Physics 9702 Doubts | Help Page 113

Physics 9702 Doubts | Help Page 113

Question 573: [Forces > Equilibrium]
Diagram shows a child’s balancing game.

Wooden rod is uniform and all the rings are of equal mass. Two rings are hung on peg 13 and one on peg 1.
On which hook must a fourth ring be hung in order to balance the rod?
A 2 B 3 C 5 D 6

Reference: Past Exam Paper – November 2011 Paper 11 Q15 & Paper 13 Q13

Solution 573:

Answer: C.

Let the weight of 1 ring be W.

For the rod to balance (equilibrium),

Clockwise moment = Anticlockwise moment

The 2 rings on peg 13 causes a clockwise moment.

Clockwise moment = 2[(13 – 8)xW] = 10W

Anticlockwise moment = (8 – 1)W + (8 – y)W

where y is the position of hook where fourth ring must be hung.

For equilibrium,

(8 – 1)W + (8 – y)W = 10W

7W + (8 – y)W = 10W

(8 – y)W = 3W

8 – y = 3 giving y = 8 – 3 = 5

Question 574: [Electric field]

(a) An α-particle and a proton are at rest a distance 20 μm apart in a vacuum, as illustrated in Fig.1.

(i) State Coulomb’s law.

(ii) The α-particle and proton may be considered to be point charges.

Calculate electric force between the α-particle and the proton.

(b)

(i) Define electric field strength.

(ii) A point P is distance x from α-particle along the line joining the α-particle to the proton (see Fig.1). Variation with distance x of electric field strength E_α due to the α-particle alone is shown in Fig.2.

Variation with distance x of the electric field strength E_P due to the proton alone is also shown in Fig.2.

1. Explain why the two separate electric fields have opposite signs.

2. On Fig.2, sketch variation with x of the combined electric field due to the α-particle and the proton for values of x from 4 μm to 16 μm.

Reference: Past Exam Paper – November 2013 Paper 41 & 42 Q4

Solution 574:

(a)

(i) Coulomb’s law states that the (electric) force between 2 point charges is proportional to the product of the 2 charges and inversely proportional to the square of their separation

(ii)

Distance between α-particle and proton, r = 20μm = 20 x 10^-6m

Charge of α-particle (contains 2 protons) = 2 x (1.6x10^-19) C

Charge of a proton, q = 1.6x10^-19 C

ϵ₀ = 8.85 x 10^-12Fm^-1

Electric force, F = Qq/4πϵ₀r²

F = 2 x (1.6x10^-19)² /{4π x 8.85 x 10^-12 x (20 x 10^-6)²} = 1.15 x 10^-18N

{Q is the charge of the α-particle which contains 2 protons and thus has charged +2e = 2 x (1.6x10^-19) C. q is the charge of the proton which is +e = (1.6x10^-19) C. The product Qq is 2 x (1.6x10^-19)²}

(b)

(i) Electric field strength is defined as the force per unit charge acting on EITHER a stationary charge OR a positive charge.

(ii)

Choose any 2:

Electric field is a vector quantity

Electric fields are in opposite directions

The 2 charges repel one another

2. In the graph, the line should always be between the given lines, crossing the x-axis between 11.0μm and 12.3μm. The curve should have a reasonable shape.

{The value of electric field strength at any value of x is the sum of the corresponding values of E_α and E_p at that point – this value would be less than E_α and greater than E_p since E_p is negative.}

[A typical drawing is shown. This can be done in more properly on paper.]

Question 575: [Dynamics > Momentum]
Diagram shows a cannon ball fired from a cannon.

Mass of the cannon is 1000 kg and mass of the cannon ball is 10 kg.
Recoil velocity of the cannon is 5 m s^–1 horizontally.
What is horizontal velocity of the cannon ball?
A 200 m s^–1 B 500 m s^–1 C 2000 m s^–1 D 5000 m s^–1

Reference: Past Exam Paper – November 2006 Paper 1 Q12

Solution 575:

Answer: B.

