Monday, January 19, 2015

Physics 9702 Doubts | Help Page 47

  • Physics 9702 Doubts | Help Page 47



Question 279: [Matter > Elasticity]
(i) What’s the difference between the elastic limit and the limit of proportionality?
(ii) After reaching the elastic limit, the wire cannot do the work done equal to the strain energy stored within it and hence cannot return to its original position right?



Solution 279:
(i)
Hooke’s law states that provided the elastic limits have not been exceeded the extension e (or compression) is proportional to the force F applied. [F = ke]
The elastic limit is the limit beyond which a material will not return to its original dimensions when the deforming force is removed.

So, when a material has reached its limit of proportionality, its extension is no longer proportional to the force applied [Hooke’s law is no longer obeyed], but when the deforming force is removed, the material returns to its original dimensions.

When the elastic limit has been reached, the material will not return to its original dimensions when the deforming force is removed.
The elastic limit is slightly beyond the limit of proportionality. Up to the elastic limit, any work done in stretching the material (wire) can be recovered completely.

To conclude, the limit of proportionality is the limit up to which Hooke’s law is obeyed while the elastic limit is the limit up to which a material remains elastic.


(ii)
Beyond the elastic limit, plastic deformation of the material occurs – that is, the material (wire) does not return completely to its original length when the deforming force is removed. This is because some of the work done in stretching the material cannot be recovered from the wire.







Question 280: [Electricity > Electric field strength]
(i) How can the equation for electric field strength E (E = Force / Charge) be derived?
(ii) As the voltage between two plates increases the Electric field strength increases as well? Is that so?
(iii) Two horizontal metal plates are shown. One plate has a voltage while the other has 0 voltage. How do we know which plate is positively charged?



Solution 280:
(i)
The electric field strength E at a point in an electric field is defined as the force per unit charge acting on a positive test charge placed at that point.
Electric field strength, E = Force / Charge

This is how the quantity ‘electric field strength’ is defined (force per unit charge …). It is a defining equation – it cannot be derived.


(ii)
The electric field strength E between two parallel conducting plates with a potential difference V and a plate separation d is given by
Electric field strength, E = V / d
(The derivation of this equation is not required for A-level.)

So, as the potential difference V across the plates (not ‘voltage’ – if the potential on both plates are increased, the potential difference [difference in potential] may still be small) is increased, the electric field strength also increases.

Similarly, a decrease in the plate separation also causes the electric field strength to increase.


(iii)
Electric force is from positive charge towards negative charge. Positive charges are found at the plate with a relatively positive potential and negative charges are found at the plate with a relatively negative potential.

This is relative. If one of the plates has a potential of zero and the plate has a positive potential, then positive charges are found at the plate with a positive potential. But if one plate has a potential of zero and the other plate has a negative potential, then positive charges are found at the plate with zero potential.

As stated first, the direction of the electric force is from positive to negative. So if the field lines are shown, look at the direction of the electric force. The direction from which the electric force is shown ‘coming from’ corresponds to the positively charged plate and the direction to which the electric field is shown ‘going to’ correspond to the negatively charged plate.








Question 281: [Matter > Pressure]
(a) Define density

(b) U-tube contains some mercury. Water is poured into one arm of the U-tube and oil is poured into other arm, as shown.

Amounts of oil and water are adjusted until surface of the mercury in the two arms is at the same horizontal level.
(i) State how it is known that pressure at the base of the column of water is the same as the pressure at the base of the column of oil.
(ii) Column of water, density 1.0 × 103kgm–3, is 53 cm high. Column of oil is 71 cm high. Calculate density of the oil. Explain your working.

Reference: Past Exam Paper – June 2006 Paper 2 Q4



Solution 281:
(a) Density is defined as the mass per unit volume

(b)
(i)
{The mercury is at the same horizontal level. If the pressure in one of the columns was greater than the other, then the mercury is one arm would be at a lower level and the mercury in the other arm would be at a higher level. But since the mercury is at the same level, it means that the pressure is the same in both arms.
Since water and oil have different densities, if the length of the water column and the length of the oil column were the same, then their pressures would be different. Pressure = hρg. Since density ρ is different for oil and water, if h is the same, this would mean than pressure is different.}
Pressure is the same at the surface of the mercury because the mercury is at the same horizontal level.

(ii)
(The pressure) hρg is the same for both columns.
{Pressure due to water column = Pressure due to oil column}
(53x10-2) x (1.0x103) x g = (71x10-2) x ρ x g
Density ρ of oil = 7.5x102kgm-3









Question 282: [Dynamics > Momentum]
Sand falls vertically on a conveyor belt at a rate of m kgs-1.

In order to keep the belt moving at constant speed v the horizontal force that must be exerted on the belt is
A mv               D ½ mv                       C mv2              D ½ mv2



Solution 282:
Answer: A.
Rate at which the sand is falling is m kgs-1. The rate at which the sand is falling is the mass of sand falling per second (this can be seen from the units). Do not confuse m to be the mass.

The unit of Force is kg ms-2.

Units of choice C and D are [kgs-1] [ms-1]2 = kg m2 s-3. This is not the unit of force. [C and D are incorrect]

Force is defined as the rate of change of momentum.
Momentum p = mv. Change in momentum = Δp
Force = Δp / t

In this case, the velocity is not changing. The velocity has a constant value of v. What is actually changing is the mass of sand on the belt. Since the sand is falling, the mass of sand on belt is increasing. This is what causes the momentum to change.
Δp = (ΔM)v
[Capital M refers to the mass]

Force F = Δp / t = (ΔM)v / t = (ΔM / t) v
ΔM / t is the rate of fall of the sand and is given to be m.
Force = mv









Question 283: [Dynamics > Resultant force]
Sky diver falls vertically from stationary balloon. She leaves balloon at time t = 0. At time t = T, she reaches terminal velocity. Beyond time shown in the graphs, she opens her parachute.
Which graph shows variation with time t of the force F due to air resistance?


Reference: Past Exam Paper – June 2013 Paper 12 Q6



Solution 283:
Answer: B.
The force F is due to air resistance. Air resistance opposes motion. Air resistance increases with speed.

At time t = 0, air resistance is zero since the sky diver is initially in a stationary balloon. Her speed is also initially zero. [D is incorrect]

As the sky diver falls, her speed increases from zero. Comparing her speed at earlier values of t to the speed at greater value of t, it is obvious that the speed during the earlier stage is less than after. Air resistance increases with speed. So, during the first few seconds, the increase in air resistance is relatively smaller compared to later time. That is, the rate of increase of air resistance is not constant (gradient of the graph shown cannot be constant – so, from time = o to time = T, the graph is not a straight line). [A is incorrect]

The sky diver falls due to the acceleration due to gravity. Therefore, until terminal velocity is reached, the speed keeps on increasing.

At the terminal velocity, the resultant force on the sky diver is zero since air resistance equals force of gravity (weight). Since the weight of the sky diver is constant, air resistance should also be constant from time T and beyond. Air resistance cannot keep on increasing after time T. [C is incorrect]






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