Physics 9702 Doubts | Help Page 6

Physics 9702 Doubts | Help Page 6

Question 35: [Gravitation > Weight]

(a) What is the difference between weight and gravity?

(b) How is g a measure of gravitational field strength?

(c) Why is g constant only near the surface of the earth and varies in all other possible dimensions?

(d) Can you explain why weight is lesser at equator than at poles?

Solution 35:

(a)

{The information given below is based on Newton’s law of gravitation which is a good approximation for calculations. I won’t discuss the modern theories here.}

Weight (W) is the gravitational force felt by a body of mass, m. W = mg where g is acceleration due to gravity. So, the weight of a body depends on its mass and the acceleration due to gravity.

Gravity (or force of gravity) is the force exerted by a body of a specific mass on another body having a mass. So, just like electric force is the force (in an electric field) between CHARGES, gravitational force is the force (in a gravitational field) between MASSES. However, electric force can be attractive or repulsive, while the force of gravity is always attractive.

{Note also that afield is a region in space where an object (charge, mass, …) experiences a force} So, if by gravity, you are referring to the gravitational field, then it’s a concept that there is afield around any mass, in which another mass would experience an attractive force towards to first mass and the first mass experiences an attractive force towards the second.

(b)

It is known that Force, F = ma. That is, a force causes a body to accelerate.

The gravitational force between 2 bodies is given by

F = GMm / r² (Newton’s law of Gravitation)

where G is the gravitational constant, M and m are the masses of the bodies and r is the separation of the bodies (assuming they are point masses).

So, the larger the mass, the greater is the force it exerts on another mass. Also, the smaller mass will be attracted to the larger one. That is, if there are 2 bodies A and B, and the mass of A is much larger than that of B, then body A would appear to be stationary while body B would be attracted towards A. This is the case with the Earth and the objects on Earth. Earth would be body A and the other objects would be body B.

The gravitational field strength at a point (in the gravitational field) is the gravitational force acting on a body of unit mass at that point. Obviously, the gravitational force is due to another mass, M.

So, gravitational field strength = Force / mass = GM/r²

Comparing with the equation F = ma where the force per unit mass, F / m = a, it can be concluded that the

Acceleration (due to gravity) = GM/r²

So, the gravitational field strength is just another name for the acceleration due to gravity.

(c)

Hence, acceleration due to gravity, g = GM / r²

For Earth, an averaged value of g is taken as g = 9.81ms^-2. Like I said, this is an averaged value, not the actual value at any location on Earth.

Why is this averaged value acceptable?

G is the gravitational constant and M is the mass of the Earth and can also be taken to be constant. r is the separation between the mass being considered and the mass of the earth, assuming that both have their masses concentrated at their centre. So, r is the sum of the radius of Earth (assuming Earth to be a sphere) and the distance of the body being considered above the surface of Earth (let’s call this x).

For bodies close to the surface of Earth, distance x << radius of Earth. So, x can be neglected, making the separation of bodies close to the surface of Earth always equal to the radius of Earth. Since all the terms: G, M and r and now constant, the acceleration due to gravity, g is also taken to be constant and equal to about 9.81ms^-2.

Why is the acceleration due to gravity of Earth not always 9.81ms^-2?

The averaged value of 9.81ms^-2 actually varies from locations to locations. Here are some possible reasons:

Earth is not perfectly spherical – it contains mountains, valleys, … So, if we do not neglect the distance x, g actually changes slightly. {So, if for an exam for example, you are required to calculate g and everyone in every country got the same exact value of 9.81ms^-2, then they have surely tricked their practical results because this actually changes. Even with the simple pendulum experiment, different values of g may be obtained but of course, they should all average at about 9.91ms^-2}

If the distance of the body is far enough from the surface (e.g satellites, …) then x cannot be neglected. The value of g will change.

The solar system, in which Earth is situated, is a complex system. Just like Earth exerts gravitational force of attraction on bodies, the same is true for the others planets. So, on a body at a distance from Earth and the planets, the acceleration on the body is due to the resultant of the all the forces acting on the body.

(d)

{There are many discussions involved here. But at A-levels, it can be understood as follows:}

First note that weight, W = mg and the mass of the body remains constant here. What will be changing is the g or ‘effective’ g.

Spinning of the Earth

The spinning of the Earth causes a centrifugal force (directed outwards) which is maximum at the equator and zero at the poles. {Centrifugal force depends on distance from axis of rotation. This is largest at the equator and zero at the poles}.