Before firing the cannon ball, the cannon + cannon ball are stationary. So, there initial momentum is zero.

As the cannon fires the ball, it moves backwards. This is the recoil velocity which is 5ms^-1 horizontally.

From the conservation of momentum, the magnitude of the momentum of the cannon ball (moving forwards) is equal to the magnitude of the momentum of the cannon (moving backwards).
Let the velocity of the cannon ball be v.

10v = 1000 (5)
Speed v = 5000 / 10 = 500 ms^-1

Question 576: [Electromagnetism > Hall Probe]

(a) State relation between magnetic flux density B and magnetic flux Φ, explaining any other symbols you use.

(b) Large horseshoe magnet has uniform magnetic field between its poles. Magnetic field is zero outside the space between poles.

A small Hall probe moved at constant speed along line XY that is midway between, and parallel to, the faces of poles of magnet, as shown in Fig.1.

An e.m.f. is produced by Hall probe when it is in the magnetic field.

Angle between plane of the probe and direction of magnetic field is not varied.

On axes of Fig.2, sketch graph to show variation with time t of the e.m.f. V_H produced by Hall probe

(c)

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

(ii) Hall probe in (b) is replaced by small flat coil of wire. Coil is moved at constant speed along line XY. Plane of coil is parallel to the faces of the poles of magnet.

On axes of Fig.3, sketch graph to show variation with time t of e.m.f. E induced in the coil

Reference: Past Exam Paper – November 2012 Paper 43 Q5

Solution 576:

(a)

EITHER

Φ = BAsinθ where A is the area (through which flux passes) and θ is the angle between magnetic flux density B and (the plane of) A

Φ = BA where A is the area normal to the magnetic flux density B

(b) For the graph, V_H is constant and non-zero between the poles and zero outside. There is a sharp increase / decrease at the ends of the magnet

{V_H is constant between the poles because the area through which flux passes is constant as the angle between the plane of the Hall probe and the direction of magnetic field is not varied. There is a sharp increase is due to the sudden cutting of the flux by the area of the plane of the Hall probe.}

(c)

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

(ii) For the graph, there is a short pulse on entering and on leaving the region between the poles. The pulses have approximately the same shape but opposite polarities. The e.m.f. is zero between the poles and outside.

{As the coil enters the field, the sharp increase in flux produces a pulse of e.m.f., which acts in such a way as to oppose the change of flux producing it (Lenz’s law). The same applies when the coil is leaving the magnetic field.

Since coil is moved a constant speed, once the whole coil has entered the field completely, the rate of change of flux is then zero (there is a flux, but it is not changing – so, the rate of change of flux is zero), causing the e.m.f. to be zero (from Faraday’s law).

The pulses have opposite polarities because as the coil enters the field, the change is an increase in flux and when it leaves the field, the change is a decrease in flux.}

Question 577: [Forces > Equilibrium]

A cylinder of weight W is placed on smooth slope. Contact force of the slope on the cylinder is R. A thread is attached to surface of the cylinder. The other end of the thread is fixed.

Which diagram shows the cylinder in equilibrium?

Reference: Past Exam Paper – November 2014 Paper 13 Q13

Solution 577:

Answer: C.

For equilibrium, the resultant force and the resultant torque on the system should be zero.

The resultant force can be identified by considering the components of the forces involved. Since the forces are not drawn to scale, we can assume that the forces act in such a way that the resultant is zero in this case.

Additionally, for equilibrium, the resultant torque should be zero.

Torque or moment = Force x perpendicular distance of line of action of the force to the pivot

Without doing any calculations, the easiest way to know if the resultant torque is zero is to make sure that all the force act on the same point – this causes the ‘perpendicular distance of line of action of the force to the pivot’ to be zero. Thus, the resultant torque would be zero. [Only C corresponds to this]