Thursday, October 15, 2015

Physics 9702 Doubts | Help Page 211

  • 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 VA due to charge A alone is shown in Fig. 3.2.


The variation with distance x of the electric potential VB 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 EK when at infinity.
Use Fig. 3.2 to determine the minimum value of EK 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 VA and VB 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






8 comments:

  1. j 15 p22 Ques 2,3
    please. thank u

    ReplyDelete
    Replies
    1. For Q2, see solution 1023 at
      http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-213.html

      Delete
  2. Please consider answering ALL of the following questions:
    4/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)

    ReplyDelete
    Replies
    1. For 41/O/N/09 Q.6(a),(b)(i),, see solution 1021 at
      http://physics-ref.blogspot.com/2015/10/physics-9702-doubts-help-page-213.html

      Delete
  3. This comment has been removed by the author.

    ReplyDelete
    Replies
    1. it’s the area of the triangle formed by the x-axis and the straight line that should be considered.

      Delete
  4. Why did we divide resultant velocity by 2?

    ReplyDelete
    Replies
    1. from the scale, 2cm represent 1.0 m/s

      Delete

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