac{6900}{6500} - 1 = \frac{4}{65} .

The required fractional increase in initial kinetic energy is therefore

\frac{Δ K}{K _{0}} = e^{2} = \frac{16}{4225} ≃ 0.00378698 = 0.378698%.

Here

Δ E = Δ K > 0

because the instantaneous kick does not change potential energy. Since the initial total orbital energy is

E_{0} = - K_{0}

, an alternative convention for the requested fractional energy change gives

\frac{Δ E}{∣ E _{0} ∣} = + 0.00378698, \frac{Δ E}{E _{0}} = - 0.00378698.

The total orbital energy becomes less negative. Explicitly specifying its denominator avoids interpreting a positive energy input as a decrease of energy.

c

Solution

Let

D

be the separation of the Earth and Moon, and let

x

be the spacecraft's distance from the Earth's centre towards the Moon. On

0 < x < D

, the two gravitational forces oppose each other. Their equality, by Newton's law of universal gravitation, is

\frac{G M _{E}}{x ^{2}} = \frac{G M _{M}}{( D - x ) ^{2}} .

Taking the positive square root gives

x = \frac{D M _{E}}{M _{E} + M _{M}} = \frac{D}{1 + M _{M} / M _{E}} .

The centre of mass of the two bodies is at

x_{G} = D M_{M} / (M_{E} + M_{M})

from the Earth's centre. Thus the gravitational field null between two point masses lies a distance

x - x_{G} = D [\frac{1}{1 + M _{M} / M _{E}} - \frac{M _{M}}{M _{E} + M _{M}}]

from their centre of mass, on the Moon's side. With

M_{E} / M_{M} = 81

x = 9 D /10

and

x_{G} = D /82

, so

x = 3.456 \times 1 0^{8} m, x - x_{G} = \frac{182}{205} D ≃ 3.40917 \times 1 0^{8} m .

The point is

3.84 \times 1 0^{7} m

from the Moon. It is an instantaneous zero of the physical gravitational field, not a co-rotating Lagrange point: a spacecraft remaining fixed relative to the Earth-Moon line would need a nonzero centripetal acceleration. With zero net force it cannot follow that circular co-rotation.

10

Solution

Invariant and collision. For a particle of rest mass

m

, special relativity gives relativistic energy and momentum

E = γm c^{2}, p = γm v, γ = \frac{1}{1 - v ^{2} / c ^{2}} .

Therefore the relativistic energy-momentum relation is

E^{2} - p^{2} c^{2} = γ^{2} m^{2} c^{4} (1 - v^{2} / c^{2}) = m^{2} c^{4} .

It is the squared Minkowski norm of the four-momentum, multiplied by

c^{2}

, and hence is preserved by every Lorentz transformation. Explicitly a boost with speed

U

along

x

gives

E^{'} = Γ (E - U p_{x})

p_{x}^{'} = Γ (p_{x} - U E / c^{2})

, with

p_{y}^{'} = p_{y}

p_{z}^{'} = p_{z}

. Expanding

E^{'2} - c^{2} p_{x}^{'2}

gives

E^{2} - c^{2} p_{x}^{2}

because

Γ^{2} (1 - U^{2} / c^{2}) = 1

. Spatial rotations preserve

p^{2}

, completing the invariance argument for all inertial frames.

For the incoming moving particle,

v = 3 c /5

and

γ = 5/4

. The total laboratory energy and momentum before the perfectly inelastic collision are

E_{i} = (\frac{5}{4} + 2) m_{0} c^{2} = \frac{13}{4} m_{0} c^{2}, p_{i} = \frac{5}{4} m_{0} \frac{3 c}{5} = \frac{3}{4} m_{0} c .

Four-momentum conservation identifies these with the final particle's energy and momentum. Its rest mass is therefore

m^{'} = \frac{E _{i}^{2} - p _{i}^{2} c ^{2}}{c ^{2}} = \frac{m _{0}}{4} 1 3^{2} - 3^{2} = 10 m_{0} .

The result exceeds the sum

3 m_{0}

of the incoming rest masses; some initial kinetic energy has become internal rest energy. Since

p c^{2} / E = v

for a massive particle,

v^{'} = \frac{p _{i} c ^{2}}{E _{i}} = \frac{3}{13} c, γ^{'} = \frac{13}{4 10} .

