Module 5: Advanced Mechanics Quiz

Test Your Understanding

Instructions

This quiz covers Module 5: Advanced Mechanics. Answer all questions, then check your answers at the bottom.

Topics covered:

  • Projectile motion
  • Circular motion
  • Gravitational fields
  • Orbital mechanics

Section A: Multiple Choice (1 mark each)

Question 1

A ball is thrown horizontally at 20 m/s from a height of 45 m. How long does it take to hit the ground?

A. 2.0 s B. 3.0 s C. 4.5 s D. 9.0 s

Question 2

An object moves in a circle at constant speed. Which statement is correct?

A. The velocity is constant B. The acceleration is zero C. The net force is directed toward the centre D. The kinetic energy is changing

Question 3

A satellite orbits Earth at altitude \(h\). If moved to altitude \(2h\), its orbital speed will:

A. Double B. Increase by a factor of \(\sqrt{2}\) C. Decrease D. Remain the same

Question 4

For a projectile launched at angle \(\theta\), maximum range occurs when:

A. \(\theta = 30°\) B. \(\theta = 45°\) C. \(\theta = 60°\) D. \(\theta = 90°\)

Question 5

The gravitational field strength at Earth’s surface is \(g\). At height \(R\) (Earth’s radius) above the surface, it is:

A. \(g/4\) B. \(g/2\) C. \(g\) D. \(2g\)

Section B: Short Answer (3-4 marks each)

Question 6

A projectile is launched at 25 m/s at 53° above horizontal.

  1. Calculate the initial horizontal and vertical velocity components. (2 marks)

  2. Calculate the maximum height reached. (2 marks)

Use: \(u_x = u\cos\theta\), \(u_y = u\sin\theta\), \(H = \frac{u_y^2}{2g}\)

Question 7

A car travels around a horizontal circular track of radius 50 m at constant speed of 15 m/s.

  1. Calculate the centripetal acceleration. (2 marks)

  2. If the car has mass 1200 kg, calculate the friction force required. (2 marks)

Use: \(a_c = \frac{v^2}{r}\), \(F_c = \frac{mv^2}{r}\)

Question 8

Calculate the orbital velocity and period of a satellite orbiting Earth at altitude 400 km.

Data: \(R_E = 6.37 \times 10^6\) m, \(M_E = 5.97 \times 10^{24}\) kg, \(G = 6.67 \times 10^{-11}\) N·m²/kg²

Use: \(v = \sqrt{\frac{GM}{r}}\), \(T = \frac{2\pi r}{v}\)

Section C: Extended Response (5-6 marks each)

Question 9

A tennis ball is served horizontally at 50 m/s from a height of 2.5 m.

  1. Calculate the time for the ball to hit the ground. (2 marks)

  2. Calculate the horizontal distance travelled. (2 marks)

  3. Calculate the velocity (magnitude and direction) when it hits the ground. (3 marks)

Question 10

Explain why astronauts in the International Space Station appear to be “weightless” even though they are still within Earth’s gravitational field. Include a quantitative comparison of gravitational field strength at ISS altitude (400 km) versus Earth’s surface. (5 marks)

Answers

  1. B - Using \(y = \frac{1}{2}gt^2\): \(45 = \frac{1}{2}(9.8)t^2\), so \(t = 3.0\) s

  2. C - Centripetal force is always directed toward the centre

  3. C - Orbital speed \(v = \sqrt{GM/r}\), larger \(r\) means smaller \(v\)

  4. B - Maximum range at 45° (from \(R = \frac{u^2\sin 2\theta}{g}\))

  5. A - At height \(R\): distance from centre is \(2R\), so \(g' = g/4\)

Q6: (a) \(u_x = 25\cos 53° = 15\) m/s, \(u_y = 25\sin 53° = 20\) m/s (b) \(H = \frac{20^2}{2 \times 9.8} = 20.4\) m

Q7: (a) \(a_c = \frac{15^2}{50} = 4.5\) m/s² (b) \(F = 1200 \times 4.5 = 5400\) N

Q8: \(r = 6.37 \times 10^6 + 400 \times 10^3 = 6.77 \times 10^6\) m \(v = \sqrt{\frac{6.67 \times 10^{-11} \times 5.97 \times 10^{24}}{6.77 \times 10^6}} = 7670\) m/s \(T = \frac{2\pi \times 6.77 \times 10^6}{7670} = 5540\) s = 92 min

Q9: (a) \(t = \sqrt{\frac{2h}{g}} = \sqrt{\frac{2 \times 2.5}{9.8}} = 0.71\) s (b) \(x = v_x t = 50 \times 0.71 = 35.7\) m (c) \(v_y = gt = 9.8 \times 0.71 = 7.0\) m/s \(v = \sqrt{50^2 + 7^2} = 50.5\) m/s \(\theta = \tan^{-1}(7/50) = 8.0°\) below horizontal

Q10: Key points: - ISS is in free fall around Earth (orbiting) - Both astronauts and ISS accelerate at same rate - No normal force between astronaut and ISS surfaces - At 400 km: \(g' = g(R/(R+h))^2 = 9.8 \times (6370/6770)^2 = 8.7\) m/s² - Still ~89% of surface gravity, so not truly “zero gravity”

Score Guide

Score Performance
90-100% Excellent - Ready for HSC
75-89% Good - Minor revision needed
60-74% Satisfactory - Review weak areas
Below 60% More study required