Questions 1
Topic -4.3 Density and pressure
A solid metal sphere has a diameter of (3.42 ± 0.02) cm and a mass of (67 ± 2)g.
(a) Calculate the density, in \(gcm^{–3}\), of the metal.
(b) Determine the percentage uncertainty in the density.
▶️Answer/Explanation
Ans
(a) ρ = m / V
\(V = (4 / 3) \times \pi \times r^3\)
= \((4 / 3) \times \pi \times (3.42 / 2)^3\)
\(( = 20.9 cm^3)\)
ρ = 67 / 20.9
= 3.2 \(g cm^{−3}\)
(b)
%ρ = %m + 3 × %d
= [(2 / 67) × 100] + [3 × (0.02 / 3.42) × 100]
= 3.0% + 3 × 0.58%
= 4.7% or 5%
Questions 2
Topic -3.3 Linear momentum and its conservation
An archer releases an arrow towards a target at a velocity of 65.0\(ms^{–1}\) at an angle of 4.30° above the horizontal, as shown in Fig. 2.1.
When released, the tip of the arrow is a horizontal distance of 70.0m from the target and 1.66m above the horizontal ground. The arrow hits the centre of the target. Assume that air resistance is negligible and that all the mass of the arrow is at its tip.
(a) Show that the time taken for the arrow to reach the target is 1.08s.
(b) Calculate the height of the centre of the target above the ground.
(c) By considering energy changes, state and explain how the final kinetic energy of the arrow as it hits the target compares with its initial kinetic energy immediately after release. A numerical calculation is not required.
▶️Answer/Explanation
Ans
(a) (time =) displacement / velocity
(time =) 70.0 / 65.0 cos 4.30° = 1.08 (s)
(b)\( s= ut + ½at^2\)
= (65 × sin 4.30° × 1.08) – (0.5 × 9.81 × 1.082)
s = −0.46 (m)
height above ground = 1.66− 0.46= 1.2m
(c) GPE has decreased
(total energy conserved so) KE has increased
Questions 3
Topic -2.1 Equations of motion
(a) Define velocity.
(b) A constant driving force of 2400N acts on a car of mass 1200kg. The car accelerates from rest in a straight line along a horizontal road. Assume that the resistive forces acting on the car are negligible.
(i) Calculate the acceleration of the car.
(ii) On Fig. 3.1, sketch a graph showing the variation with time t of the velocity v of the car for the first 20 seconds of its motion. Label this line A
(c) In reality, a resistive force due to air resistance acts on the car in (b). This resistive force increases with speed until it becomes equal in magnitude to the driving force at time t = 12s.
(i) On Fig. 3.1, sketch a second line to show the variation with time t of the velocity v of the car for the first 20 seconds of its motion. Label this line B.
(ii) At time t = 20s, the driving force is increased to 3000N and remains constant at this value. Describe how the velocity of the car changes due to this increase in the driving force.
▶️Answer/Explanation
Ans
(a) (velocity =) change in displacement / time
(b)(i) a = 2400 / 1200= 2.0\(ms^{−2}\)
(ii) straight line from the origin with positive gradient (labelled A)
ending at (20, 40)
(c)(i) line starting at origin (with the same gradient as A) and beneath A at all points gradient decreasing to zero
straight horizontal line from t = 12 s and ending at t = 20 s (and labelled B)
(ii) the velocity/speed will increase
to a new terminal/constant/maximum velocity/speed
or
the car has an acceleration
to a new (higher) terminal/constant/maximum velocity/speed
Questions 4
Topic -5.1 Energy conservation
(a) A mass m moves a vertical distance Δh in a uniform gravitational field and gains gravitational potential energy \(ΔE_P\). The acceleration of free fall is g. Use the concept of work done to show that
\(ΔE_P = mgΔh\).
(b) A 0.60kg mass is attached to a string which is wrapped around the wheel of a generator, as shown in Fig. 4.1.
The mass is held stationary above the floor. When released, the mass initially accelerates and then falls at a steady speed and spins the wheel. The generator causes a current in a resistor. Air resistance is negligible. State the main energy change when the mass is falling at a steady speed.
(c) When falling at a steady speed, the mass in (b) falls through a vertical distance of 1.4m in a time of 4.0s. This causes a current of 90mA in the resistor. The resistance of the resistor is
47Ω. Calculate:
(i) the rate of work done by the falling mass
(ii) the power dissipated in the resistor
(iii) the efficiency of the generator.
▶️Answer/Explanation
Ans
(a) work (done) = force × displacement
(force = mg and distance = Δh)
(so) work (done) = mgΔh and work =\(ΔE_{(P)}\) (so, \(ΔE_{(P)}\) = mgΔh)
(b) gravitational potential (energy) to heat/thermal (energy)
(c)(i) P = mg(Δ)h / (Δ)t or Fv
P = (0.60 × 9.81 × 1.4) / 4.0 or 0.60 × 9.81 × (1.4 / 4.0)
= 2.1W
(ii) \(P = I^2R\) or IV or \(V^2 / R\)
= \(0.09^2 × 47\) or 0.09 × 4.23 or 4.232 / 47
= 0.38W
(iii) \(efficiency = P_{out} / P_{in} (× 100) or \(E_{out} / E_{in}\) (× 100)\)
= 0.38 / 2.1 ( × 100) or 0.38 × 4.0 / 2.1 × 4.0 ( × 100)
= 0.18 or 18%
Questions 5
Topic -7.4 Electromagnetic spectrum
(a) Parallel light rays from the Sun are incident normally on a magnifying glass. The magnifying glass directs the light to an area A of radius r, as shown in Fig. 5.1.
