AS · Topics 1–11

What is the Young modulus expressed in SI base units?

1. Akg m s⁻²
2. Bkg m⁻¹ s⁻²
3. Ckg m⁻¹ s⁻¹
4. Dkg m² s⁻²

A stone is thrown horizontally at 12 m s⁻¹ from the top of a cliff 45 m high. Air resistance is negligible. How far from the base of the cliff does it land?

1. A27 m
2. B36 m
3. C45 m
4. D110 m

A ball of mass 0.20 kg strikes a wall at right angles with a speed of 8.0 m s⁻¹ and rebounds along the same line at 6.0 m s⁻¹. The ball is in contact with the wall for 0.050 s. What is the magnitude of the average force exerted on the ball?

1. A8.0 N
2. B24 N
3. C56 N
4. D112 N

A stationary wave is set up on a stretched string of length 1.2 m fixed at both ends. Four loops are seen between the fixed ends and the frequency is 150 Hz. What is the speed of the progressive waves on the string?

1. A22.5 m s⁻¹
2. B45 m s⁻¹
3. C90 m s⁻¹
4. D180 m s⁻¹

A battery of e.m.f. 9.0 V and internal resistance 1.5 Ω is connected to a resistor of resistance 6.0 Ω. What is the potential difference across the 6.0 Ω resistor?

1. A1.8 V
2. B6.0 V
3. C7.2 V
4. D9.0 V

Which combination of quarks forms a neutron?

1. Aup, up, down
2. Bup, down, down
3. Cup, up, strange
4. Ddown, down, down

A skydiver of total mass 85 kg falls from rest from a stationary balloon. The drag force F on the skydiver is related to her speed v by F = kv², where k is a constant. She reaches a terminal velocity of 55 m s⁻¹ before opening her parachute.

1. (a)

   State State what is meant by terminal velocity.

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   1. 1

      the constant (maximum) velocity reached by a falling body

   2. 2

      when the drag force has grown to equal the weight, so the resultant force and the acceleration are zero

2. (b)

   Explain Explain, with reference to Newton's laws, why the acceleration of the skydiver decreases as her speed increases.

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   1. 1

      the weight is constant, but the drag force increases as the speed increases

   2. 2

      so the resultant force, weight minus drag, decreases

   3. 3

      and by Newton's second law a = F/m, so the acceleration decreases while the speed still increases

3. (c)

   Determine Determine the value of k, and state its SI base units.

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   1. 1

      at terminal velocity, kv² = mg, so k = 85 × 9.81 / 55²

   2. 2

      Substituting the terminal values into k = mg/v² gives k = 0.276.

   3. 3

      From k = F/v², the SI base units of k are kg m⁻¹.

4. (d)

   Calculate Calculate the acceleration of the skydiver at the instant when her speed is 30 m s⁻¹.

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   1. 1

      drag F = 0.276 × 30² = 248 N

   2. 2

      resultant = 834 − 248 = 586 N

   3. 3

      a = 586 / 85 = 6.9 m s⁻²

5. (e)

   Explain The skydiver opens her parachute while travelling at the terminal velocity of 55 m s⁻¹. Explain what happens to her motion immediately afterwards, and describe the new terminal velocity she reaches.

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   1. 1

      opening the parachute increases k sharply, so at 55 m s⁻¹ the drag now greatly exceeds the weight

   2. 2

      the resultant force acts upwards, so she decelerates

   3. 3

      the drag falls as she slows until it again equals the weight, giving a new, much lower terminal velocity

A vertical steel wire of original length 2.50 m and diameter 0.56 mm hangs from a fixed support. A load of 45 N is attached to its lower end. The Young modulus of steel is 2.0 × 10¹¹ Pa and the wire does not exceed its limit of proportionality.

1. (a)

   Define Define the Young modulus.

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   1. 1

      the ratio of tensile stress to tensile strain

   2. 2

      for a material obeying Hooke's law / below the limit of proportionality

2. (b)

   Calculate Calculate the extension of the wire produced by the 45 N load.

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   1. 1

      A = π(0.28 × 10⁻³)² = 2.46 × 10⁻⁷ m²

   2. 2

      stress = F/A = 45 / 2.46 × 10⁻⁷ = 1.83 × 10⁸ Pa

   3. 3

      strain = stress / E = 1.83 × 10⁸ / 2.0 × 10¹¹ = 9.14 × 10⁻⁴

   4. 4

      e = strain × L = 9.14 × 10⁻⁴ × 2.50 = 2.3 × 10⁻³ m (2.3 mm)

3. (c)

   Calculate Calculate the elastic potential energy stored in the wire under this load.

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   1. 1

      E = ½Fe = ½ × 45 × 2.3 × 10⁻³

   2. 2

      E = 0.051 J

4. (d)

   Determine The diameter is measured with a micrometer as 0.56 ± 0.01 mm. Determine the percentage uncertainty this contributes to the calculated extension.

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   1. 1

      percentage uncertainty in d = (0.01 / 0.56) × 100 = 1.8%

   2. 2

      e is inversely proportional to d², so the percentage uncertainty is doubled

   3. 3

      percentage uncertainty in e = 3.6%

5. (e)

   Explain The load is increased until the wire is stretched beyond its elastic limit and then removed. Explain how the wire's behaviour differs from that described above.

