• 9702 June 2014 Paper 22 Worked Solutions | A-Level Physics

Paper 22

Question 1

(a)

Show SІ base units of power are kg m² s^–3:

Power = work done / time      OR energy / time

EITHER Units of force: kg ms^-2

OR Units of kinetic energy (½mv²): kg (ms^-1)²

(Unit distance: m        and (time)^-1: s^-1)

And hence Unit of power:  kgms^-2 m s^-1 = kgm²s^-3

(b)

Rate of flow of thermal energy Q /t in material is given by Q / t = CAT / x where A is cross-sectional area of material, T is temperature difference across thickness of material, x is thickness of material, C is a constant.

SI base units of C:

Units of Q / t: kgm²s^-3

Units of area, A: m²    distance, x: m              and temperature, T: K

Correct substitution into C = Qx / tAT

So, units of C: kg ms^-3 K^-1

Question 2

Coin is made in shape of thin cylinder, as shown. Fig. shows measurements made in order to determine density ρ of material used to make coin.

Quantity                      measurement               Uncertainty

Mass                            9.6g                             ±0.5g

Thickness                    2.00mm                       ±0.01mm

Diameter                     22.1mm                       ±0.1mm

(a)

Density ρ in kgm^-3:

ρ = m / V

Volume, V = (πd²/4) t = 7.67x10^-7m³

ρ = (9.6x10^-3) / [π({22.1/2}x10^-3)² (2.00x10^-3)] = 12513kgm^-3

(b)

(i)

Percentage uncertainty in ρ:

Δρ/ρ x 100% = [Δm/m + Δt/t + 2(Δd/d)] x 100%

Δρ/ρ x 100% = 5.21% + 0.50% + 0.905% = 6.6%     (6.61%)

(ii)

ρ with its actual uncertainty:

ρ = 12500 ± 800 kgm^-3 

Question 3

(a)

Newton’s first law of motion states that a body / mass / object continues (at rest or) at constant / uniform velocity unless acted on by a resultant force.

(b)

Box slides down slope, as shown. Angle of slope to horizontal is 20°. Box has mass of 65 kg. Total resistive force R acting on box is constant as it slides down slope.

(i)

Names and directions of other 2 forces acting on box:

The weight: vertically down

The normal / reaction / contact (force): perpendicular / normal to the slope

(ii)

Variation with time t of velocity v of box as it moves down slope is shown.

1.

Use data from Fig to show that acceleration of box is 2.6ms^-2:

Acceleration of box = gradient OR (v – u) / t            OR Δv/t

Acceleration of box = (6.0 – 0.8) / (2.0 – 0.0) = 2.6ms^-2

2.

Resultant force on box:

Resultant force, F = ma = 65 x 2.6 = 169N (allow to 2 or 3 sf)

3.

Resistive force R on box:

Component of weight along slope = mgsinθ (= 218N)

218 – R = 169

Resistive force, R = 49N

Question 4

(a)

Gravitational potential energy is the energy of a mass due to its position in a gravitational field

Kinetic energy is the energy (a mass has) due to its motion / speed / velocity

(b)

Ball of mass 400g is thrown with initial velocity of 30.0 m s^–1 at angle of 45.0° to horizontal, as shown. Air resistance negligible. Ball reaches maximum height H after a time of 2.16 s.

(i)

1.

Initial kinetic energy of ball:

Initial Kinetic energy = ½ mv² = ½ (0.4)(30)² = 180J

2.

Maximum height H of ball:

s = (30sin45)(2.16) + ½ (-9.81)(2.16)² = 22.94 (22.9) m

OR

s = (30sin45)² / (2 x 9.81) = 22.94 (22.9) m

3.

Gravitational potential energy of ball at height H:

Gravitational potential energy, GPE = mgh = 0.4(9.81)(22.88) = 89.8 (90) J

(ii)

1.

Kinetic energy of ball at maximum height:

KE at maximum height = initial KE – GPE = 180 – 90 = 90J

2.

Why kinetic energy of ball at maximum height not zero:

The (horizontal) velocity is not zero / (object) is still moving / answer explained in terms of conservation of energy

Question 5

(a)

Young modulus is defined as the ratio of stress to strain

(b)

2 wires P and Q of same material and same original length l_o are fixed so that they hang vertically, as shown. Diameter of P is d and diameter of Q is 2d. Same force F is applied to lower end of each wire.

(i)

Ratio of stress in P to stress in Q:

Stress = Force / Area

In P, stress = F / (πd²/4)

In Q, stress = F / (πd²)

Ratio of stress in P to stress in Q = 4 (or 4:1)

(ii)

Ratio of strain in P to strain in Q:

Young modulus, E is the same for both wires (as they are of the same material) [e.g. E_P = E_Q]

Strain = stress / E

So, ratio of strain in P to strain in Q = 4 (or 4:1)

Question 6

Battery is connected in series with resistors X and Y, as shown. Resistance of X is constant. Resistance of Y is 6.0 Ω. Battery has electromotive force (e.m.f.) 24 V and zero internal resistance. Variable resistor of resistance R is connected in parallel with X.

Current І from battery is changed by varying R from 5.0 Ω to 20 Ω. Variation with R of І is shown.

(a)

Why potential difference (p.d.) between points A and C is 24 V for all values of R:

There are no lost volts / energy in the battery OR There are no lost volts / energy lost in the internal resistance.

(b)

Use Fig to explain variation of p.d. across resistor Y as R is increased:

The current / I decreases (as R increases). So, the p.d. across resistor Y decreases (as R increases).

OR

The parallel resistance (of X and R) increases. So, the p.d. across the parallel resistors increases and so, the p.d. (across Y) decreases.

(c)

For R = 6.0 Ω,

(i)

Show that p.d. between points A and B is 9.6V:

(from graph) For R = 6.0 Ω, Current = 2.4A

So, p.d. across points A and B = 24 – (2.4 x 6) = 9.6V

OR

Total resistance = (24V / 2.4A =)10 Ω

(parallel resistance = (10 – 6 =) 4Ω), p.d. = 2.4 x (4/10) = 9.6V  

(ii)

Resistance of X:

Resistance through AB, R = 9.6 / 2.4 = 4.0Ω

1/6 + 1/X = 1/4

So, resistance of X = 12Ω

OR

Current through R, I_R = 9.6 / 6.0 = 1.6A

Current through X, I_X = 2.4 – 1.6 = 0.8A

Resistance of X = (9.6 / 0.8 =) 12Ω

(iii)

Power provided by battery:

Power provided by battery = VI or EI or V²/R or E²/R or I²/R

(total resistance through circuit = 6 + [1/12 + 1/6]^-1 = 10)

Power = (24 x 2.4) or (24)²/10 or (2.4)²(10) = 57.6W

(d)

Explain qualitatively how power provided by battery changes as resistance R is increased:

The power provided by the battery decreases since the e.m.f. is constant or power = 24 x current, and the current decreases OR the e.m.f. is constant or power = 24² / resistance, and the resistance increases.

Question 7
{Detailed explanations for this question is available as Solution 679 at Physics 9702 Doubts | Help Page 137 - http://physics-ref.blogspot.com/2015/05/physics-9702-doubts-help-page-137.html}