# Physics 9702 Doubts | Help Page 211

__Question 1012: [Measurements]__**(a)**The distance between the Sun and the Earth is 1.5 × 10

^{11}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 = 1×10

^{9}m}
1.5 × 10

^{11}m = 1.5 × 10^{2}× 10^{9}m = 150 or 1.5×10^{2}Gm**(b)**

{Distance from equator to satellite = (42.3 – 6.38) ×
10

^{6}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) × 10

^{6}(= 7.184 × 10^{7}m)
{Speed = distance / time. Microwaves are EM waves and
travel at the speed of light.}

(Time =) 7.184 × 10

^{7}/ (3.0 × 10^{8}) = 0.24 (0.239) s**(c)**

{Pressure P = Force / Area}

Units of pressure P: kg m s

^{–2}/ m^{2}= 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 = v

^{2}ρ / P units: (m^{2}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*

__Solution 1014:__**Answer: D.**

Particular care is needed with this type of question, which requires
interpretation of graphs.

Remember that the wave formed is a transverse wave – the direction of
vibration (of the particles on the rope) is perpendicular to the direction of
propagation (which is to the right, as indicated by the arrow).

The particles can only vibrate vertically – that is, they move up and
down. Thus, particle P will only move up and down, it is not moving
horizontally.

Consider the wave some instant later.

After some time, the vertical part (on the right) will be at P –
particle would move according to the form of the wave.

So, at the earliest time there is a sudden jump up, followed by a
constant high value of displacement s and finishing with the negative peak.

The answer is therefore D.

Note that s is the displacement. Positive values of s means that P has
moved upwards and negative values correspond to its downward movement.

__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}
j 15 p22 Ques 2,3

ReplyDeleteplease. thank u

For Q2, see solution 1023 at

Deletehttp://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-213.html

Please consider answering ALL of the following questions:

ReplyDelete4/O/N/02 Q.5(b),Q.6(c)(i)

6/O/N/02 Q.11(a)(b)

6/O/N/03 Q.9

04/M/J/04 Q.8(a),(b)(i),(ii)1.

06/M/J/04 Q.9(b)(iii),Q.11(b)

06/O/N/04 Q.3(b)(i)1.,Q.4(a),Q.6(b),Q.8(a)(i)

04/M/J/05 Q.7(a)

06/O/N/05 Q.8(b),Q.10(a)

04/M/J/06 Q.6(a),(c),Q.7(b)

06/M/J/06 Q.14(b)

04/O/N/06 Q.3(c)

06/O/N/06 Q.3(b)

05/M/J/07 Q.2(d)

04/O/N/07 Q.7(b)(i),(c),Q.10(c)

04/M/J/08 Q.5(b),Q.9(b)

41/O/N/09 Q.6(a),(b)(i),Q.10

51/M/J/10 Q.2(d)

For 41/O/N/09 Q.6(a),(b)(i),, see solution 1021 at

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ReplyDeleteit’s the area of the triangle formed by the x-axis and the straight line that should be considered.

DeleteWhy did we divide resultant velocity by 2?

ReplyDeletefrom the scale, 2cm represent 1.0 m/s

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