Two-body decay. Work first in the parent's rest frame, denoted by a star. Four-momentum conservation requires equal and opposite daughter momenta. Their equal rest masses give equal energies:

E_{*} = \frac{m ^{'} c ^{2}}{2} = \frac{10}{2} m_{0} c^{2}, p_{*} c = E_{*}^{2} - (\frac{m _{0} c ^{2}}{2})^{2} = \frac{3}{2} m_{0} c^{2} .

Thus their rest-frame velocities are

v_{*\pm} = \pm 3 c / 10

. Applying the relativistic velocity-addition formula with boost

v^{'} = 3 c /13

gives the two laboratory velocities:

\frac{v _{\pm}}{c} = \frac{3/13 \pm 3/ 10}{1 \pm 9/ ( 13 10 )} .

Their laboratory energies follow directly from the Lorentz transformation of four-momentum:

E_{\pm} = γ^{'} (E_{*} \pm v^{'} p_{*}) = (\frac{13}{8} \pm \frac{9}{8 10}) m_{0} c^{2} .

The corresponding signed laboratory momenta are

p_{\pm} = (\frac{3}{8} \pm \frac{39}{8 10}) m_{0} c .

Their sums recover

E_{i}

and

p_{i}

, and each satisfies

E_{\pm}^{2} - p_{\pm}^{2} c^{2} = (m_{0} c^{2} /2)^{2}

. One daughter moves forwards and the other backwards in the laboratory; both are collinear with the original flight axis. Numerically,

v_{+} ≃ 0.967615 c

v_{-} ≃ - 0.919138 c

E_{+} ≃ 1.980756 m_{0} c^{2}

and

E_{-} ≃ 1.269244 m_{0} c^{2}

2021 (A10287W1)

Source paper: www.whatdotheyknow.com/request/oxford_physics_past_papers_2/response/2340696/attach/9/Physics%20Papers%202020.pdf?cookie_passthrough=1

The cover identifies the ordinary sitting of 7 June 2021, despite the public bundle's filename. Solutions cover all ten questions, including every alternative in Section B.

1

Solution

Neglecting the planets' masses compared with the Sun's mass, Newton's law of universal gravitation supplies the centripetal acceleration of each circular orbit:

\frac{G M _{⊙}}{r ^{2}} = \frac{4 π ^{2} r}{T ^{2}} .

Thus Kepler's third law, calibrated by the Earth's orbital period of one year and orbital radius of one astronomical unit, gives

\frac{T}{1 yr} = (\frac{r}{1 AU})^{3/2}, T_{J} = 5. 2^{3/2} = 11.86 yr, T_{S} = 9. 5^{3/2} = 29.28 yr .

Both orbits are taken to have the same direction. Their relative angular speed is

2 π (1/ T_{J} - 1/ T_{S})

, so the Sun-centred interval between successive conjunctions is the synodic period

T_{syn} = \frac{1}{1/ T _{J} - 1/ T _{S}} = 19.93 yr .

Using the supplied rounded orbital radii explains the small difference from the commonly quoted approximately twenty-year interval.

From the Earth, the two lines of sight are displaced differently because Jupiter and Saturn are at different distances. The Earth's annual motion along its orbit therefore shifts the dates of apparent conjunctions by amounts of order months. The broad recurrence remains about twenty years, but successive apparent conjunctions need not be equally spaced: apparent retrograde motion can produce a group of three conjunctions separated by months during one encounter. The Earth's daily rotation adds a much smaller change of viewpoint. This is a geometrical comparison in the stipulated coplanar model; daylight and the horizon can prevent a particular conjunction from being visible at a chosen surface location.

2

Solution

Take downward distance

s

and downward speed

v = d s / d t \geq 0

, with constant gravitational acceleration

g > 0

and quadratic drag coefficient

k > 0

. The printed vector square has no standard vector meaning; the directionally consistent drag force is

F_{d} = - mk ∣ v ∣ v

. For this downward fall, Newton's second law becomes

\frac{d v}{d t} = g - k v^{2} .

The terminal velocity is obtained by setting the net force to zero:

v_{\infty} = g / k downward .

From rest, the speed increases monotonically towards this limiting value. In fact

v (t) = v_{\infty} tanh (g k t)

; thus it is a supremum approached at infinite time, and a tower of finite height need not permit it to be reached.

For the required distance, use the chain rule

d v / d t = v d v / d s

. Writing

w = v^{2}

gives a linear ordinary differential equation:

\frac{d w}{d s} + 2 k w = 2 g, w (0) = 0, v^{2} (s) = \frac{g}{k} (1 - e^{- 2 k s}) .

