Year 12 Physics Module 5: Advanced Mechanics Full Module 60 min

Module 5 Quiz: Advanced Mechanics

Comprehensive assessment covering all of Module 5: projectile motion, circular motion, gravitational fields, orbital mechanics, and energy in orbits. This is a full module examination.

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What This Quiz Covers

IQ1: Projectile Motion

Lessons 1–5

Vector components of launch velocity
Time of flight, range, max height
Launching from a height
Complementary angles for equal range
Effect of changing $g$ on trajectory
Parabolic trajectory derivation

IQ2: Circular Motion

Lessons 6–11

Centripetal acceleration and force
Conical pendulum
Banked curves — design speed
Vertical circles — minimum speeds
Orbital velocity and geostationary orbits
Energy in bound orbits; escape velocity

IQ3: Gravitational Fields

Lessons 12–16

Newton’s Law of Universal Gravitation
Gravitational field strength at altitude
Gravitational potential energy ($U = -GMm/r$)
Work done in gravitational fields
Gravitational potential and potential gradient
Kepler’s Third Law; Schwarzschild radius

Complete Formula Reference — Module 5

Projectile Motion (IQ1)

$v_x = v\cos\theta$ Horizontal velocity component (m/s)

$v_y = v\sin\theta$ Vertical velocity component (m/s)

$v = \sqrt{v_x^2 + v_y^2}$ Resultant speed

$t_{\text{flight}} = \dfrac{2v\sin\theta}{g}$ Time of flight (from level ground)

$R = \dfrac{v^2\sin(2\theta)}{g}$ Horizontal range (m)

$H_{\text{max}} = \dfrac{v^2\sin^2\theta}{2g}$ Maximum height (m)

$y = x\tan\theta - \dfrac{gx^2}{2v^2\cos^2\theta}$ Trajectory equation (parabola)

$\theta_{\text{above}} + \theta_{\text{below}} = 90°$ Complementary angles for equal range

Circular Motion (IQ2)

$a_c = \dfrac{v^2}{r} = \omega^2 r = \dfrac{4\pi^2 r}{T^2}$ Centripetal acceleration (m/s²)

$F_c = \dfrac{mv^2}{r} = m\omega^2 r$ Centripetal force (N)

$\tan\theta = \dfrac{v^2}{rg}$ Banked curve / conical pendulum angle

$v_{\text{banked}} = \sqrt{rg\tan\theta}$ Design speed on banked curve (m/s)

$v_{\text{top}} = \sqrt{gr}$ Min speed at top of vertical circle

$T = 2\pi\sqrt{\dfrac{L\cos\theta}{g}}$ Period of conical pendulum (s)

$v_{\text{orbit}} = \sqrt{\dfrac{GM}{r}}$ Orbital speed (m/s); $r = R + h$

$T^2 = \dfrac{4\pi^2}{GM}\,r^3$ Orbital period squared (Kepler’s 3rd Law)

$r_{\text{geo}}^3 = \dfrac{GMT^2}{4\pi^2}$ Geostationary orbital radius (m); $T = 24\,$h

Gravitational Fields (IQ3)

$F = G\dfrac{Mm}{r^2}$ Newton’s Law of Universal Gravitation (N)

$g = G\dfrac{M}{r^2}$ Gravitational field strength (N/kg); $r = R + h$

$U = -G\dfrac{Mm}{r}$ Gravitational potential energy (J) — negative!

$E = KE + U = \dfrac{1}{2}mv^2 - G\dfrac{Mm}{r}$ Total mechanical energy in orbit (J)

$E_{\text{bound}} = -G\dfrac{Mm}{2r}$ Total energy for circular orbit (J)

$W = \Delta U = GMm\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right)$ Work done moving mass between radii (J)

$V = -G\dfrac{M}{r}$ Gravitational potential (J/kg)

$g = -\dfrac{dV}{dr}$ Field strength = negative potential gradient

$v_e = \sqrt{\dfrac{2GM}{r}}$ Escape velocity (m/s) — factor of 2!

$R_s = \dfrac{2GM}{c^2}$ Schwarzschild radius (m)

Multiple Choice Questions

25 questions — select the best answer for each.

Projectile Motion

Questions 1–8 — IQ1

Q1. A projectile is launched at 25 m/s at 30° above the horizontal. What is the vertical component of its initial velocity?

A $21.7 \text{ m/s}$

B $12.5 \text{ m/s}$

C $10.8 \text{ m/s}$

D $25.0 \text{ m/s}$

✓ Correct! $v_y = v\sin\theta = 25 \times \sin 30° = 25 \times 0.5 = 12.5 \text{ m/s}$.