The magnifying glass is circular in cross‑section with a radius of 5.5cm. The intensity of the light from the Sun incident on the magnifying glass is \(1.3kWm^{–2}\). Assume that all of the light incident on the magnifying glass is transmitted through it.
(i) Calculate the power of the light from the Sun incident on the magnifying glass.
(ii) The value of r is 1.5mm. Calculate the intensity of the light on area A.
(b) A laser emits a beam of electromagnetic waves of frequency \(3.7 × 10^{15}Hz\) in a vacuum.
(i) Show that the wavelength of the waves is \(8.1 × 10^{–8}m\).
(ii) State the region of the electromagnetic spectrum to which these waves belong.
(iii) The beam from the laser now passes through a diffraction grating with 2400 lines per millimetre. A detector sensitive to the waves emitted by the laser is moved through an arc of 180° in order to detect the maxima produced by the waves passing through the grating, as shown in Fig. 5.2.
Calculate the number of maxima detected as the detector moves through 180° along the line shown in Fig. 5.2. Show your working.
(iv) The laser is now replaced with one that emits electromagnetic waves with a wavelength of 300nm. Explain, without calculation, what happens to the number of maxima now detected. Assume that the detector is also sensitive to this wavelength of electromagnetic waves.
▶️Answer/Explanation
Ans
(a)(i) power = intensity × area
= 1.3 × \(10^3\) × (π × \(0.055^2\))
= 12 W
(ii) intensity = power / area
= 12 / (π × \(0.0015^2\))
= \(1.7\times 10^6 Wm^{−2}\)
(b)(i) (\(\lambda\)=) v / f or c / f
(\(\lambda\) =) × \(10^8 / 3.7 × 10^{15} = 8.1 × 10^{−8}\) (m)
(ii) ultraviolet
(iii) d sin(θ) = nλ or (1 / N) × sin(θ) = nλ
d = 1 / 2400 ×\( 10^3\) (m)
= \(4.2\times 10^{–7}\) (m)
or
N = 2400 ×\( 10^3 (m^{–1})\)
n = 4.2 × \(10^{−7}\) × sin 90° / 8.1 × 10−8 or sin 90° / \(2400 × 10^3 × 8.1 × 10^{–8}\)
n = 5.2 or 5.1
or
when n = 5,
θ = 76.4° and when n = 6, sin θ > 1
(so) n = 5
number of maxima = (2 × 5) + 1
= 11
(iv) the wavelength has increased
(so) number of maxima decreases
Questions 6
Topic -10.1 Practical circuits
(a) (i) On Fig. 6.1, sketch the I–V characteristic of a filament lamp.
(ii) Explain the shape of the line in (a)(i).
(b) A conducting wire has length 5.8m and cross‑sectional area \(3.4 × 10^{–8}m^2\). The resistivity of the metal of the wire is \(5.6 × 10^{–8}Ωm\). Calculate the resistance of the wire.
(c) A resistor of resistance R is placed in a circuit with a cell of negligible internal resistance, two switches \(S_1\) and \(S_2\), a second resistor of resistance 2R and three ammeters X, Y and Z. The circuit is shown in Fig. 6.2.
The reading on X is 1.0A when \(S_1\) is open and \(S_2\) is closed. Complete Table 6.1.
▶️Answer/Explanation
Ans
(a)(i) line passes through (0,0) and is in first and third quadrants only
gradient of line becoming less steep in both quadrants and roughly symmetrical
(ii) (asI increases) the temperature (of the filament wire/lamp) increases
(as I / temperature / V increases) the resistance (of wire/lamp) increases
(as I / temperature / V increases the graph curves because) ratio V / I increases or ratio I / V decreases
(b) R= = ρL / A
= \((5.6 \times 10^{−8} \times 5.8) / 3.4 \times 10^{−8}\)
= 9.6 Ω
Questions 7
Topic -11.2 Fundamental particles
(a) Fluorine‑18 \((_9^{18}F)\) is an isotope that decays to an isotope of oxygen (O) by the emission of a β+ particle.
(i) Complete the nuclear equation for the decay, including all the particles involved.
(ii) A quark in the fluorine‑18 nucleus changes flavour during the decay. State this change of flavour.
(b) A hadron has a charge of –2e, where e is the elementary charge.
(i) State and explain whether the hadron is a meson or a baryon
(ii) State a possible quark composition for the hadron
▶️Answer/Explanation
Ans
(a)(i) 
(ii) up quark to down quark
(b)(i) must be three (anti)quarks, as largest (negative) quark charge is (–)2/3 (e)
or
mesons can only have a charge of 0 or ±1(e)
(so hadron is) a baryon
(ii) any combination of three from:
antiup (quark) / up antiquark
and/or anticharm (quark) / charm antiquark
and/or antitop (quark) / top antiquark