# Physics 9702 Doubts | Help Page 140

__Question 691: [Waves > Superposition]__**(a)**Explain the principle of superposition.

**(b)**Sound waves travel from source S to a point X along two paths SX and SPX, as shown in Fig.1.

(i) State phase difference between
these waves at X for this to be the position of

1. a minimum,

2. a maximum.

(ii) Frequency of the sound from S
is 400 Hz and speed of sound is 320 m s

^{–1}. Calculate wavelength of the sound waves.
(iii) Distance SP is 3.0 m and
distance PX is 4.0 m. Angle SPX is 90°. Suggest whether a maximum or a minimum
is detected at point X. Explain your answer.

**Reference:**

*Past Exam Paper – June 2013 Paper 23 Q5*

__Solution 691:__**(a)**When waves overlap / meet, the resultant displacement is the sum of the individual displacements of the waves.

**(b)**

(i)

1. Phase difference for minimum =
180 º / (n + ½) 360 º

2. Phase difference for maximum = 0
/ 360 º / (n360 º)

(ii)

Speed v = f λ

Wavelength λ = 320 / 400 = 0.80 m

(iii)

{Path difference = 7 – 5 =
2m. Wavelength λ = 0.80m. In terms of λ, a path difference of 2m
is equal to 2 × (λ /
0.80) = 2.5 λ.}

Path difference = 7 – 5 = 2 (m) =
2.5 λ

Hence a minimum is detected at point
X.

OR maximum if phase change at P is
suggested

{If a phase change
occurred at point P, then a maximum could be obtained at point X. The condition
for destructive interference: 180 º / (n + ½) 360 º is for coherent sources (waves
from these sources are in phase)}

__Question 692: [Kinematics + Dynamics]__
A trolley moves down a slope, as
shown in Fig.1.

Slope makes an angle of 25° with the
horizontal. A constant resistive force F

_{R}acts up the slope on trolley.
At time t = 0, trolley has velocity
v = 0.50 m s

^{−1}down the slope.
At time t = 4.0 s, v = 12 m s

^{−1}down the slope.**(a)**

(i) Show that acceleration of the
trolley down the slope is approximately 3 m s

^{−2}.
(ii) Calculate distance x moved by
the trolley down the slope from time t = 0 to t = 4.0 s.

(iii) On Fig.2, sketch variation
with time t of distance x moved by the trolley.

**(b)**Mass of the trolley is 2.0 kg.

(i) Show that component of the
weight of the trolley down the slope is 8.3 N.

(ii) Calculate resistive force F

_{R}.**Reference:**

*Past Exam Paper – November 2014 Paper 21 Q4*

__Solution 692:__**(a)**

(i)

Acceleration a = (v – u) / t or (12
– 0.5) / 4

Acceleration a = (12 – 0.5) / 4 =
2.9 (2.875) (= approximately 3 m s

^{–2})
(ii) {Distance
travelled = Average speed × time}

Distance x = (u + v) t / 2 = [(12 +
0.5) × 4] / 2 = 25 m

(iii) A line with increasing
gradient is drawn with a non-zero gradient at origin.

{The gradient of a distance-time graph gives the velocity.

There is a constant acceleration on the trolley, so its speed
increases with time. That is, the distance travelled in a unit time becomes
greater with time. This is represented by an increase in gradient of the graph.

Now, initially, the speed of the trolley is 0.5ms

^{-1}. So, the gradient cannot be zero at the origin.}(graph can be drawn better)

**(b)**

(i) Weight down slope = 2 × 9.81 ×
sin 25° = 8.29 / 8.3 N

(ii)

(Resultant force F = ma)

8.3 – F

_{R}= 2 × 2.9
Resistive force F

_{R}= 2.5 N (2.3N if 3 used for a)

__Question 693: [Gravitation]__**(a)**Define

*gravitational potential*at a point.

**(b)**Gravitational potential φ at distance r from point mass M is given by expression

φ = – GM / r

where G is the gravitational
constant.

Explain significance of the negative
sign in this expression.

**(c)**A spherical planet may be assumed to be an isolated point mass with its mass concentrated at its centre. Small mass m is moving near to, and normal to, surface of the planet. Mass moves away from the planet through a short distance h.

State and explain why change in
gravitational potential energy ΔE

_{P}of the mass is given by the expression
ΔE

_{P}= mgh
where g is the acceleration of free
fall.

**(d)**Planet in (c) has mass M and diameter 6.8 × 10

^{3}km. The product GM for this planet is 4.3 × 10

^{13}N m

^{2}kg

^{–1}.

A rock, initially at rest a long
distance from planet, accelerates towards the planet. Assuming that the planet
has negligible atmosphere, calculate speed of the rock as it hits the surface
of the planet.

**Reference:**

*Past Exam Paper – June 2012 Paper 41 & 43 Q1*

__Solution 693:__**(a)**The gravitational potential at a point is the work done in bringing unit mass from infinity (to the point).