Consequently,

\frac{v ^{2}}{v _{\infty}^{2}} = \frac{1}{4} = 1 - e^{- 2 k s}, s_{1/2} = \frac{1}{2 k} lo g \frac{4}{3} .

Here

k

has units of inverse length, so the result has the required units of distance and is independent of the falling mass.

3

Solution

First take

a \neq = 0

, put

A = ∣ a ∣

, and suppose

k > 0

. The force is the negative derivative of the potential energy:

U (x) = - \frac{k x}{A ^{2} + x ^{2}}, F (x) = - U^{'} (x) = \frac{k ( A ^{2} - x ^{2} )}{( A ^{2} + x ^{2} ) ^{2}} .

The potential energy is an odd function, passes through

U (0) = 0

, and approaches zero from above as

x \to - \infty

and from below as

x \to + \infty

. Its stationary points are

x = \pm A

U (- A) = \frac{k}{2 A}, U (A) = - \frac{k}{2 A}, U^{''} (- A) = - \frac{k}{2 A ^{3}}, U^{''} (A) = \frac{k}{2 A ^{3}} .

Thus the sole stable equilibrium is the local minimum at

x = A

; the local maximum at

x = - A

is an unstable equilibrium. The requested values at the stable equilibrium are

x_{stable} = A, U_{stable} = - \frac{k}{2 A}, F_{stable} = 0.

The force is an even function, positive for

∣ x ∣ < A

, negative for

∣ x ∣ > A

, and zero at

x = \pm A

. Its central maximum is

F (0) = k / A^{2}

. Since

F^{'} (x) = \frac{2 k x ( x ^{2} - 3 A ^{2} )}{( A ^{2} + x ^{2} ) ^{3}},

its two minima occur at

x = \pm 3 A

, with

F = - k / (8 A^{2})

, and

F \to 0^{-}

at both ends. These values fix the shapes of both requested sketches.

Figure 1.
Dimensionless potential energy and force for positive k and nonzero a, with the stable and unstable equilibria marked
. **Details**.

In one dimension, attraction towards the origin means

x F (x) < 0

, while repulsion from the origin means

x F (x) > 0

. With that explicitly stated convention,

attraction: repulsion: - A < x < 0 or x > A, x < - A or 0 < x < A .

A positive force alone means rightward, not necessarily outward. At the origin

F (0) \neq = 0

, so it is not an equilibrium; the inward/outward distinction is undefined exactly there.

The sign assumptions matter because the paper calls

a, k

merely constants. For

k < 0

both curves reverse sign, the stable equilibrium is

x = - A

, its potential energy is

- ∣ k ∣/ (2 A)

, and the attraction/repulsion intervals interchange. For

k = 0

the force vanishes everywhere on the nonsingular domain and there is no isolated potential energy minimum. If

a = 0

and

k \neq = 0

, then

U = - k / x

for

x \neq = 0

and

F = - k / x^{2}

, with no finite equilibrium.

4

Solution

Choose the vertical

z

axis downward, and measure the polar angle

θ

from it. With azimuth

ϕ

and fixed rod length

r

, the Cartesian coordinates are

x = r sin θ cos ϕ, y = r sin θ sin ϕ, z = r cos θ .

The massless rod has no kinetic energy. Differentiating these coordinates and adding the squared components of velocity yields

v^{2} = \overset{x}{˙}^{2} + \overset{y}{˙}^{2} + \overset{z}{˙}^{2} = r^{2} (\dot{θ}^{2} + sin^{2} θ \dot{ϕ}^{2}) .

The gravitational potential energy is

V = - m g z = - m g r cos θ

, up to an arbitrary constant. The Lagrangian of this spherical pendulum is therefore

L = \frac{1}{2} m r^{2} (\dot{θ}^{2} + sin^{2} θ \dot{ϕ}^{2}) + m g r cos θ .

Applying the Euler-Lagrange equation to the two generalized coordinates gives

m r^{2} \ddot{θ} - m r^{2} sin θ cos θ \dot{ϕ}^{2} + m g r sin θ = 0, \frac{d}{d t} (m r^{2} sin^{2} θ \dot{ϕ}) = 0.

Equivalently, away from the coordinate singularities at the poles,

\ddot{θ} = sin θ cos θ \dot{ϕ}^{2} - \frac{g}{r} sin θ, \ddot{ϕ} + 2 cot θ \dot{θ} \dot{ϕ} = 0.