✗ Incorrect. Use $v_y = v\sin\theta$. $\sin 30° = 0.5$, so $v_y = 25 \times 0.5 = 12.5 \text{ m/s}$. Option A uses $\cos 30°$, option C incorrectly halves the angle.

Q2. A ball is kicked from level ground at 18 m/s, 45° above horizontal. How long is it in the air? ($g = 9.8 \text{ m/s}^2$)

A $1.30 \text{ s}$

B $1.84 \text{ s}$

C $2.60 \text{ s}$

D $3.67 \text{ s}$

✓ Correct! $t = \frac{2v\sin\theta}{g} = \frac{2 \times 18 \times \sin 45°}{9.8} = \frac{25.46}{9.8} = 2.60 \text{ s}$.

✗ Incorrect. $t = \frac{2v\sin\theta}{g}$. $\sin 45° = 0.707$, so $2 \times 18 \times 0.707 = 25.46$, divided by 9.8 gives 2.60 s. B is half the correct answer (one-way time only).

Q3. A stone is thrown at 15 m/s, 60° above horizontal on Earth ($g = 9.8 \text{ m/s}^2$). What is its horizontal range?

A $9.75 \text{ m}$

B $19.9 \text{ m}$

C $39.8 \text{ m}$

D $11.5 \text{ m}$

✓ Correct! $R = \frac{v^2\sin(2\theta)}{g} = \frac{15^2 \times \sin 120°}{9.8} = \frac{225 \times 0.866}{9.8} = 19.9 \text{ m}$.

✗ Incorrect. $R = \frac{v^2\sin(2\theta)}{g}$. $\sin(120°) = 0.866$, not 0.5. Option A uses $\sin 60°$ instead of $\sin 120°$.

Q4. What is the maximum height reached by a projectile launched at 30 m/s, 50° above horizontal? ($g = 9.8 \text{ m/s}^2$)

A $44.1 \text{ m}$

B $27.0 \text{ m}$

C $53.9 \text{ m}$

D $70.4 \text{ m}$

✓ Correct! $H_{\max} = \frac{v^2\sin^2\theta}{2g} = \frac{30^2 \times \sin^2 50°}{2 \times 9.8} = \frac{900 \times 0.587}{19.6} = 27.0 \text{ m}$.

✗ Incorrect. $H_{\max} = \frac{v^2\sin^2\theta}{2g}$. $\sin 50° = 0.766$, so $\sin^2 50° = 0.587$. Option A omits the factor of 2 in the denominator.

Q5. A ball is thrown horizontally at 12 m/s from a 45 m high cliff. How far horizontally does it travel before hitting the ground? ($g = 9.8 \text{ m/s}^2$, ignore air resistance)

A $36.4 \text{ m}$

B $24.0 \text{ m}$

C $18.2 \text{ m}$

D $48.5 \text{ m}$

✓ Correct! First find fall time: $s = \frac{1}{2}gt^2 \Rightarrow 45 = 4.9t^2 \Rightarrow t = 3.03 \text{ s}$. Then $x = v_x t = 12 \times 3.03 = 36.4 \text{ m}$.

✗ Incorrect. Find time from vertical: $45 = \frac{1}{2}(9.8)t^2$, giving $t = 3.03$ s. Then $x = 12 \times 3.03 = 36.4$ m. Option B incorrectly uses $\sqrt{45/9.8}$.

Q6. Two projectiles are launched at the same speed from level ground. One at 30° has the same range as one launched at what other angle?

A $45°$

B $15°$

C $60°$

D $75°$

✓ Correct! Complementary angles satisfy $\theta_1 + \theta_2 = 90°$, so $30° + 60° = 90°$. Both give the same range because $\sin(2 \times 30°) = \sin(60°) = \sin(120°) = \sin(2 \times 60°)$.

✗ Incorrect. Complementary angles sum to 90°. $\sin(2\theta_1) = \sin(2\theta_2)$ when $\theta_1 + \theta_2 = 90°$. So $90° - 30° = 60°$.

Q7. A projectile is launched on a planet where $g$ is half that of Earth. If launched at the same speed and angle, how do the range and maximum height compare?

A Range halves; max height stays the same

B Range stays the same; max height doubles

C Range doubles; max height doubles

D Range quadruples; max height doubles

✓ Correct! Both $R = \frac{v^2\sin(2\theta)}{g}$ and $H_{\max} = \frac{v^2\sin^2\theta}{2g}$ are inversely proportional to $g$. Halving $g$ doubles both.

✗ Incorrect. Both range and max height have $g$ in the denominator, so both are inversely proportional to $g$. Half $g$ means double both.

Q8. The trajectory equation $y = x\tan\theta - \frac{gx^2}{2v^2\cos^2\theta}$ describes what shape?

