HSC Physics Module 5: Advanced Mechanics flashcards

Question 1

The orbital speed of a satellite in a circular orbit depends on …

Accepted Answer

the mass of the planet and the orbital radius. v = √(GM/r). The satellite's own mass cancels when equating gravitational force to centripetal force.

Question 2

The escape velocity from the surface of a planet of radius R is v. What would the escape velocity be from a planet of the same mass but radius 2R?

Accepted Answer

v/√2. v_esc = √(2GM/r). Doubling r divides v by √2.

Question 3

A 1200 kg car rounds a curve of radius 50 m at 15 m s^-1. What centripetal force is required?

Accepted Answer

5400 N. F = mv²/r = 1200 × 225/50 = 5400 N.

Question 4

What is the orbital speed of a satellite in a circular orbit of radius 7.0 × 10^6 m around Earth? (GM_E = 4.0 × 10^14 m³ s^-2) …

Accepted Answer

7.6 km s^-1. v = √(GM/r) = √(4.0 × 10^14/7.0 × 10^6) = √(5.7 × 10^7) = 7.6 × 10³ m s^-1.

Question 5

A 1000 kg satellite orbits Earth at radius 7.0 × 10^6 m. What is its gravitational potential energy? (GM_E = 4.0 × 10^14 m³ s^-2) …

Accepted Answer

−5.7 × 10^10 J. U = −GMm/r = −(4.0 × 10^14 × 1000)/(7.0 × 10^6) = −5.7 × 10^10 J. Potential energy is negative because zero is set at infinity.

Question 6

A ball is thrown upward at an angle and lands at the same height. Compared with its launch velocity, the landing velocity has …

Accepted Answer

the same magnitude, with the vertical component reversed. Horizontal velocity is unchanged. By symmetry the vertical component returns to the same size but points downward.

Question 7

A projectile takes 2.0 s to reach its maximum height. What was its initial vertical velocity?

Accepted Answer

19.6 m s^-1. v_y = 0 at the top: 0 = u_y − gt, so u_y = 9.8 × 2.0 = 19.6 m s^-1.

Question 8

One ball is dropped from a height h. At the same instant another is thrown horizontally from the same height. Ignoring air resistance …

Accepted Answer

they land at the same time. Vertical motion is independent of horizontal motion. Both start with zero vertical velocity and the same acceleration, so they fall for the same time.

Question 9

What is the escape velocity from Earth's surface? (GM_E = 4.0 × 10^14 m³ s^-2, R_E = 6.371 × 10^6 m) …

Accepted Answer

11.2 km s^-1. v_esc = √(2GM/R) = √(2 × 4.0 × 10^14/6.371 × 10^6) = √(1.26 × 10^8) = 1.12 × 10^4 m s^-1.

Question 10

A planet has twice Earth's mass and twice Earth's radius. Its surface gravitational field strength is …

Accepted Answer

4.9 m s^-2. g = GM/R². Mass × 2 and R² × 4 gives g × ½ = 4.9 m s^-2.

Question 11

A projectile is launched at 40 m s^-1 at 45° on level ground. What is its range?

Accepted Answer

163 m. R = u² sin2θ/g = 1600 × sin90°/9.8 = 163 m.

Question 12

In a conical pendulum, the centripetal force on the bob is provided by …

Accepted Answer

the horizontal component of the tension. T sinθ points towards the centre of the horizontal circle. T cosθ balances the weight.

Question 13

A satellite is moved from a circular orbit of radius r to one of radius 2r. Its kinetic energy …

Accepted Answer

halves. KE = GMm/2r, so doubling r halves the kinetic energy. (Total energy becomes less negative; potential energy rises.)

Question 14

A stone is thrown horizontally at 10 m s^-1 from a 20 m high cliff. What is its speed just before it lands?

Accepted Answer

22 m s^-1. v_y² = 2gh = 392, v_y = 19.8 m s^-1. Speed = √(10² + 19.8²) = 22 m s^-1.

Question 15

At the highest point of its flight, a projectile has …

Accepted Answer

horizontal velocity and downward acceleration of 9.8 m s^-2. Only the vertical velocity is zero at the top. Gravity acts throughout, so acceleration is always 9.8 m s^-2 downward.