Since

L

is independent of

ϕ

, the azimuth is a cyclic coordinate. Its conjugate momentum

p_{ϕ} = m r^{2} sin^{2} θ \dot{ϕ}

is the conserved component of angular momentum about the downward vertical axis. The polar angle is not a cyclic coordinate. Using an upward

z

axis instead reverses the sign of the cosine term in

L

and of the corresponding gravity term; it describes the same motion with the opposite polar angle convention.

5

Solution

Orient the

x

axis along the displacement from the earlier event to the later one, so

Δ x = d = 1000 m

and

Δ t = 1 0^{- 6} s

. For a parallel Lorentz transformation, simultaneity in the second inertial frame requires

Δ t^{'} = γ (Δ t - \frac{v Δ x}{c ^{2}}) = 0, v = \frac{c ^{2} Δ t}{d} = 8.99 \times 1 0^{7} m s^{- 1} ≃ 0.300 c .

The second inertial frame moves towards the later event's location. The positive value

d^{2} - c^{2} Δ t^{2}

establishes spacelike separation, which permits this change of simultaneity. For a general velocity

V

, the condition is

V \cdot Δ r = c^{2} Δ t

: only its component along the separation is fixed at the displayed value, and transverse components are allowed provided

∣ V ∣ < c

. Thus the parallel velocity has the smallest possible speed.

A physical inertial frame must have

∣ v ∣ < c

∣Δ t ∣ < \frac{d}{c} = 3.336 μ s .

There is no attained maximum:

d / c

is the supremum. For the concrete boundary case

Δ t = 1000 m / c

, the events have null separation and the formula would require

v = c

; no inertial frame can make them simultaneous.

6

Solution

Use the Earth's inertial frame, with the Earth at

x = 0

and the star at

x = D = 4.2

light years, neglecting their relative motion. Take the stated relative velocity of

B

to be forward along the same direction as

A

. The relativistic velocity addition law gives

v_{B} = \frac{0.25 c + 0.50 c}{1 + ( 0.25 ) ( 0.50 )} = \frac{2}{3} c .

In units of light years and years,

c = 1

and the spacecraft worldlines are

x_{A} = t /4

and

x_{B} = 2 t /3

. Their passages at the star occur at

t_{B *} = \frac{4.2}{2/3} = 6.3 yr, t_{A *} = \frac{4.2}{1/4} = 16.8 yr .

The return radio wave from

B

follows the leftward light worldline

x = 4.2 - (t - 6.3) = 10.5 - t .

Intersecting this with

A

's worldline gives

t /4 = 10.5 - t

, hence

t_{A \leftarrow B} = 8.4 yr, x_{A \leftarrow B} = 2.1 light years .

This is the Earth-frame coordinate time of reception on

A

, as requested; it is not the reading of

A

's own clock. If an observer on the Earth watched that reception through a returning light signal, they would see it later.

Each transmission from the star takes another

D / c = 4.2

years to reach the Earth. Therefore

t_{\oplus \leftarrow B} = 10.5 yr, t_{\oplus \leftarrow A} = 21.0 yr .

All times are measured from the common passage by the Earth; the spacecraft continue at their stated velocities during the calculation.

7

Solution

Inertia is resistance to a change of motion. Translational inertia is measured by mass in Newton's second law; rotational inertia about a specified fixed axis is measured by the moment of inertia

I = \int r_{⊥}^{2} d m,

where

r_{⊥}

is perpendicular distance from that axis. For a thin uniform rod, take axes perpendicular to the rod. With linear mass density

λ = m / l

I_{CM} = \frac{m}{l} \int_{- l /2}^{l /2} x^{2} d x = \frac{m l ^{2}}{12}, I_{end} = I_{CM} + m (\frac{l}{2})^{2} = \frac{m l ^{2}}{3} .

The second result uses the parallel axis theorem; direct integration over

0 \leq x \leq l

gives the same value. The axis specification is essential: an ideal thin rod has zero moment of inertia about its own longitudinal axis.

During the short elastic collision, neglect the impulse from gravity. The pivot can exert an external impulse, so total linear momentum of rod plus particle need not be conserved. Its torque about the pivot is zero, so conservation of angular momentum about that point applies. Let

w

be the particle's signed outgoing horizontal velocity, and

ω

the rod's outgoing angular velocity, positive towards the incident motion. Then

m l v_{0} = m lw + I_{end} ω, \frac{1}{2} m v_{0}^{2} = \frac{1}{2} m w^{2} + \frac{1}{2} I_{end} ω^{2} .