A A circle

B A parabola

C A hyperbola

D An ellipse

✓ Correct! The trajectory equation has $y$ as a quadratic function of $x$ (contains $x^2$ term), which describes a parabola.

✗ Incorrect. $y$ is a quadratic function of $x$: $y = Ax - Bx^2$. This is the equation of a parabola opening downward.

Circular Motion

Questions 9–17 — IQ2

Q9. A car drives around a circular track at constant speed. The direction of the centripetal acceleration is:

A Tangential to the circle, in the direction of motion

B Towards the centre of the circle

C Away from the centre of the circle

D Zero, because speed is constant

✓ Correct! Centripetal acceleration always points towards the centre of the circular path, perpendicular to the velocity.

✗ Incorrect. Centripetal acceleration is always directed towards the centre. Even though speed is constant, velocity changes direction, so acceleration is non-zero.

Q10. A 1200 kg car rounds a flat curve of radius 60 m at 20 m/s. What is the magnitude of the centripetal force required?

A $4000 \text{ N}$

B $8000 \text{ N}$

C $12000 \text{ N}$

D $24000 \text{ N}$

✓ Correct! $F_c = \frac{mv^2}{r} = \frac{1200 \times 20^2}{60} = \frac{1200 \times 400}{60} = 8000 \text{ N}$.

✗ Incorrect. $F_c = \frac{mv^2}{r} = \frac{1200 \times 400}{60} = 8000 \text{ N}$. Check that you squared the velocity and divided by radius.

Q11. In a conical pendulum, what provides the horizontal component of tension that produces centripetal force?

A The weight of the bob

B The horizontal component of tension in the string

C A centrifugal force pushing the bob outward

D Air resistance on the bob

✓ Correct! The horizontal component of tension ($T\sin\theta$) provides the centripetal force. The vertical component ($T\cos\theta$) balances the weight. Note: centrifugal force is a fictitious force, not a real force.

✗ Incorrect. The horizontal component of tension provides the centripetal force. Option C references centrifugal force, which is fictitious — there is no real outward force in the inertial frame.

Q12. A curve is banked at 15° with radius 100 m. What is the design speed? ($g = 9.8 \text{ m/s}^2$)

A $12.8 \text{ m/s}$

B $16.2 \text{ m/s}$

C $26.2 \text{ m/s}$

D $32.4 \text{ m/s}$

✓ Correct! $v = \sqrt{rg\tan\theta} = \sqrt{100 \times 9.8 \times \tan 15°} = \sqrt{100 \times 9.8 \times 0.268} = \sqrt{262.6} = 16.2 \text{ m/s}$.

✗ Incorrect. $v = \sqrt{rg\tan\theta}$. $\tan 15° = 0.268$, so $v = \sqrt{100 \times 9.8 \times 0.268} = 16.2$ m/s. Option C forgets the square root.

Q13. A 0.5 kg mass on a 1.2 m string swings in a vertical circle. What is the minimum speed at the top of the circle to maintain circular motion? ($g = 9.8 \text{ m/s}^2$)

A $3.43 \text{ m/s}$

B $4.85 \text{ m/s}$

C $5.88 \text{ m/s}$

D $11.8 \text{ m/s}$

✓ Correct! At the top, minimum speed occurs when tension $\to 0$ and weight alone provides centripetal force: $mg = \frac{mv^2}{r}$, so $v = \sqrt{gr} = \sqrt{9.8 \times 1.2} = \sqrt{11.76} = 3.43 \text{ m/s}$.

✗ Incorrect. At minimum speed, $mg = mv^2/r$, so $v = \sqrt{gr}$. The mass cancels. $v = \sqrt{9.8 \times 1.2} = 3.43$ m/s. Option B incorrectly includes the mass.

Q14. A satellite orbits Earth at altitude 400 km. What is its orbital speed? ($M_E = 5.97 \times 10^{24} \text{ kg}$, $R_E = 6.37 \times 10^6 \text{ m}$, $G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$)

A $6.67 \text{ km/s}$

B $7.22 \text{ km/s}$

C $7.67 \text{ km/s}$

D $11.2 \text{ km/s}$

✓ Correct! $r = R_E + h = 6.37 \times 10^6 + 0.400 \times 10^6 = 6.77 \times 10^6 \text{ m}$. Then $v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.77 \times 10^6}} = \sqrt{5.88 \times 10^7} = 7670 \text{ m/s} = 7.67 \text{ km/s}$.

✗ Incorrect. Remember $r = R_E + h$ (centre-to-centre). $r = 6.37 \times 10^6 + 0.40 \times 10^6 = 6.77 \times 10^6$ m. Then $v = \sqrt{GM/r}$. Option A uses $r = R_E$ only.

