• Physics 9702 Doubts | Help Page 211

Question 1012: [Measurements]

(a) The distance between the Sun and the Earth is 1.5 × 10¹¹ m. State this distance in Gm.

(b) The distance from the centre of the Earth to a satellite above the equator is 42.3 Mm. The radius of the Earth is 6380 km.

A microwave signal is sent from a point on the Earth directly below the satellite.

Calculate the time taken for the microwave signal to travel to the satellite and back.

(c) The speed v of a sound wave through a gas of density ρ and pressure P is given by

v = √(CP / ρ)

where C is a constant.

Show that C has no unit.

(d) Underline all the scalar quantities in the list below.

acceleration                 energy             momentum                  power              weight

(e) A boat travels across a river in which the water is moving at a speed of 1.8 m s^–1.

The velocity vectors for the boat and the river water are shown to scale in Fig. 1.1.

In still water the speed of the boat is 3.0 m s^–1. The boat is directed at an angle of 60° to the river bank.

(i) On Fig. 1.1, draw a vector triangle or a scale diagram to show the resultant velocity of the boat.

(ii) Determine the magnitude of the resultant velocity of the boat.

Reference: Past Exam Paper – June 2015 Paper 23 Q1

Solution 1012:

(a)

{1Gm = 1×10⁹m}

1.5 × 10¹¹ m = 1.5 × 10² × 10⁹ m = 150 or 1.5×10² Gm

(b)

{Distance from equator to satellite = (42.3 – 6.38) × 10⁶ m

We need to consider twice this distance as the signal travels from ‘equator to satellite’ and from ‘satellite to equator’.}

Distance = 2 × (42.3 – 6.38) × 10⁶ (= 7.184 × 10⁷ m)

{Speed = distance / time. Microwaves are EM waves and travel at the speed of light.}

(Time =) 7.184 × 10⁷ / (3.0 × 10⁸) = 0.24 (0.239) s

(c)

{Pressure P = Force / Area}

Units of pressure P: kg m s^–2 / m² = kg m^–1 s^–2

{Density = mass / volume}

Units of density ρ: kg m^–3       and speed v: m s^–1

simplification for units of C: C = v² ρ / P       units: (m² s^–2 kg m^–3) / kg m^–1 s^–2

and cancelling to give no units for C

(d) acceleration                       energy             momentum                  power              weight

(e)

(i)

The vector triangle should be of a correct orientation

three arrows for the velocities in the correct directions

(ii)

{Scale: 2cm represents 1.0ms^-1}

Length measured from scale diagram = 5.2 ± 0.2 cm                         OR components of boat speed determined parallel and perpendicular to river flow

Velocity {= 5.2 / 2} = 2.6 m s^–1

Question 1013: [Electric field]

(a) Define electric potential at a point.

(b) Two point charges A and B are separated by a distance of 20 nm in a vacuum, as illustrated in Fig. 3.1.

A point P is a distance x from A along the line AB.

The variation with distance x of the electric potential V_A due to charge A alone is shown in Fig. 3.2.

The variation with distance x of the electric potential V_B due to charge B alone is also shown in Fig. 3.2.

(i) State and explain whether the charges A and B are of the same, or opposite, sign.

(ii) By reference to Fig. 3.2, state how the combined electric potential due to both charges may be determined.

(iii) Without any calculation, use Fig. 3.2 to estimate the distance x at which the combined electric potential of the two charges is a minimum.

(iv) The point P is a distance x = 10 nm from A.

An α-particle has kinetic energy E_K when at infinity.

Use Fig. 3.2 to determine the minimum value of E_K such that the α-particle may travel from infinity to point P.

Reference: Past Exam Paper – November 2013 Paper 43 Q3

Solution 1013:

(a) Electric potential at a point is the work done bringing unit positive charge from infinity (to the point)

(b)

(i)

EITHER Both potentials are positive / same sign, so they have the same sign

OR The gradients are positive & negative (so fields in opposite directions), so they have the same sign

(ii) The individual potentials are summed

(iii) Allow value of x between 10 nm and 13 nm

{This corresponds to a distance where BOTH V_A and V_B have small values of potentials, so that their sum will also be small. So, this cannot correspond to the extreme sections of the graph. E.g. 100+1 = 101 (the sum is big even if one of them is small) while 3+4 = 7 (this sum is relatively small even if both ‘3 and 4’ are bigger than the ‘1’ in the previous case).

Consider point of intersection: sum ≈ 0.218V + 0.218V = 0.436V

Consider points at x = 12nm: sum = 0.18V + 0.24V = 0.42V (this is smaller)}

(iv)

{Minimum potential,}V = 0.43 V (allow 0.42 V → 0.44 V)

{Energy = qV. Alpha particle is a helium nucleus, so its charge q is +2e where e is the charge of an electron.}

Energy = 2 × 1.6 × 10^–19 × 0.43

Energy = 1.4 × 10^–19 J

Question 1014: [Waves]

A wave pulse moves along a stretched rope in the direction shown.

Which diagram correctly shows the variation with time t of the displacement s of the particle P in the rope?

Reference: Past Exam Paper – June 2015 Paper 13 Q26 & March 2018 Paper 12 Q21

Solution 1014:

Go to
A wave pulse moves along a stretched rope in the direction shown. Which diagram correctly shows the variation with time t of the displacement s of the particle P in the rope?

Question 1015: [Kinematics > Linear motion]

An experiment is conducted on the surface of the planet Mars.

A sphere of mass 0.78 kg is projected almost vertically upwards from the surface of the planet. The variation with time t of the vertical velocity v in the upward direction is shown in Fig. 2.1.

The sphere lands on a small hill at time t = 4.0 s.

(a) State the time t at which the sphere reaches its maximum height above the planet’s surface.

(b) Determine the vertical height above the point of projection at which the sphere finally comes to rest on the hill.

(c) Calculate, for the first 3.5 s of the motion of the sphere,

(i) the change in momentum of the sphere,

(ii) the force acting on the sphere.

(d) Using your answer in (c)(ii),

(i) state the weight of the sphere,

(ii) determine the acceleration of free fall on the surface of Mars.

Reference: Past Exam Paper – June 2009 Paper 22 Q2

Solution 1015:

(a) Time t = 2.4s

(b)

{In (b) and (c), allow answers as (+) or (-) }

Distance travelled is the area under the graph line

{Take (area of graph above x-axis) – (area of graph under x-axis)}

Height = (½ × 2.4 × 9.0) – (½× 1.6 × 6.0)

Height = 6.0m

(c)

(i)

Change in momentum = 0.78 × (9.0 + 4.2)     (allow 4.2 ± 0.2)

Change in momentum = 10.3Ns         (allow 10Ns)

(ii)

Force = Δp / Δt            (or mΔv / Δt)

Force = 10.3 / 3.5

Force = 2.9N  

(d)

(i) Weight = 2.9N

(ii)

Acceleration of free fall g = weight / mass

Acceleration of free fall g = 2.9 / 0.78 = 3.7ms^-2