Question 16

A 500 kg car travels at 20 m s^-1 around a curve of radius 80 m. The centripetal force is …

Accepted Answer

2500 N. F = mv²/r = 500 × 400/80 = 2500 N.

Question 17

A 50 N force is applied perpendicular to a spanner 0.25 m from the nut. The torque is …

Accepted Answer

12.5 N m. τ = rF sin90° = 0.25 × 50 = 12.5 N m.

Question 18

A moon orbits a planet at radius 4.0 × 10^8 m with period 1.0 × 10^5 s. What is the planet's mass?

Accepted Answer

3.8 × 10^27 kg. M = 4π²r³/(GT²) = 39.5 × 6.4 × 10^25/(6.67 × 10^-11 × 1.0 × 10^10) = 3.8 × 10^27 kg.

Question 19

Both masses in a two-body system are doubled and the distance between them is also doubled. The gravitational force …

Accepted Answer

stays the same. F ∝ m_1m_2/r². Numerator × 4, denominator × 4: unchanged.

Question 20

Two projectiles are launched at the same speed, one at 30° and the other at 60°. Compared with the 30° projectile, the 60° projectile has …

Accepted Answer

the same range and a longer flight time. Complementary angles give the same range. The 60° launch has a larger vertical component, so it goes higher and stays in the air longer.

Question 21

A cyclist leans inward while turning a corner on a flat road. The centripetal force is provided by …

Accepted Answer

friction between the tyres and the road. On a flat road the only horizontal force available is friction. Leaning keeps the net torque zero so the cyclist does not topple.

Question 22

A satellite has orbital radius r and period T. Its orbital speed is …

Accepted Answer

2πr/T. It travels the circumference 2πr in one period.

Question 23

A package is released from a plane flying horizontally at 100 m s^-1 at an altitude of 500 m. How far horizontally does it travel before hitting the ground?

Accepted Answer

1000 m. t = √(2 × 500/9.8) = 10.1 s. x = 100 × 10.1 = 1010 m ≈ 1000 m.

Question 24

Compared with the ideal (no air resistance) case, a projectile experiencing air resistance has …

Accepted Answer

a smaller range and smaller maximum height, with an asymmetric path. Drag opposes motion in both directions, reducing height and range. The descent is steeper than the ascent, so the path is not symmetric.

Question 25

A roller-coaster car goes through a vertical loop of radius 10 m. What is the minimum speed at the top of the loop to stay on the track?

Accepted Answer

9.9 m s^-1. At minimum speed the normal force is zero: v = √(gr) = √(9.8 × 10) = 9.9 m s^-1.

Question 26

An object completes one revolution every 0.50 s. Its angular velocity is …

Accepted Answer

12.6 rad s^-1. ω = 2π/T = 2π/0.50 = 12.6 rad s^-1.

Question 27

The Moon has mass 7.35 × 10^22 kg and radius 1.74 × 10^6 m. The gravitational field strength at its surface is …

Accepted Answer

1.6 m s^-2. g = GM/R² = 6.67 × 10^-11 × 7.35 × 10^22/(1.74 × 10^6)² = 1.6 m s^-2.

Question 28

A satellite in low Earth orbit loses energy to atmospheric drag. As its orbit decays, its speed …

Accepted Answer

increases because it moves to a lower orbit. Total energy falls, so r decreases. Since v = √(GM/r), the satellite speeds up even though drag removes energy — the lost energy comes from potential energy.

Question 29

A projectile is launched with an initial vertical velocity of 14 m s^-1. Its maximum height is …

Accepted Answer

10 m. H = u_y²/2g = 196/19.6 = 10 m.

Question 30

For an object launched horizontally, the time of flight depends only on …

Accepted Answer

the height of launch. The vertical motion starts from zero vertical velocity, so t = √(2h/g). Launch speed only changes how far it goes horizontally.

Question 31

A 1500 kg car rounds a flat curve of radius 60 m. The maximum friction force is 7500 N. What is the maximum speed without skidding?

Accepted Answer

17 m s^-1. F = mv²/r, so v = √(Fr/m) = √(7500 × 60/1500) = √300 = 17 m s^-1.