Q15. What is the approximate altitude of a geostationary orbit above Earth’s surface? ($M_E = 5.97 \times 10^{24} \text{ kg}$, $R_E = 6.37 \times 10^6 \text{ m}$)

A $20\,200 \text{ km}$

B $26\,400 \text{ km}$

C $35\,800 \text{ km}$

D $42\,200 \text{ km}$

✓ Correct! $T = 24 \text{ h} = 86400 \text{ s}$. From $r^3 = \frac{GMT^2}{4\pi^2}$, $r = \left(\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 86400^2}{4\pi^2}\right)^{1/3} = 4.22 \times 10^7 \text{ m}$. Altitude $h = r - R_E = 4.22 \times 10^7 - 6.37 \times 10^6 \approx 35\,800 \text{ km}$.

✗ Incorrect. $r^3 = GMT^2/(4\pi^2)$ with $T = 86400$ s gives $r = 4.22 \times 10^7$ m. The altitude is $h = r - R_E = 4.22 \times 10^7 - 6.37 \times 10^6 \approx 35\,800$ km. Option D gives the orbital radius, not altitude.

Q16. A satellite in circular orbit has kinetic energy $KE$. What is its total mechanical energy?

A $+KE$

B $-KE$

C $-2KE$

D $+2KE$

✓ Correct! For a circular orbit, $KE = \frac{GMm}{2r}$ and $U = -\frac{GMm}{r} = -2KE$. Therefore $E = KE + U = KE - 2KE = -KE$. The total energy equals $-KE$.

✗ Incorrect. For circular orbits, $KE = \frac{GMm}{2r}$ and $U = -\frac{GMm}{r}$, so $U = -2KE$. Then $E = KE + U = KE - 2KE = -KE$.

Q17. What is the relationship between escape velocity $v_e$ and orbital velocity $v_o$ at the same radius?

A $v_e = v_o$

B $v_e = 2v_o$

C $v_e = \sqrt{2}\,v_o$

D $v_e = \sqrt{3}\,v_o$

✓ Correct! $v_o = \sqrt{GM/r}$ and $v_e = \sqrt{2GM/r} = \sqrt{2} \times \sqrt{GM/r} = \sqrt{2}\,v_o \approx 1.41\,v_o$.

✗ Incorrect. $v_e = \sqrt{2GM/r}$ while $v_o = \sqrt{GM/r}$. The ratio is $\frac{v_e}{v_o} = \sqrt{2}$. The factor of 2 in the escape velocity formula is under the square root.

Gravitational Fields

Questions 18–25 — IQ3

Q18. Two 5.0 kg masses are placed 2.0 m apart. What is the gravitational force between them? ($G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$)

A $8.34 \times 10^{-11} \text{ N}$

B $4.17 \times 10^{-10} \text{ N}$

C $1.67 \times 10^{-9} \text{ N}$

D $3.34 \times 10^{-10} \text{ N}$

✓ Correct! $F = G\frac{Mm}{r^2} = \frac{6.67 \times 10^{-11} \times 5.0 \times 5.0}{2.0^2} = \frac{1.668 \times 10^{-9}}{4} = 4.17 \times 10^{-10} \text{ N}$.

✗ Incorrect. $F = G\frac{m_1 m_2}{r^2}$. Remember to square the distance: $r^2 = 4.0$, not 2.0. Option C forgets to divide by $r^2$.

Q19. What is the gravitational field strength at an altitude of 2000 km above Earth’s surface? ($M_E = 5.97 \times 10^{24} \text{ kg}$, $R_E = 6.37 \times 10^6 \text{ m}$)

A $5.69 \text{ m/s}^2$

B $6.53 \text{ m/s}^2$

C $7.84 \text{ m/s}^2$

D $9.80 \text{ m/s}^2$

✓ Correct! $r = R_E + h = 6.37 \times 10^6 + 2.00 \times 10^6 = 8.37 \times 10^6 \text{ m}$. $g = \frac{GM}{r^2} = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{(8.37 \times 10^6)^2} = \frac{3.98 \times 10^{14}}{7.00 \times 10^{13}} = 5.69 \text{ m/s}^2$.

✗ Incorrect. Remember $r = R_E + h$ (centre-to-centre). $r = 6.37 \times 10^6 + 2.00 \times 10^6 = 8.37 \times 10^6$ m. Option B uses $h$ instead of $r$.

Q20. Which statement about gravitational potential energy $U = -GMm/r$ is correct?

A The negative sign is a mathematical convention with no physical meaning

B The negative sign indicates that gravity is a binding, attractive force

C $U$ becomes positive when $r < 1$ m

D The negative sign can be dropped when calculating work done

✓ Correct! The negative sign in $U = -GMm/r$ is physically meaningful: it reflects that gravity is attractive and work must be done against the field to move a mass to infinity (where $U = 0$).