Centrifugal acceleration is outwards while the gravitational attractive is inwards. So, the weight is the product of mass and the effective g resulted from the 2 accelerations. {Weight is lowered by about 0.4% due to this effect}

Equatorial Bulge

Another reason for this change in weight is due to the bulge found at the equator. The Earth is actually not a perfect sphere and has a bulge at the equator. So, the distance of a body there (from the centre of Earth) is greater than at the poles and since g = GM / r², this causes g to be smaller. Hence, weight = mg is also smaller at the equator.

[So, even if the Earth was not spinning, the weight would be lower due to the equatorial bulge. But then again, the equatorial bulge may itself be due to the spinning of the Earth]

{Weight is lowered by about 0.1% due to this effect}

So, again, the mass of a body is unchanged even in the presence of these effects. It is only the weight that changes.

Here are some keywords involved in this explanation:

Weight, Mass, Force of gravity, Acceleration due to gravity, Gravitational field strength, Gravity, Gravitational Field

If you are still having doubts, try to read again or ask through the comment box

Question 36: [Kinematics > Braking force]

Why is this method of stopping more comfortable for the passenger than a more usual method where the braking force is kept constant during deceleration?

Method: Car is stopped by varying the breaking force. The braking force is increased to a maximum and then reduced at the same rate to zero.

Solution 36:

When the brakes are applied, the minimum distance that the car travels before coming to rest is determined by the effective coefficient of friction between the tires of the car and the road.

The force of friction (of the road) opposes motion and should do enough work on the car so that its kinetic energy is reduced to zero (velocity becomes zero).

There are 2 types of friction: Static friction and Kinetic friction

The force of friction is given by

F_R = μmg

where μ is the coefficient of friction {the coefficient of friction is different for static and kinetic friction. That of static is typically larger than that of kinetic}, m is the mass of the car and g is the acceleration due to gravity.

For the car to stop, the work done by the force of friction should be equal in magnitude to the kinetic energy of the car when the brake is applied.

Work done by force of friction F_R = (μmg)d = ½mv²

where d is the distance travelled by the car before coming to rest, v is the velocity of the car when the brake is applied.

Stopping distance, d = v² / 2μg

Note that this is independent of the mass and depends on the square of velocity.

To understand the static and kinetic forces of friction, consider a box at rest on a surface. A person applies a force F to move towards the right.

If the force is not enough, the box will remain at rest. So, there should be a force that is opposing the force applied by the person. This force is equal in magnitude and opposite in direction to the force applied (from Newton’s 3^rd law). This force is called the force of static friction. If the force applied by the person is increased, when the box is on the verge of slipping, the force of static friction is maximum.

If the force applied is large enough to cause the box to move, then there will still be a force opposing the motion. This is the force of kinetic friction.

(Reference for graph: page 106 of College Physics, 9^th edition, by Raymond A. Serway, Chris Vuille)

As seen from the graph, static friction is greater than kinetic friction.

Static friction is operating if the wheels of the car keep rolling (without locking up) when the brake is applied {this is because the bottom point of the tire is instantaneously at rest with respect to the roadway (it is not slipping)}, while kinetic friction takes over if the wheels are locked when the brakes are applied {in this case, there is a relative motion between the tire and the roadway in the form of slipping}. In the latter case, there is relative slipping between the wheel and the road.

The presence of static frictions results in a maximum braking force (and thus, minimum stopping distance) while the relative slipping of the tire and the road causes the presence of kinetic friction which reduces the braking force (so, the stopping distance is greater).

If there is a significant difference because the forces of static and kinetic frictions, there will be more braking force.

So, the method of firstly increasing the braking force to a maximum allows the static friction to have a maximum value (as seen from the graph). Since the velocity of the car is being reduced, by reducing the braking force at a constant rate, the locking of the wheel is prevented. This causes the friction force to be always static and equal to the forward force on the car. Thus, the resultant force on the car (and thus, the passengers) is zero. The passengers will not be jerked forward. By preventing the wheels from locking up, the driver can have a greater control on the vehicle.

There are systems on vehicles that prevent the wheels from locking up when the brakes are applied. These are called Anti-lock braking systems (ABS). This minimizes the amount of time required for the vehicles to come to rest.

Question 37: [Resistance]

(a) Output of heater is 2.5 kW when connected to 220 V supply.

(i) Calculate resistance of heater.

(ii) Heater is made from wire of cross-sectional area 2.0 × 10^–7 m² and resistivity 1.1 × 10^–6Ω m. Use answer in (i) to calculate length of wire.

(b) Supply voltage is changed to 110 V.

(i) Calculate power output of heater at this voltage, assuming there is no change in resistance of wire.

(ii) State and explain quantitatively one way that wire of heater could be changed to give same power as in (a).