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   1. 1

      the wire no longer returns to its original length when the load is removed

   2. 2

      it retains a permanent (plastic) extension, because layers of atoms have slipped past one another rather than simply being pulled further apart

Coherent light of wavelength 590 nm is incident normally on a diffraction grating with 500 lines per millimetre. The diffracted light is observed on a screen a long way from the grating.

1. (a)

   State State what is meant by coherent.

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   1. 1

      the waves have a constant phase difference

   2. 2

      which requires them to have the same frequency (and, in practice, the same wavelength)

2. (b)

   Calculate Calculate the angle between the second-order maximum and the straight-through direction.

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   1. 1

      d = 1 / (500 × 10³) = 2.0 × 10⁻⁶ m

   2. 2

      sin θ = nλ/d = 2 × 590 × 10⁻⁹ / 2.0 × 10⁻⁶ = 0.590

   3. 3

      Taking the inverse sine of the second-order ratio gives θ = 36.2°.

3. (c)

   Determine Determine the highest order of maximum that can be observed with this grating and this light.

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   1. 1

      the maximum possible value of sin θ is 1, so n ≤ d/λ

   2. 2

      d/λ = 2.0 × 10⁻⁶ / 590 × 10⁻⁹ = 3.39

   3. 3

      n must be a whole number, so the highest order observed is the third

4. (d)

   Explain The grating is replaced by one with 300 lines per millimetre. Explain, without calculation, how this changes the pattern observed.

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   1. 1

      the slit separation d is larger

   2. 2

      so for each order sin θ = nλ/d is smaller and every maximum moves closer to the centre

   3. 3

      and because d/λ is larger, more orders can be seen

A battery of e.m.f. 12.0 V and internal resistance r is connected in series with a variable resistor R and an ammeter of negligible resistance. When R is set to 4.0 Ω, the ammeter reads 2.0 A.

1. (a)

   Define Define the electromotive force of a source.

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   1. 1

      the energy transferred from chemical (or other non-electrical) form to electrical form

   2. 2

      per unit charge driven through the source

2. (b)

   Determine Determine the internal resistance r of the battery.

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   1. 1

      E = I(R + r), so 12.0 = 2.0 × (4.0 + r)

   2. 2

      4.0 + r = 6.0

   3. 3

      r = 2.0 Ω

3. (c)

   Calculate Calculate the ratio of the power dissipated inside the battery to the total power produced by the battery.

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   1. 1

      power in r = I²r = 2.0² × 2.0 = 8.0 W

   2. 2

      total power = EI = 12.0 × 2.0 = 24 W

   3. 3

      ratio = 8.0 / 24 = 0.33 (33%)

4. (d)

   Determine R is now increased to 12.0 Ω. Determine the new terminal potential difference of the battery.

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   1. 1

      I = 12.0 / (12.0 + 2.0) = 0.857 A

   2. 2

      V = IR = 0.857 × 12.0

   3. 3

      V = 10.3 V

5. (e)

   Explain Explain why the terminal potential difference increased when R was increased, even though the e.m.f. of the battery did not change.

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   1. 1

      increasing R decreases the current in the circuit

   2. 2

      so the lost volts Ir across the internal resistance are smaller, and V = E − Ir is closer to E

A student investigates how the period T of the vertical oscillations of a mass m hanging from a spring depends on m. The student suggests that T and m are related by T = km^p, where k and p are constants. She measures T for five values of m and calculates lg(T/s) and lg(m/kg).

1. (a)

   State State the independent variable, the dependent variable, and two quantities that must be kept constant in this investigation.

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   1. 1

      independent variable: the mass m hanging from the spring

   2. 2

      dependent variable: the period T of the oscillations

   3. 3

      constants: the same spring (same spring constant) and the same amplitude of oscillation — accept also the same support and no additional damping

2. (b)

   Describe Describe how the student should measure T so that the uncertainty in each value is as small as possible.

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   1. 1

      time a number of complete oscillations — at least 10 or 20 — and divide by that number

   2. 2

      start and stop the timing at the centre of the oscillation, where the mass moves fastest, using a fiducial marker

   3. 3

      repeat the timing and take a mean

3. (c)

   Explain Explain why a graph of lg T against lg m tests the suggested relationship, and state what the gradient and the y-intercept of that graph represent.

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   1. 1

      taking logarithms of T = km^p gives lg T = p lg m + lg k

   2. 2

      this has the form y = mx + c, so if the relationship holds the points lie on a straight line

   3. 3

      the gradient of the line is p

   4. 4

      the y-intercept is lg k, so k = 10^(intercept)

4. (d)

   Determine Use the first and last data points to determine values for p and for k. Give k to two significant figures.

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   1. 1

      gradient p = (0.051 − (−0.401)) / (−0.097 − (−1.000)) = 0.452 / 0.903

   2. 2

      The logarithmic graph gradient gives the exponent p = 0.50.

   3. 3

      intercept: lg k = −0.401 − 0.50 × (−1.000) = 0.099

   4. 4

      k = 10^0.099 = 1.3 (accept 1.26)

5. (e)

   Deduce Theory predicts T = 2π√(m/k_s), where k_s is the spring constant. Deduce a value for k_s and comment on whether the student's results support the theory.

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   1. 1

      the theory predicts p = 0.5, which matches the gradient found, so the form of the relationship is supported

   2. 2

      comparing constants, k = 2π/√k_s, so k_s = (2π/k)² = (2π/1.26)²

   3. 3

      k_s = 25 N m⁻¹