✗ Incorrect. The negative sign is physically significant — it means energy is required to escape the gravitational field, and $U = 0$ is defined at infinity. It cannot be dropped in calculations.

Q21. How much work is required to move a 100 kg mass from Earth’s surface ($r = R_E$) to an altitude where $r = 2R_E$? ($g_0 = 9.8 \text{ m/s}^2$, $R_E = 6.37 \times 10^6 \text{ m}$)

A $3.13 \times 10^6 \text{ J}$

B $3.13 \times 10^9 \text{ J}$

C $6.24 \times 10^9 \text{ J}$

D $9.80 \times 10^8 \text{ J}$

✓ Correct! $W = GMm\left(\frac{1}{R_E} - \frac{1}{2R_E}\right) = GMm \times \frac{1}{2R_E} = \frac{g_0 R_E^2 m}{2R_E} = \frac{g_0 R_E m}{2} = \frac{9.8 \times 6.37 \times 10^6 \times 100}{2} = 3.13 \times 10^9 \text{ J}$.

✗ Incorrect. $W = GMm(\frac{1}{r_1} - \frac{1}{r_2})$. Using $GM = g_0 R_E^2$, $W = mg_0 R_E^2(\frac{1}{R_E} - \frac{1}{2R_E}) = \frac{mg_0 R_E}{2} = 3.13 \times 10^9$ J. Option A has a power-of-ten error.

Q22. The gravitational potential at a point in Earth’s field is $-40 \times 10^6 \text{ J/kg}$. What does this mean?

A A 1 kg mass at this point has $-40 \times 10^6 \text{ J}$ of kinetic energy

B $40 \times 10^6 \text{ J}$ of work is done by the field per kg moved to this point

C $40 \times 10^6 \text{ J/kg}$ of work must be done against the field to move a mass from this point to infinity

D The gravitational field strength at this point is $40 \times 10^6 \text{ N/kg}$

✓ Correct! Gravitational potential is the work done per unit mass against the field to move from that point to infinity (where $V = 0$). A negative potential means work must be supplied to escape.

✗ Incorrect. Gravitational potential $V = -GM/r$ is the work per unit mass to move to infinity. $V = -40 \times 10^6$ J/kg means $40 \times 10^6$ J of work per kg is needed to reach infinity.

Q23. The relationship between gravitational field strength and gravitational potential is:

A $g = -\frac{dV}{dr}$

B $g = \frac{dV}{dr}$

C $g = -\frac{V}{r}$

D $g = \frac{d^2V}{dr^2}$

✓ Correct! $g = -\frac{dV}{dr}$. The negative sign indicates that the field points in the direction of decreasing potential (towards the mass). The field strength equals the negative potential gradient.

✗ Incorrect. The correct relationship is $g = -\frac{dV}{dr}$. The negative sign is essential: gravity points toward decreasing $r$, where $V$ becomes more negative. Without the minus sign, the direction would be wrong.

Q24. According to Kepler’s Third Law, if a planet’s orbital radius is doubled, its period becomes:

A $\sqrt{2}$ times the original

B $2\sqrt{2}$ times the original

C $2$ times the original

D $4$ times the original

✓ Correct! $T^2 \propto r^3$, so $T \propto r^{3/2}$. If $r$ doubles, $T$ becomes $2^{3/2} = 2\sqrt{2} \approx 2.83$ times the original.

✗ Incorrect. $T^2 \propto r^3$ means $T \propto r^{3/2}$. Doubling $r$: $T \to (2)^{3/2} \times T = 2\sqrt{2}\,T$. Option D confuses $T^2$ with $T$.

Q25. What is the Schwarzschild radius of a black hole with mass $3 \times 10^{31}$ kg? ($G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$, $c = 3.00 \times 10^8 \text{ m/s}$)

A $4.45 \times 10^3 \text{ m}$

B $2.22 \times 10^4 \text{ m}$

C $4.45 \times 10^4 \text{ m}$

D $8.90 \times 10^4 \text{ m}$

✓ Correct! $R_s = \frac{2GM}{c^2} = \frac{2 \times 6.67 \times 10^{-11} \times 3 \times 10^{31}}{(3 \times 10^8)^2} = \frac{4.00 \times 10^{21}}{9.00 \times 10^{16}} = 4.45 \times 10^4 \text{ m} = 44.5 \text{ km}$.

✗ Incorrect. $R_s = \frac{2GM}{c^2}$. Remember the factor of 2. Option B omits the factor of 2. Make sure to square $c$ in the denominator.

Short Answer Questions

8 questions — show all working for full marks.

