Questions 1
Topic – 2.1
The drag force \(F_D\) acting on an object falling through air is given by
where A is the cross-sectional area of the object,
v is the velocity of the object in the air,
ρ is the density of the air and
C is a constant called the drag coefficient.
(a) Use SI base units to show that the drag coefficient has no units.
(b) Fig. 1.1 shows a sphere falling at terminal velocity in air. 
Assume that the upthrust on the sphere is negligible. On Fig. 1.1, draw and label arrows to show the directions of the two forces acting on the sphere.
(c) The mass of the sphere is 49g. Calculate the drag force \(F_D\) acting on the sphere.
(d) The sphere is falling in air at a terminal velocity of 25 in SI base units. The density of the air is 1.2 in SI base units. The diameter of the sphere is 0.060 in SI base units. Use your answer in (c) to calculate the drag coefficient C for the sphere.
▶️Answer/Explanation
Ans
(a) units of \(F_D\): \(kgms^{–2}\)
units of \(\rho\) \(kg m^{–3}\)
and
units of A: \(m^2\)
and
units of v: \(m s^{−1}\) or units of \(v^2: m^2 s^{–2}\)
\(kgms^{–2} =C kgms^{–2}\) and comment ‘(so) C has no units’ / unit terms cancelled
or
C = \(kg m s^{−2} / (kg m^{–3} m^2 m^2 s^{–2}\)) and comment ‘(so) C has no units’ / unit terms cancelled
(b) one arrow vertically downward labelled weight to within 10° of the vertical
one arrow vertically upwards labelled drag / drag force / \(F_D\) / air resistance / viscous force to within 10° of the vertical
(c) (at terminal velocity)
\(F_D\) = mg
\(F_D = 0.049\times 9.81\)
= 0.48 N
(d) area = \(\pi \times (0.060 / 2)^2\)
0.48 = \(½ \times C \times 1.2 \times \pi \times (0.060 / 2)^2 \times 25^2\)
C = 0.45
Questions 2
Topic – 3.2
(a) Define velocity.
(b) A student throws a ball over a vertical wall of height h, as shown in Fig. 2.1.
The ball leaves the hand of the student at a height of 1.2m above the horizontal ground. The ball has an initial velocity of \(22ms^{–1}\) at an angle of 40° to the horizontal. The wall is a horizontal distance of 36m from where the student releases the ball. Air resistance is negligible.
(i) Determine the time taken for the ball to reach the wall.
(ii) Calculate the vertical component u of the initial velocity of the ball.
(iii) The ball just goes over the wall. Calculate the height h of the wall.
▶️Answer/Explanation
Ans
(a) change in displacement / time (taken)
(b)(i) horizontal velocity =\( 22\times cos 40°\)
time taken = \(36 / (22\times cos 40°)\)
= 2.1 s
(ii) u = \(22\times sin 40°\)
= \(14ms^{−1}\)
(iii) \(s = ut + ½ at^2\)
= \((14\times 2.1) + (½\times−9.81\times 2.1^2)\)
= 7.8 (m)
(therefore) height of wall = 7.8 + 1.2
= 9.0 m
Questions 3
Topic – 3.1
(a) State the principle of conservation of momentum.
(b) An object of mass 2m is travelling at a speed of \(5.0ms^{–1}\) in a straight line. It collides with an object of mass 3m which is initially stationary, as shown in Fig. 3.1.
After the collision, the object of mass 2m moves with velocity v at an angle of 30° to its original direction of motion. The object of mass 3m moves with velocity w also at an angle of 30°, as shown in Fig. 3.2. 
By considering the conservation of momentum in two dimensions, calculate the magnitudes of v and w.
(c) An object of mass 4.2kg is travelling in a straight line at a speed of \(6.0ms^{–1}\). The object is brought to rest in a distance of 0.050m by a constant force. Calculate the magnitude of this force.
▶️Answer/Explanation
Ans
(a) sum / total momentum before (a collision) = sum / total momentum after (a collision)
or
sum / total momentum (of a system) is constant
if no (resultant) external force (acts) / for an isolated system
(b) along direction of motion:
10 m = 2mv cos 30° + 3mw cos 30°
perpendicular to direction of motion:
2mv cos 60° = 3mw cos 60°
(v = 3w / 2)
v = \(2.9ms^{–1}\)
w = \(1.9ms^{–1}\)
(c)
EK =\( ½ \times m \times v^2\)
( \(= ½\times 4.2\times 6.0^2)\)
( = 76 J)
force = work done / distance
force = 76 / 0.050
= 1500N
Questions 4
Topic – 6.1
(a) Define strain.
