• 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 × 10³ km. The product GM for this planet is 4.3 × 10¹³ N m² 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 / r²

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 r₁ to r₂.

1/r₁ – 1/r₂ = (r₂ – r₁) / r₁r₂}

ΔE_P = mΔφ = GMm(1/r₁ – 1/r₂) = GMm(r₂ – r₁) / r₁r₂

if r₂ ≈ r₁ (r₂ is approximately equal to r₁ – that h is small), then (r₂ – r₁) = h and r₁r₂ = r² (since r₁ is assumed to be approximately equal to r₂, the product is either r₁² or r₂². Let call it r².)

{ΔE_P = GMm h / r²}

let g = GM / r²

ΔE_P = mgh

(d)

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

½ mv² = mΔφ

{Distance r is the radius, not the diameter. So, the diameter should be halved.}

v² = 2 × GM/r = (2 × 4.3 × 10¹³) / (3.4 × 10⁶)

Speed v = 5.0 × 10³ 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.