Written Response

Questions 26–33 — working required

Apply Band 4–5 3 marks

Q26. A ball is thrown at $20 \text{ m/s}$ at $40°$ above the horizontal from the top of a $30 \text{ m}$ high cliff. Find the horizontal range from the base of the cliff and the speed at which it hits the ground below. ($g = 9.8 \text{ m/s}^2$)

Apply Band 5 3 marks

Q27. A $0.5 \text{ kg}$ bob on a $0.8 \text{ m}$ string makes 2 revolutions per second in a horizontal circle. Find the tension in the string and the angle the string makes with the vertical.

Analyse Band 5–6 4 marks

Q28. A $2000 \text{ kg}$ car rounds a banked curve of radius $80 \text{ m}$ banked at $12°$.

(a) Calculate the design speed at which no friction is required. ($g = 9.8 \text{ m/s}^2$)

(b) If $\mu_s = 0.30$, calculate the maximum speed the car can travel without slipping up the bank.

Apply Band 5 4 marks

Q29. A satellite of mass $800 \text{ kg}$ orbits Earth at $r = 7.2 \times 10^6 \text{ m}$ from Earth’s centre. Calculate:

(a) the orbital speed $v$ and period $T$;

(b) the kinetic energy $KE$, gravitational potential energy $U$, and total energy $E$.

($M_E = 5.97 \times 10^{24} \text{ kg}$, $G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$)

Apply Band 4–5 3 marks

Q30. Calculate the gravitational field strength at an altitude of $1500 \text{ km}$ above Earth’s surface. A $2.0 \text{ kg}$ mass is placed at this point — what is its weight? ($M_E = 5.97 \times 10^{24} \text{ kg}$, $R_E = 6.37 \times 10^6 \text{ m}$)

Apply Band 5 3 marks

Q31. Find the minimum energy required to move a $1000 \text{ kg}$ satellite from Earth’s surface to an orbital radius of $1.0 \times 10^7 \text{ m}$. ($M_E = 5.97 \times 10^{24} \text{ kg}$, $R_E = 6.37 \times 10^6 \text{ m}$)

Analyse Band 6 4 marks

Q32. Starting from $V = -\dfrac{GM}{r}$, derive the relationship $g = -\dfrac{dV}{dr}$. Explain the physical meaning of the negative sign in this equation.

Analyse Band 5–6 4 marks

Q33. The star Proxima Centauri has mass $0.12\,M_{\odot}$ ($M_{\odot} = 1.99 \times 10^{30} \text{ kg}$). A planet orbits with period $11.2 \text{ days}$.

(a) Calculate the orbital radius of the planet.

(b) The habitable zone for this star is roughly $0.023\text{--}0.054 \text{ AU}$ ($1 \text{ AU} = 1.50 \times 10^{11} \text{ m}$). Assess whether liquid water could exist on this planet.

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Multiple Choice Answers

Q1: B — $v_y = v\sin\theta = 25 \times 0.5 = 12.5 \text{ m/s}$

Q2: C — $t = \frac{2 \times 18 \times \sin 45°}{9.8} = 2.60 \text{ s}$

Q3: B — $R = \frac{15^2 \times \sin 120°}{9.8} = 19.9 \text{ m}$

Q4: B — $H_{\max} = \frac{30^2 \times \sin^2 50°}{2 \times 9.8} = 27.0 \text{ m}$

Q5: A — $t = \sqrt{\frac{2 \times 45}{9.8}} = 3.03 \text{ s}$; $x = 12 \times 3.03 = 36.4 \text{ m}$

Q6: C — Complementary angles: $90° - 30° = 60°$

Q7: C — Both $R$ and $H_{\max}$ are inversely proportional to $g$

Q8: B — Quadratic in $x$ describes a parabola

Q9: B — Centripetal acceleration always points toward the centre

Q10: B — $F_c = \frac{1200 \times 20^2}{60} = 8000 \text{ N}$

Q11: B — Horizontal component of tension provides $F_c$

Q12: B — $v = \sqrt{100 \times 9.8 \times \tan 15°} = 16.2 \text{ m/s}$

Q13: A — $v_{\min} = \sqrt{gr} = \sqrt{9.8 \times 1.2} = 3.43 \text{ m/s}$

Q14: C — $v = \sqrt{\frac{GM}{6.77 \times 10^6}} = 7.67 \text{ km/s}$

Q15: C — Geostationary altitude $\approx 35\,800 \text{ km}$

Q16: B — $E = KE + U = KE - 2KE = -KE$

Q17: C — $v_e = \sqrt{2GM/r} = \sqrt{2}\,v_o$

Q18: B — $F = \frac{6.67 \times 10^{-11} \times 25}{4} = 4.17 \times 10^{-10} \text{ N}$