Question 32

The work done by the centripetal force on an object in uniform circular motion is …

Accepted Answer

zero. The force is always perpendicular to the displacement, so W = Fs cos90° = 0. This is why the speed stays constant.

Question 33

A satellite orbits Earth at a radius of 4.2 × 10^7 m. Its period is closest to (GM_E = 4.0 × 10^14 m³ s^-2) …

Accepted Answer

24 hours. T = √(4π²r³/GM) = √(39.5 × 7.4 × 10^22/4.0 × 10^14) = 8.5 × 10^4 s ≈ 24 h. This is the geostationary radius.

Question 34

Gravitational potential energy of a mass in a planet's field is taken as …

Accepted Answer

zero at infinity and negative elsewhere. U = −GMm/r. Choosing U = 0 at infinite separation makes bound objects have negative potential energy.

Question 35

Two balls are launched from the ground at the same speed, one at 45° and the other at 30°. Just before landing …

Accepted Answer

both have the same speed. Energy is conserved and both land at launch height, so both return to the launch speed.

Question 36

An object moves in a circle of radius 2.0 m with a period of 4.0 s. Its speed is …

Accepted Answer

3.1 m s^-1. v = 2πr/T = 2π × 2.0/4.0 = 3.1 m s^-1.

Question 37

A 20 N force is applied to the edge of a door 0.80 m wide, at 45° to the door. The torque about the hinges is …

Accepted Answer

11 N m. τ = rF sinθ = 0.80 × 20 × sin45° = 11.3 N m.

Question 38

The mass of a satellite is doubled while its orbital radius is unchanged. Its orbital speed …

Accepted Answer

is unchanged. v = √(GM/r) depends only on the central mass and radius, not on the satellite's mass.

Question 39

Two projectiles are launched at the same speed, one at 30° and one at 60°. The ratio of their maximum heights (30° : 60°) is …

Accepted Answer

1 : 3. H ∝ sin²θ. sin²30° = 0.25, sin²60° = 0.75. Ratio 1 : 3.

Question 40

A projectile reaches a maximum height of 20 m. The time taken to fall from the top back to launch height is …

Accepted Answer

2.0 s. From rest vertically: h = ½gt², t = √(2 × 20/9.8) = 2.0 s.

Question 41

On a frictionless banked curve at the design speed, the relationship between banking angle θ, speed v and radius r is …

Accepted Answer

tanθ = v²/(rg). N sinθ = mv²/r and N cosθ = mg. Dividing gives tanθ = v²/(rg).

Question 42

The Moon orbits at 3.84 × 10^8 m with a period of 2.36 × 10^6 s. Its centripetal acceleration is about …

Accepted Answer

2.7 × 10^-3 m s^-2. a = 4π²r/T² = 39.5 × 3.84 × 10^8/(2.36 × 10^6)² = 2.7 × 10^-3 m s^-2.

Question 43

The gravitational force between two 1000 kg masses 1.0 m apart is …

Accepted Answer

6.7 × 10^-5 N. F = Gm_1m_2/r² = 6.67 × 10^-11 × 10^6/1 = 6.7 × 10^-5 N.

Question 44

Planet X has four times the mass of Earth. A satellite orbiting X at the same radius as one orbiting Earth would have an orbital speed …

Accepted Answer

twice as large. v = √(GM/r). Four times the mass gives √4 = 2 times the speed.

Question 45

A mass is whirled in a horizontal circle on a string. If the string suddenly breaks, the mass …

Accepted Answer

moves in a straight line along the tangent. With no force, Newton's first law applies: it keeps its instantaneous velocity, which is tangential.

Question 46

A car doubles its speed around the same curve. The centripetal force needed …

Accepted Answer

quadruples. F = mv²/r. Doubling v multiplies F by 4.

Question 47

The orbital radius of a satellite is increased by a factor of 9. Its period increases by a factor of …

Accepted Answer

27. T² ∝ r³. 9³ = 729, √729 = 27.

Question 48

How much work is needed to lift 500 kg from Earth's surface to a height of one Earth radius? (GM_E = 4.0 × 10^14, R_E = 6.371 × 10^6 m) …

Accepted Answer

1.6 × 10^10 J. ΔU = GMm(1/R − 1/2R) = GMm/2R = 4.0 × 10^14 × 500/(2 × 6.371 × 10^6) = 1.6 × 10^10 J. (Using mgh would be wrong — g is not constant over this range.)