Substituting

ω = 3 (v_{0} - w) / l

into the kinetic energy equation gives

v_{0}^{2} = w^{2} + 3 (v_{0} - w)^{2}, (v_{0} - w) (v_{0} - 2 w) = 0.

Discarding the zero-impulse branch

w = v_{0}

, the actual collision gives

w = \frac{v _{0}}{2} forward, ω = \frac{3 v _{0}}{2 l} .

The rod's lower end initially moves forward at

l ω = 3 v_{0} /2

, so the bodies separate. Afterwards the free particle undergoes projectile motion. With the pivot as origin and

z

downward, its trajectory is

x = (v_{0} /2) t

z = l + g t^{2} /2

, where

t = 0

at the collision. The rod instead swings about the fixed pivot in a vertical plane as a physical pendulum.

The rod's initial rotational kinetic energy is

I_{end} ω^{2} /2 = 3 m v_{0}^{2} /8

. Reaching the horizontal raises its centre of mass by

l /2

, increasing gravitational potential energy by

m g l /2

. Consequently,

v_{0}^{2} \geq \frac{4 g l}{3} to reach the horizontal, v_{0} = \frac{4 g l}{3} to just reach it at rest .

For the threshold case, measure the rod's angle

θ

from the downward vertical and reset

t = 0

at its first horizontal turning point, so

θ (0) = π /2

and

\dot{θ} (0) = 0

. The restoring torque about the pivot gives the exact equation of motion

\ddot{θ} + \frac{3 g}{2 l} sin θ = 0.

At the horizontal,

\ddot{θ} = - 3 g / (2 l)

, so the rod immediately swings back. Conservation of mechanical energy yields

\frac{1}{2} \frac{m l ^{2}}{3} \dot{θ}^{2} + \frac{m g l}{2} (1 - cos θ) = \frac{m g l}{2}, \dot{θ}^{2} = \frac{3 g}{l} cos θ .

On the initial descent its explicit time dependence is specified by the definite integral

t = \frac{l}{3 g} \int_{θ (t)}^{π /2} \frac{d α}{cos α} .

It continues through the vertical to

- π /2

, reverses, and repeats without damping. An alternative expression valid throughout the oscillation uses the Jacobi elliptic sine. Put

k = 1/ 2

ω_{r} = 3 g / (2 l)

, and let

K (k)

denote the complete elliptic integral of the first kind. Then

θ (t) = 2 arcsin [k sn (K (k) - ω_{r} t, k)], T_{r} = 4 \frac{2 l}{3 g} K (\frac{1}{2}) .

(10)

This is not simple harmonic motion at an amplitude of

π /2

. Its oscillation period exceeds its own small-amplitude value

2 π 2 l / (3 g)

by the factor

2 K (1/ 2) / π ≃ 1.18034

A simple pendulum of length

l /2

obeys

\ddot{θ} + (2 g / l) sin θ = 0

. At the same amplitude

π /2

, its oscillation period is

T_{p} = 4 l / (2 g) K (1/ 2)

, so

\frac{T _{r}}{T _{p}} = \frac{2}{3} > 1.

(11)

The rod swings more slowly. Its equivalent simple pendulum length is

I_{end} / [m (l /2)] = 2 l /3

, not the distance

l /2

to its centre of mass. The same oscillation period ratio holds in the common small-amplitude limit, but replacing the actual 90-degree motion by that limit would underestimate both oscillation periods.

8

Solution

For a satellite of mass

m

moving in the Newtonian gravitational field of a much larger, effectively fixed mass

M

, put

μ = GM

. Its conserved mechanical energy, with gravitational potential energy zero at infinity, is

E = \frac{1}{2} m v^{2} - \frac{GM m}{r} = - \frac{GM m}{2 a} .

Here

r

is the instantaneous centre-to-centre distance,

v

the satellite's speed,

G

the gravitational constant, and

a

the semi-major axis of its bound elliptic orbit. The kinetic energy and potential energy vary around the orbit, while their sum remains constant. Dividing by

m

and solving for

v

gives the vis-viva equation,

v^{2} = μ (2/ r - 1/ a)

For this Hohmann transfer orbit, the periapsis and apoapsis distances are

r_{1} = 1 AU

and

r_{2} = 1.52 AU

, so

a = \frac{r _{1} + r _{2}}{2} = 1.26 AU, v_{1} = \frac{μ}{r _{1}} .