Q19: A — $g = \frac{GM}{(8.37 \times 10^6)^2} = 5.69 \text{ m/s}^2$

Q20: B — Negative sign indicates binding, attractive force

Q21: B — $W = mg_0R_E/2 = 3.13 \times 10^9 \text{ J}$

Q22: C — Work against field to reach infinity = $40 \times 10^6$ J/kg

Q23: A — $g = -dV/dr$ (negative potential gradient)

Q24: B — $T \propto r^{3/2}$, so $T \to 2^{3/2}T = 2\sqrt{2}\,T$

Q25: C — $R_s = \frac{2GM}{c^2} = 4.45 \times 10^4 \text{ m}$

Q26 (3 marks) — Projectile from Cliff

Given: $v = 20 \text{ m/s}$, $\theta = 40°$, $h = 30 \text{ m}$, $g = 9.8 \text{ m/s}^2$

$v_x = 20\cos 40° = 15.32 \text{ m/s}$; $v_y = 20\sin 40° = 12.86 \text{ m/s}$ (upward)

Fall time: Downward displacement is $-30$ m. Using $s = ut + \frac{1}{2}at^2$:

$-30 = 12.86t - 4.9t^2 \Rightarrow 4.9t^2 - 12.86t - 30 = 0$

$t = \frac{12.86 + \sqrt{12.86^2 + 4 \times 4.9 \times 30}}{9.8} = \frac{12.86 + 19.42}{9.8} = 3.29 \text{ s}$ (1 mark)

Range: $x = v_x t = 15.32 \times 3.29 = \mathbf{50.4 \text{ m}}$ (1 mark)

Impact speed: $v_y(\text{final}) = 12.86 - 9.8 \times 3.29 = -19.38 \text{ m/s}$

$v = \sqrt{15.32^2 + (-19.38)^2} = \sqrt{234.7 + 375.6} = \sqrt{610.3} = \mathbf{24.7 \text{ m/s}}$ (1 mark)

Q27 (3 marks) — Conical Pendulum

Given: $m = 0.5 \text{ kg}$, $L = 0.8 \text{ m}$, $f = 2 \text{ Hz}$, $\omega = 2\pi f = 4\pi = 12.57 \text{ rad/s}$

Radius of horizontal circle: $r = L\sin\theta$

Vertical: $T\cos\theta = mg$; Horizontal: $T\sin\theta = m\omega^2 r = m\omega^2 L\sin\theta$

From horizontal: $T = m\omega^2 L = 0.5 \times (12.57)^2 \times 0.8 = 0.5 \times 157.9 \times 0.8 = \mathbf{63.2 \text{ N}}$ (1 mark)

From vertical: $\cos\theta = \frac{mg}{T} = \frac{0.5 \times 9.8}{63.2} = \frac{4.9}{63.2} = 0.0775$

$\theta = \cos^{-1}(0.0775) = \mathbf{85.6°}$ (1 mark for method, 1 mark for answer)

Q28 (4 marks) — Banked Curve

(a) Design speed (2 marks):

$v_d = \sqrt{rg\tan\theta} = \sqrt{80 \times 9.8 \times \tan 12°} = \sqrt{80 \times 9.8 \times 0.2126}$

$v_d = \sqrt{166.7} = \mathbf{12.9 \text{ m/s}}$ ($\approx 46.5 \text{ km/h}$)

(b) Maximum speed with friction (2 marks):

At maximum speed, friction acts down the slope. Resolving:

$N\cos\theta = mg + \mu_s N\sin\theta$ and $N\sin\theta + \mu_s N\cos\theta = \frac{mv^2}{r}$

$v_{\max}^2 = rg\left(\frac{\tan\theta + \mu_s}{1 - \mu_s\tan\theta}\right) = 80 \times 9.8 \times \left(\frac{0.2126 + 0.30}{1 - 0.30 \times 0.2126}\right)$

$v_{\max}^2 = 784 \times \frac{0.5126}{0.9362} = 784 \times 0.5475 = 429.2$

$v_{\max} = \sqrt{429.2} = \mathbf{20.7 \text{ m/s}}$ ($\approx 74.5 \text{ km/h}$)

Q29 (4 marks) — Satellite Orbit Energy

(a) Orbital speed and period (2 marks):

$v = \sqrt{\frac{GM}{r}} = \sqrt{\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{7.2 \times 10^6}} = \sqrt{\frac{3.98 \times 10^{14}}{7.2 \times 10^6}} = \sqrt{5.53 \times 10^7}$

$v = \mathbf{7.43 \times 10^3 \text{ m/s}}$ ($7.43 \text{ km/s}$)

$T = \frac{2\pi r}{v} = \frac{2\pi \times 7.2 \times 10^6}{7.43 \times 10^3} = \mathbf{6.09 \times 10^3 \text{ s} = 101.5 \text{ min}}$

(b) Energies (2 marks):

$KE = \frac{1}{2}mv^2 = 0.5 \times 800 \times (7.43 \times 10^3)^2 = \mathbf{2.21 \times 10^{10} \text{ J}}$

$U = -\frac{GMm}{r} = -\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 800}{7.2 \times 10^6} = \mathbf{-4.42 \times 10^{10} \text{ J}}$

$E = KE + U = 2.21 \times 10^{10} - 4.42 \times 10^{10} = \mathbf{-2.21 \times 10^{10} \text{ J}}$

Note: $E = -KE$ and $U = -2KE$ as expected for a circular orbit.

