• Physics 9702 Doubts | Help Page 5

Question 29: [Gravitation]

(a) Define gravitational field strength.

(b) Spherical planet has diameter 1.2 × 10⁴ km. Gravitational field strength at surface of planet is 8.6 N kg^–1. Planet may be assumed to be isolated in space and to have its mass concentrated at its centre. Calculate mass of planet.

(c) Gravitational potential at point X above surface of planet in (b) is – 5.3 × 10⁷ J kg^–1. For point Y above surface of planet, gravitational potential is – 6.8 × 10⁷ J kg^–1.

(i) State, with reason, whether point X or point Y is nearer to planet.

(ii) Rock falls radially from rest towards planet from one point to other. Calculate final speed of rock.

Reference: Past Exam Paper – June 2009 Paper 4 Q1

Solution 29:

(a)

Gravitational field strength is defined as the (gravitational) force per unit mass

(b)

Gravitational field strength, g = GM / R²

{Radius = diameter / 2. Also, distance should be converted to metre}

8.6 x (0.6x10⁷)² = M x (6.67x10^-11)

Mass of planet, M = 4.6x10²⁴kg

(c)

(i)

EITHER The potential decreases as the distance from the planet decreases {This is similar to potential energy (mgh). Here the planet is the surface. So, as the distance (height) from the planet increases, the potential energy increases and vice versa.}

OR The potential is zero at infinity and point X is closer to zero {hence close to infinity}

OR The potential is proportional to – 1/r and point Y is more negative {the more negative the potential, the smaller is r}

So, the point Y is closer to the planet

(ii)

{Change in potential energy = change in kinetic energy. Change in potential energy = mΔϕ where m is the mass of the rock. Since the rock is at rest, initial velocity = 0. So, change in kinetic energy = ½mv². m can be cancelled on both sides, leading to}

The idea of (change in potential,) Δϕ = ½ v²

(6.8 – 5.3)x10⁷ = ½ v²

Final speed, v = 5.5x10³ms^-1

Question 30: [Circular motion & Simple Harmonic motion]

Vertical peg is attached to edge of horizontal disc of radius r, as shown.

Disc rotates at constant angular speed ω. Horizontal beam of parallel light produces shadow of peg on screen, as shown.

At time zero, peg is at P, producing shadow on screen at S. At time t, disc has rotated through angle θ. Peg is now at R, producing shadow at Q.

(a) Determine,

(i) in terms of ω and t, the angle θ

(ii) in terms of ω, t and r, the distance SQ.

(b) Use answer to (a)(ii) to show that shadow on screen performs simple harmonic motion:

(c) Disc has radius r of 12 cm and is rotating with angular speed ω of 4.7 rad s^–1. Determine, for shadow on screen,

(i) Frequency of oscillation

(ii) Its maximum speed

Reference: Past Exam Paper – June 2009 Paper 4 Q4

Solution 30:

(a)

(i) Angle, θ = ωt

(ii) Distance SQ = r sin ωt

(b)

This (r sin ωt) is the solution of the equation a = – ω²x. a = – ω²x is the (defining) equation of simple harmonic motion

(c)

(i) Frequency, f = ω / 2π = 4.7 / 2π = 0.75Hz

(ii) Speed, v = rω = 4.7 x 12 = 56cms^-1

Question 31: [Hooke’s law]

(a) Metal wire has unstretched length L and area of cross-section A. When wire supports load F, wire extends by amount ΔL. Wire obeys Hooke’s law. Write down expressions, in terms of L, A, F and ΔL, for

(i) Applied stress.

(ii) Tensile strain in wire.

(iii) Young modulus of material of wire.

(b) Steel wire of uniform cross-sectional area 7.9 × 10^–7 m² is heated to temperature of 650 K. It is then clamped between two rigid supports, as shown.

Wire is straight but not under tension and length between supports is 0.62 m. Wire is then allowed to cool to 300 K. When wire is allowed to contract freely, a 1.00 m length of wire decreases in length by 0.012 mm for every 1 K decrease in temperature.

(i) Show that change in length of wire, if it were allowed to contract as it cools from 650 K to 300 K, would be 2.6 mm.

(ii) Young modulus of steel is 2.0 × 10¹¹ Pa. Calculate tension in wire at 300 K, assuming that wire obeys Hooke’s law.

(iii) Ultimate tensile stress of steel is 250MPa. Use this information and answer in (ii) to suggest whether wire will, in practice, break as it cools.

Reference: Past Exam Paper – November 2004 Paper 2 Q5

Solution 31:

(a)

(i) Applied stress = F / A

(ii) Tensile strain in wire = ΔL / L

(iii) Young modulus (= stress / strain) = FL / AΔL

(b)

 (i)

{Decrease in temperature = 650 – 300 = 350K

For a 1.00m length of wire, (since the wire decreases in length by 0.012mm for every 1K decrease in temperature) this would correspond to a decrease of 0.012 x 350.