**(b)**The gravitational force is (always) attractive.

EITHER as the distance r decreases, the
object/mass/body does work

OR work is done by masses as they
come together

**(c)**

EITHER

Force on the mass = mg

(where g is the acceleration of free
fall /gravitational field strength)

Gravitational field strength, g = GM
/ r

^{2}
if r is at h {that is, the distance of the mass m is at a distance h from the point
mass M – h is a distance close to the surface.}, g is constant

{This is the same as the
acceleration of free fall on Earth. Close to the surface of Earth, g is about
9.81ms

^{-2}. However, as we go further away from the Earth into the space, the value changes. For g to be constant, we should be close to the surface of the Earth.
As stated in the question
above, h is a short distance from the surface of the planet. So, the distance r
should also be a short distance from the surface of the planet for g to be
constant. If r is at h, then it is close to the surface of the planet.}

ΔE

_{P}= force × distance moved = mgh
OR

{The following formula
relates the gravitational potential energy to the gravitational potential. This
is obvious from the definition of the gravitational potential at a point.}

Change in gravitational potential
energy, ΔE

_{P}= mΔφ
{Consider the distance of
the small mass changing from r

_{1}to r_{2}.
1/r

_{1}– 1/r_{2}^{ }= (r_{2}– r_{1}) / r_{1}r_{2}}
ΔE

_{P}= mΔφ = GMm(1/r_{1}– 1/r_{2}) = GMm(r_{2}– r_{1}) / r_{1}r_{2}
if r

_{2}≈ r_{1}(r_{2}is approximately equal to r_{1}– that h is small), then (r_{2}– r_{1}) = h and r_{1}r_{2}= r^{2}(since r_{1}is assumed to be approximately equal to r_{2}, the product is either r_{1}^{2}or r_{2}^{2}. Let call it r^{2}.)
{ΔE

_{P}= GMm h / r^{2}}
let g = GM / r

^{2}
ΔE

_{P}= mgh**(d)**

{From the conservation of
energy, the potential energy is converted to kinetic energy.}

½ mv

^{2}= mΔφ
{Distance r is the radius,
not the diameter. So, the diameter should be halved.}

v

^{2}= 2 × GM/r = (2 × 4.3 × 10^{13}) / (3.4 × 10^{6})
Speed v = 5.0 × 10

^{3}m s^{–1}

__Question 694: [Dynamics > Equilibrium]__
Climber is supported by a rope on a
vertical wall, as shown in Fig.1.

Weight W of the climber is 520 N.
The rope, of negligible weight, is attached to climber and to a fixed point P
where it makes an angle of 18° to the vertical. Reaction force R acts at
right-angles to the wall.

The climber is in equilibrium.

**(a)**State conditions necessary for the climber to be in equilibrium.

**(b)**Complete Fig.2 by drawing a labelled vector triangle to represent the forces acting on the climber.

**(c)**Resolve forces or use vector triangle to calculate

(i) tension T in the rope,

(ii) reaction force R.

**(d)**Climber moves up the wall and the angle the rope makes with the vertical increases. Explain why magnitude of the tension in the rope increases.

**Reference:**

*Past Exam Paper – June 2011 Paper 23 Q2*

__Solution 694:__**(a)**

Resultant moment = zero / sum of
clockwise moments = sum of anticlockwise moments

Resultant force = 0

**(b)**

shape and orientation correct and
forces labelled and arrows correct

angles correct / labelled

{Tension T in rope =
Weight W + Reaction R}

**(c)**

(i) {The
vertical forces should be balanced.}

T cos18° = W

Tension T = 520 / cos18° = 547 N (Scale diagram: ± 20 N)

(ii) {The
horizontal forces should be balanced.}

Reaction force R = T sin18° = 169 N (Scale diagram: ± 20 N)

**(d)**The angle θ is larger, hence cos θ is smaller. Tension T = W / cos θ. Hence the tension T is larger.

How to do question 8 c ii? I don't get why you subtract the 17.7 but you add the binding ebergy per nucleon of helium?

ReplyDeleteTo which paper are you referring to?

Deletenovember 2009 paper 41. sorry, forgot to include that

DeleteSee question 710 at

Deletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-144.html

11/O/N/10 Q.1,24,31

ReplyDeleteFor Q1, see solution 697 at

Deletehttp://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-141.html

For Q24, see solution 440 at

http://physics-ref.blogspot.com/2015/03/physics-9702-doubts-help-page-84.html

s13_13 question number 5(iii) how are you able to obtain 7-5?? how did the 7 get there for the phase difference that is! please help!

ReplyDeletePossible paths from S to X.

Delete1st path: S to X directly – from Pythagoras’ theorem, this is (√[3^2 + 4^2] =) 5m.

2nd path: (S to P) + (P to X). This is 3 + 4 = 7m.

thankyou so much!! for the reply for this website! really appreciated! ^_^

DeleteQ 692 what does it mean by grad cant be 0 initially?

ReplyDeletethe gradient gives the speed. In the question, it is stated that the initial speed is 0.5m/s, so it cannot be zero.

Delete