The vis-viva equation gives the endpoint speeds

v_{p} = v_{1} 2 - \frac{1}{1.26} = v_{1} \frac{76}{63}, v_{a} = v_{1} \frac{2}{1.52} - \frac{1}{1.26} .

As a check, conservation of angular momentum at the two endpoints gives

r_{1} v_{p} = r_{2} v_{a}

because each velocity is perpendicular to its radius. Using

1 AU = 1.49598 \times 1 0^{11} m

and

μ = 1.32712 \times 1 0^{20} m^{3} s^{- 2}

gives

v_{1} = 29.78 km s^{- 1}

and

v_{p} = 32.71 km s^{- 1}, v_{a} = 21.52 km s^{- 1} .

The departure impulse must increase the speed in the direction of the initial circular orbit:

Δ v_{1} = v_{p} - v_{1} = 2.93 km s^{- 1} .

These are Sun-centred speeds; leaving the Earth's own gravitational field would require an additional calculation beyond this idealized solar orbit transfer.

r_{2}

, the desired circular orbit speed is

v_{2} = μ / r_{2} = 24.16 km s^{- 1}

. The arrival impulse is therefore

Δ v_{2} = v_{2} - v_{a} = 2.64 km s^{- 1} .

The required thrust on the spacecraft is forward, tangent to its orbit, to accelerate it. The rocket ejects exhaust backwards, opposite to the spacecraft's velocity; this distinguishes the direction of thrust from the direction in which the exhaust is fired.

The Hohmann transfer time is half the orbital period of the transfer ellipse. From Kepler's third law, relative to the Earth's one-year orbital period,

t_{H} = \frac{1}{2} (1.26)^{3/2} yr = 0.7072 yr ≃ 258 days .

Meanwhile Mars's orbital period is

T_{M} = 1.5 2^{3/2} yr = 1.8740 yr

. The spacecraft arrives

π

radians from its departure position. For a rendezvous, Mars must reach that same position after advancing through

2 π t_{H} / T_{M}

. Thus its initial lead angle over the departing spacecraft is

ϕ_{0} = π - \frac{2 π t _{H}}{T _{M}} = π [1 - (\frac{1.26}{1.52})^{3/2}] = 44.1 5^{\circ} .

Mars must be ahead by this angle in the common direction of motion around their orbits. At departure the spacecraft and the Earth share the inner orbital position; arrival occurs on the opposite side of the Sun.

9

Solution

Take

+ y

upwards in the source diagram. Write

w

for

B

's incoming speed, so the initial velocities are

u_{A} = (- u, 0), u_{B} = (\frac{3}{2} w, \frac{1}{2} w) .

A 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

B

's outgoing

x

component of velocity. Conservation of momentum and conservation of kinetic energy give

- 3 u + \frac{3}{2} w = b, 3 u^{2} + w^{2} = b^{2} + \frac{1}{4} w^{2} .

Put

q = (3 /2) w

. The second equation becomes

3 u^{2} + q^{2} = (q - 3 u)^{2}

, so

q = u

. Therefore

w = \frac{2 u}{3}, v_{A} = (0, 0), v_{B} = (- 2 u, \frac{u}{3}) .

The outgoing kinetic energies are

K_{A} = 0, K_{B} = \frac{1}{2} m (4 u^{2} + \frac{u ^{2}}{3}) = \frac{13}{6} m u^{2} .

The outgoing speed of

B

u 13/3

, directed above the negative

x

axis at angle

tan^{- 1} [1/ (23)] ≃ 16. 1^{\circ}

. The initial kinetic energy is also

(3/2 + 2/3) m u^{2} = 13 m 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 gives

V = \frac{3 m u _{A} + m u _{B}}{4 m} = (- \frac{u}{2}, \frac{u}{4 3}), ∣ V ∣ = \frac{13}{4 3} u .

Its direction is again

16. 1^{\circ}

above the negative

x

axis. The separate translational kinetic energies of the constituents and their total are

K_{A} = \frac{13}{32} m u^{2}, K_{B} = \frac{13}{96} m u^{2}, K_{final} = \frac{13}{24} m u^{2} .

The loss of translational kinetic energy is

K_{initial} - K_{final} = \frac{13}{8} m u^{2} .

This 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

(0, h)

. Then conservation of momentum determines

v_{B} = (\frac{3}{2} w - 3 u, \frac{1}{2} w - 3 h),

and conservation of kinetic energy gives only

w (3 u + h) = 2 u^{2} + 4 h^{2} .