But this is not the same for a 0.62m wire. This can be worked out by proportion. For a 1.oom, wire, the overall decrease = 0.012 x 350. So, for a 0.62m wire, the corresponding decrease = 0.62 (0.012 x 350)}

Change in length, ΔL = 0.012 x 0.62 x 350 = 2.6mm

(ii)

{Young modulus = stress / strain = FL / AΔL}

2.0x10¹¹ = F (0.062) / [(7.9x10^-7) (2.6x10^-3)]

Tension in the wire, F = 660N

(iii)

{This question can be answered in different ways. Below, if you are considering the EITHER method, then you should consider it on each line and if you are considering the OR method, then it continues after the OR on each line

In the EITHER case, we try to determine the stress when the wire is cold at 300K (assuming it does not snap) and compare it with the ultimate tensile stress. If it is larger, then the wire snaps since the ultimate tensile stress is the maximum stress of the wire before it breaks. In the OR case, we try to calculate the equivalent tension at ultimate tensile stress and compare it with the tension obtained above. If it is less, then the wire snaps since the maximum tension that can occur in the wire is that at ultimate tensile stress}

{Force = tension. EITHER: Stress = tension / area = 660 / (7.9 x 10^-7). OR: Tension = stress x area = 250MPa x (7.9x10^-7)}

EITHER when cold, stress = 660 / (7.9 x 10^-7) = 840 MPa    OR Tension at ultimate tensile stress = 198 N

EITHER This (the stress when cold) is greater than the ultimate tensile stress of steel         OR The tension at ultimate tensile stress is less then tension in the wire (F = 660N) given in part b(ii)

So, the wire will snap.

Question 32: [Current of Electricity]

Fig shows variation with applied potential difference V of current I in electrical component C.

(a)

(i) State, with reason, whether resistance of component C increases or decreases with increasing potential difference.

(ii) Determine resistance of component C at potential difference of 4.0 V.

(b) Component C is connected in parallel with resistor R of resistance 1500 Ω and battery of e.m.f. E and negligible internal resistance, as shown.

(i) On Fig, draw line to show variation with potential difference V of current I in resistor R.

(ii) Hence, or otherwise, use Fig to determine current in battery for e.m.f. of 2.0 V.

(c) Resistor R of resistance 1500 Ω and component C are now connected in series across a supply of e.m.f. 7.0 V and negligible internal resistance. Using information from Fig, state and explain which component, R or C, will dissipate thermal energy at greater rate.

Reference: Past Exam Paper – November 2004 Paper  Q6

Solution 32:

(a)

(i)

The resistance is given by the ratio of V / I (at a point)

EITHER The gradient increases OR Current, I increases more rapidly than V

{So, the resistance decreases.

Note that the resistance is not equal to the gradient of the line, but to the co-ordinates of the point on the line}

(ii)

{From graph, when p.d. = 4.0V} Current = 2.00mA

Resistance of component C (= V / I) = 2000Ω 

(b)

(i)

{As deduced from the graph, the resistance of component C decreases as p.d. increases. This causes the equivalent resistance of the parallel combination to decrease as the p.d. increases. So, current from battery increases. From Kirchhoff’s law, there is the same p.d. across both C and R. So, as p.d. increases, current across R increases linearly. The ratio of V / I of points on the line should always give R = 1500Ω. So, at p.d. = 6.0, I = 4.0mA}

The line drawn is a straight line starting from the origin and passing through point (6.0V, 4.0mA)

(ii)

{Component C and resistor R are connected in parallel, so the currents flowing through them are different. Consider the currents on the resistor R and component C at p.d. = 2.0V from the graph}

The individual currents are 0.75mA {in component C} and 1.33mA {in resistor R}

Current in the battery (= 0.75 + 1.33) = 2.1mA

(c)

{Rate of dissipation of thermal energy = power}

The same current would flow in both resistor R and in component C. The potential difference across component C is larger than that across resistor R {For series connection, the p.d. across a component is proportional to its resistance}. Since power = VI, the power is greater in C.

 

Question 33: [Dynamics > Momentum]

A steel ball of mass 250kg is suspended from the jib of a crane, as illustrated in Fig 3.1. 

In order to demolish a wall, the ball is pulled away from the wall and then released. The ball swings down and hits the wall.

The variation with time t of the speed v of the ball is shown in Fig 3.2.

(a) Using Fig 3.2, determine

(i)  the magnitude of the acceleration of the ball at time t = 0.8 s.

(ii) the distance moved by the ball before it hits the wall, that is, from time t = 0 s to          t = 1.6 s.

(b) Calculate the magnitude of

(i) the change in momentum of the ball during its collision with the wall

(ii) the average force exerted on the wall during the collision.