Question 49

A projectile has initial horizontal velocity 15 m s^-1 and initial vertical velocity 20 m s^-1. Its launch speed and angle are …

Accepted Answer

25 m s^-1 at 53°. Speed = √(15² + 20²) = 25 m s^-1. tanθ = 20/15, θ = 53°.

Question 50

A 0.50 kg bob swings as a conical pendulum with the string at 30° to the vertical. The tension is …

Accepted Answer

5.7 N. T cosθ = mg, T = 0.50 × 9.8/cos30° = 5.7 N.

Question 51

A passenger in a turning car feels pushed outward. This sensation is best explained by …

Accepted Answer

the passenger's inertia — their body tends to continue in a straight line while the car turns. No outward force acts. The seat and door must push the passenger inward to make them follow the curve; what they feel is their own inertia.

Question 52

For a satellite in circular orbit of radius r, the kinetic energy is …

Accepted Answer

GMm/2r. mv²/r = GMm/r², so ½mv² = GMm/2r.

Question 53

The escape velocity from a planet …

Accepted Answer

is independent of the mass of the escaping object. ½mv² = GMm/R; m cancels, giving v = √(2GM/R).

Question 54

An object is launched horizontally at 20 m s^-1 from a height of 80 m. Its range is …

Accepted Answer

81 m. t = √(2 × 80/9.8) = 4.04 s. Range = 20 × 4.04 = 81 m.

Question 55

A projectile launched on the Moon (g = 1.6 m s^-2) at the same speed and angle as on Earth would have a range about …

Accepted Answer

6 times larger. R = u² sin2θ/g. Range ∝ 1/g, and 9.8/1.6 ≈ 6.

Question 56

An 800 kg car takes a flat curve of radius 40 m at 12 m s^-1. The friction force required is …

Accepted Answer

2880 N. F = mv²/r = 800 × 144/40 = 2880 N.

Question 57

The angular velocity ω of a satellite moving at speed v in an orbit of radius r is …

Accepted Answer

v/r. v = rω, so ω = v/r.

Question 58

A mass is moved from Earth's surface (radius R) to a distance 2R from the centre. Its gravitational potential energy …

Accepted Answer

becomes half as negative. U = −GMm/r. Doubling r halves the magnitude, so U goes from −GMm/R to −GMm/2R.

Question 59

The Moon's orbital radius is 3.84 × 10^8 m. Its orbital speed is about (GM_E = 4.0 × 10^14 m³ s^-2) …

Accepted Answer

1.0 × 10^3 m s^-1. v = √(GM/r) = √(4.0 × 10^14/3.84 × 10^8) = √(1.04 × 10^6) = 1.0 × 10³ m s^-1.

Question 60

A projectile launched at 30° reaches a maximum height of 10 m. Its launch speed was …

Accepted Answer

28 m s^-1. u_y = √(2gH) = √196 = 14 m s^-1. u = u_y/sin30° = 28 m s^-1.

Question 61

A projectile launched at 45° on level ground has a range of 100 m. Its launch speed was …

Accepted Answer

31 m s^-1. R = u²/g at 45°, so u = √(Rg) = √980 = 31 m s^-1.

Question 62

A 1.0 kg mass on a 2.0 m string moves in a vertical circle. At the top its speed is 6.0 m s^-1. The tension at the top is …

Accepted Answer

8.2 N. At the top T + mg = mv²/r, so T = 1.0 × 36/2.0 − 9.8 = 8.2 N.

Question 63

For an object in rotational equilibrium …

Accepted Answer

the net torque is zero. Rotational equilibrium means no net turning effect: the sum of clockwise torques equals the sum of anticlockwise torques.

Question 64

Kepler's second law (equal areas in equal times) implies that a planet in an elliptical orbit moves …

Accepted Answer

fastest when closest to the Sun. To sweep the same area in the same time when the radius is short, the planet must cover more arc, so it moves faster near perihelion.