Reference: Past Exam Paper – June 2012 Paper 21 Q4

Solution 37:

(a)

(i)

Resistance, R = V² / P or P = IV and V = IR

Resistance, R = (220)² / 2500 = 19.4Ω

(ii)

Resistance, R = ρl / A

Length of wire, l = 19.4 (2.0x10^-7) / (1.1x10^-6) = 3.53m

(b)

(i)

P = V²/R = 110² / 19.4 = 623.71W. OR Since P is proportional to V², when V = 110V, P = [110² / 220²] x 2500 = 625W}

Power output, P = 625, 620 or 630K

(ii)

{P = V²/R. To increase power P, R needs to be reduced}

Resistance R needs to be reduced.

{Compare the 2 power output: 2500 / 625 = 4. So, the power should become 4 times greater. So, the resistance should be ¼ of the value in (a). R = ρl/ A. Since the same wire is used, resistivity is the same. So, l and A can be changed, or the diameter d, that would affect A.}

EITHER The length can be taken to be ¼ of the original length OR The area should be 4 times greater OR The diameter should be 2 times greater

Question 38: [Resistors > Potential difference]

A (potential difference) p.d. of 12 V is connected between P and Q.

What is the p.d. between X and Y?

Reference: Past Exam Paper – June 2006 Paper 1 Q33

Solution 38:

“p.d. of 12 V is connected between P and Q”: this means that the difference in potential between P and Q is 12V. So, any potential can be assigned to P and Q as long as there difference is 12V. {Unlike current, which flows in a specific direction from the +ve terminal of the supply, p.d. is a measure of the potential difference between 2 POINTS in the circuit. We cannot say “the p.d. at a point” [we can say “potential at a point”], but we say “the p.d. across a component” or “p.d. between 2 points”. But we can say “the current at a point is”}

Let’s assign a potential of 0V at P and 12V at Q. So, when we go from P to Q, the potential should increase and point X would be at a potential between 0 and 12V. The same is true for point Y. Hence, it can be concluded at points P, Q, X and Y are all at different potentials.

{Note that OV could be assigned at Q and 12V at P, but the process discussed would have to be reversed}

The circuit consists of a parallel combination of resistors. So, total p.d. across the upper and lower parts of the circuit are both 12V. Now, both the upper and lower part of the circuit each consists of 2 resistors in series.

For series connection of 2 resistors R₁ and R₂, the p.d. across R₁ is given by V = [R₁ / (R₁+R₂)] E where E is the e.m.f (potential supplied).

Potential at point X = p.d. across 500Ω in the upper part

Potential at point X = [500 / (500+1000)] x 12 = 4V

Similarly, Potential at point Y = p.d. across 2000Ω in the upper part

Potential at point Y = [2000 / (2000+1000)] x 12 = 8V

So, p.d. between X and Y is the difference in potential at point X and point Y. This is 8 – 4 = 4V

Question 39: [Forces]

(a) Distinguish between the mass and weight.

(b) Object O of mass 4.9 kg is suspended by rope A that is fixed at point P. Object is pulled to one side and held in equilibrium by second rope B, as shown.

Rope A is at angle θ to horizontal and rope B is horizontal. Tension in rope A is 69 N and tension in rope B is T.

(i) On Fig, draw arrows to represent directions of all forces acting on object O.

(ii) Calculate

1. Angle θ

2. Tension T

Reference: Past Exam Paper – June 2013 Paper 21 Q2

Solution 39:

Mass is the property of a body resisting changes in motion / mass is the quantity of matter in a body / measure of the inertia to changes in motion

Weight is the force due to the gravitational field / the force due to gravity OR gravitational force

(b)

(i)

An arrow vertically down through object O {This is for the weight. Weight always act downwards (towards surface) and acts ON the object. So, arrow should start from O}. The tension forces should be in correct direction on the rope {Tension forces act along the rope and act towards the points from which they are fixed [away from the weight] (not towards O, in which case, equilibrium would not be reached). Also, do not resolve into the horizontal and vertical components here. That is not what has been asked in this part.}

(ii)

{These 2 parts may also be answered graphically, but I’m not doing it in this way here because it’s a problem to draw on the computer and additionally, calculations would give more accurate results (with less errors) and is less time-consuming.}

{For equilibrium, the vertical forces should be equal. Weight is vertically downwards. The tension in rope A has an upwards vertical component of 69sinθ. Then tension in rope B has no vertical component since it is horizontal. So, by equating 69sinθ to the weight, the angle θ can be calculated}

Weight of object O = mg = 4.9 x 9.81 (=48.07N)

69sinθ = mg

Angle θ = 44.(1)^o

{Similarly, for equilibrium, the horizontal components should be equal. Weight has no horizontal component since it acts vertically downwards. The horizontal component of the tension in rope A is 69cosθ and this should be equal to the tension in rope B. So, T = 69cosθ}

Tension T = 69cosθ = 49.6 / 50N