(c) When the ball hits the wall, 15% of the kinetic energy of the ball is converted to thermal energy in the ball. Calculate the mean temperature rise of the ball. The specific heat capacity of steel is 450 Jkg^-1K^-1.

Reference: D02/P2/Q3

Solution 33:

(a)

(i)

{First, let’s try interpreting the graph. The graph is that of velocity against time. So, gradient represent acceleration and area under graph represents the distance travelled. The ball starts from rest (speed = 0) and its speed increases until time t = 1.6s. The gradient decreases as time increases, so the acceleration decreases until it is zero at t = 1.6s. At t = 1.6s to t = 1.8s, there is a sharp decrease in velocity until it becomes zero. This is the time of the collision. The ball comes to rest after hitting the wall}

Magnitude of acceleration of ball (at time = 0.8s) = gradient of graph at t = 0.8s

{First draw a tangent at t = 0.8s. The line obtained should also pass (depending on how the line is drawn) through points (0.35, 2.6) and (1.20, 5.2). You can use these points to calculate the gradient}

Gradient = (5.2 – 2.6) / (1.20 – 0.35) = 3.06ms^-2        [= 3.1ms^-2]

(ii)

Distance moved by ball before it hits wall = Area under graph between t = 0.0s and t = 1.6s

{We need to count the number of squares. There are different methods to do this. You could actually count the number of squares and multiply it by the area of the smallest square. On x-axis, the smallest division corresponds to 0.05s. On y-axis, the smallest division corresponds to 0.2ms^-1. So, the area of the smallest square = 0.05 x 0.2 = 0.01m.

Alternatively, you could use trapezium rule to obtain the area directly. Choose the method which is best and quickest for you.}

The distance should be about 5.56m

[note that others obtained the distance as 4.9m. I obtained it close to 5.56m]

(b)

(i)

Mass of ball = 250kg

Speed of ball before collision, v_i = 5.2ms^-1

Speed of ball after collision, v_f = 0ms^-1

Change in momentum = mΔv = m(v_f – v_i) = 250 (0 – 5.2) = (–) 1300kgsm^-1

{The magnitude may be given alone}

(ii)

{From Newton’s 2^nd law: Force is equal to the rate of change of momentum}

Magnitude of average force on wall during collision = magnitude of average force on ball

{This is from Newton’s 3^rd law. The 2 forces are equal in magnitude and opposite in direction. The momentum calculated above is actually for the ball}

Average force = Δp / Δt = 1300 / (1.80 – 1.60) = 6500N

(c)

Kinetic energy when ball hits the wall = ½mv² = ½ (250) (5.2²) J

Thermal energy = mcΔθ

where m is the mass of the steel ball, c is the specific capacity of steel and Δθ is the change in temperature

So, mcΔθ = 0.15 (½mv²)

Δθ = 0.15 (½ v²) / c = 0.15 [0.5 x 5.2²] / 450 = 4.51x10^-3K

Mean temperature rise = 4.51x10^-3K

Question 34: [Electromagnetism]

(a) Define magnetic flux density.

(b)

Flat coil consists of N turns of wire and has area A. Coil is placed so that its plane is at angle θ to uniform magnetic field of flux density B, as shown.

Using symbols A, B, N and θ and making reference to magnetic flux in coil, derive expression for magnetic flux linkage through coil.

(c)

(i) State Faraday’s law of electromagnetic induction.

(ii) Magnetic flux density B in coil is now made to vary with time t as shown. Sketch variation with time t of e.m.f. E induced in coil.

Reference: Past Exam Paper – November 2004 Paper 4 Q6

Solution 34:

(a)

Magnetic flux density is defined as being (numerically equal to) the force per unit length on a straight conductor carrying unit current normal to the field

(b)

Magnetic flux through the coil = BAsinθ

Magnetic flux linkage through the coil = BANsinθ

(c)

(i)

Faraday’s law of electromagnetic induction states that the (induced) e.m.f. is proportional to the rate of change of flux (linkage)

(ii)

{As stated by Faraday’s law of electromagnetic induction, the (induced) e.m.f. is proportional to the rate of change of flux (linkage). For the first section, from t = 0 to t = T, B is constant. So, no e.m.f. is induced.

From t = T to T = 2T, B increases at a constant rate (gradient is constant). So, the e.m.f. induced is constant. (The direction of e.m.f. induced can be taken arbitrarily.)

From t = 2T to t = 3T, B changes at a constant rate (gradient is constant) but in the opposite direction. So, the e.m.f. is constant and is induced in the opposite direction. Additionally, the change is greater than from t= T to t = 2T, so the e.m.f. induced is greater}

The graph consists of 2 square sections in the correct positions {as explained above}, and is zero elsewhere. The pulses are in opposite directions. The amplitude of the second pulse is about twice the amplitude of the first.