Q30 (3 marks) — $g$ at Altitude

$r = R_E + h = 6.37 \times 10^6 + 1.50 \times 10^6 = 7.87 \times 10^6 \text{ m}$ (1 mark)

$g = \frac{GM}{r^2} = \frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{(7.87 \times 10^6)^2} = \frac{3.98 \times 10^{14}}{6.19 \times 10^{13}}$

$g = \mathbf{6.43 \text{ m/s}^2}$ (1 mark)

$W = mg = 2.0 \times 6.43 = \mathbf{12.9 \text{ N}}$ (1 mark)

Note: Weight is $\approx 31\%$ less than at the surface ($19.6 \text{ N}$), showing $g$ decreases with altitude.

Q31 (3 marks) — Work to Raise Satellite

$W = \Delta U = GMm\left(\frac{1}{R_E} - \frac{1}{r_2}\right)$ (1 mark for formula)

$W = 6.67 \times 10^{-11} \times 5.97 \times 10^{24} \times 1000 \times \left(\frac{1}{6.37 \times 10^6} - \frac{1}{1.0 \times 10^7}\right)$

$W = 3.98 \times 10^{17} \times (1.570 \times 10^{-7} - 1.000 \times 10^{-7})$

$W = 3.98 \times 10^{17} \times 5.70 \times 10^{-8} = \mathbf{2.27 \times 10^{10} \text{ J}}$ (2 marks)

Alternatively: $W = mg_0R_E^2(\frac{1}{R_E} - \frac{1}{r_2}) = 1000 \times 9.8 \times (6.37 \times 10^6)^2 \times 5.70 \times 10^{-8}$

Q32 (4 marks) — Derive $g = -dV/dr$

Derivation (2 marks):

Starting with $V = -\dfrac{GM}{r} = -GMr^{-1}$

$\dfrac{dV}{dr} = -GM \times (-1)r^{-2} = \dfrac{GM}{r^2}$

Since $g = \dfrac{GM}{r^2}$, we have $\dfrac{dV}{dr} = g$

Therefore $\mathbf{g = -\dfrac{dV}{dr}}$

Physical meaning (2 marks):

The gravitational field $g$ points in the direction of decreasing potential (toward the mass). Since potential increases with $r$ (becomes less negative), $\frac{dV}{dr} > 0$, so the negative sign ensures $g$ points inward (radially inward). The negative sign tells us that the field is directed toward regions of lower potential energy.

Q33 (4 marks) — Proxima Centauri Planet

(a) Orbital radius (2 marks):

$M = 0.12 \times 1.99 \times 10^{30} = 2.39 \times 10^{29} \text{ kg}$

$T = 11.2 \text{ days} = 11.2 \times 24 \times 3600 = 9.68 \times 10^5 \text{ s}$

From Kepler’s Third Law: $r^3 = \dfrac{GMT^2}{4\pi^2}$

$r^3 = \dfrac{6.67 \times 10^{-11} \times 2.39 \times 10^{29} \times (9.68 \times 10^5)^2}{4\pi^2}$

$r^3 = \dfrac{1.50 \times 10^{29}}{39.48} = 3.80 \times 10^{27}$

$r = \sqrt[3]{3.80 \times 10^{27}} = \mathbf{1.56 \times 10^9 \text{ m}}$ (1.56 million km)

In AU: $r = \frac{1.56 \times 10^9}{1.50 \times 10^{11}} = \mathbf{0.0104 \text{ AU}}$

(b) Assessment (2 marks):

The planet orbits at $0.0104 \text{ AU}$, which is inside the habitable zone of $0.023\text{--}0.054 \text{ AU}$. (1 mark)

Since it orbits closer than the inner edge of the habitable zone, the planet would likely be too hot for liquid water to exist on its surface. It would receive significantly more stellar flux than Earth, making surface temperatures high enough to boil water. Therefore, liquid water is unlikely to exist on this planet. (1 mark for reasoned conclusion)

Mark Module 5 Quiz as Complete

I have completed this full module assessment and reviewed my answers.

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