(b) A copper wire of length 4.0m has a uniform cross-sectional area of \(4.5 × 10^{–7}m^2\). A tensile force of 18N is applied to the wire. This causes the wire to extend by 1.4mm up to its limit of proportionality.
(i) Calculate the Young modulus of the wire.
(ii) On Fig. 4.1, draw a line to show how the stress varies with the strain for the wire up to its limit of proportionality.
(c) A second copper wire has the same length as the wire in (b) but a larger diameter. Both wires are subjected to a tensile force of 18N. By placing a tick (3) in each row, complete Table 4.1 to compare the stress and strain of the two wires. 
▶️Answer/Explanation
Ans
(a) extension / original length
(b)(i) Young modulus = stress / strain
=\( (18 / 4.5\times 10^{–7}) / (1.4 \times 10^{–3} / 4.0)\)
= \((4.0 \times 10^7) / (3.5 \times 10^{–4})\)
= \(1.1\times 10^{11}\) Pa
(ii) straight line through the origin
ending at the point (3.5, 4.0)
(c)
Questions 5
Topic – 7.1
A stretched string PQ has length 1.2m. One end of the string is attached to a vibration generator and the other end is attached to a wall, as shown in Fig. 5.1.
The vibration generator is switched on and a stationary wave is formed on the string. The string is shown at one instant of time in Fig. 5.2. 
(a) Explain how a stationary wave is formed between the vibration generator and the wall.
(b) Calculate the wavelength of the stationary wave shown in Fig. 5.2.
(c) Fig. 5.3 shows the stationary wave at time t = 0 when all points on the wave are at their maximum displacements.
The period of the wave is 0.16s. On Fig. 5.3, sketch the shape of the stationary wave at time t = 0.24s.
(d) Points R and T on the string are a horizontal distance of 0.30m apart and in the positions shown in Fig. 5.4. 
State the phase difference between the oscillations of points R and T.
(e) Calculate the speed of the progressive waves on the stretched string.
▶️Answer/Explanation
Ans
(a) wave(s) (travel along string and) reflect at fixed point / wall / Q / end / vibration generator / P
incident and reflected waves superpose
(b) 0.80 m
(c) same wavelength as original throughout and passing through intersection of solid and dashed lines
reflected in dashed line and of same amplitude
(d) 180°
(e) \(v = f \lambda \) and f =1 / T
or
v =\(\lambda \)/ T
v = \(6.25 \times 0.80\) or 0.80 / 0.16
= 5.0 \(ms^{–1}\)
Questions 6
Topic – 10.2
(a) State Kirchhoff’s first law.
(b) A cell with internal resistance r is connected to two resistors of resistances \(R_1\) and \(R_2\) as shown in Fig. 6.1.
The potential differences (p.d.s) across \(R_1\) and \(R_2\) are \(V_1\) and \(V_2\) respectively. The terminal p.d. across the cell is V. The current in the circuit is I. Use Kirchhoff’s laws to show that the total resistance \(R_T\) of the external circuit is given by
\(R_T = R_1 + R_2\).
(c) The electromotive force (e.m.f.) of the cell in Fig. 6.1 is 1.50V. The values of \(R_1\) and \(R_2\) are 10Ω and 15Ω respectively. The terminal p.d. of the cell is 1.35V. Calculate the internal resistance r of the cell.
(d) A resistor of resistance \(R_3\) is added to the circuit in Fig. 6.1, so that the circuit is as shown in Fig. 6.2.
State and explain the effect, if any, of this change on:
(i) the current in the cell
(ii) the terminal p.d. of the cell.
▶️Answer/Explanation
Ans
(a) sum of current(s) entering a junction = sum of current(s) leaving (the same junction)
or
(algebraic) sum of current (s) at a junction is zero
(b) (by Kirchhoff’s second law) \(V =V_1 + V_2\)
so
\(IR_T =IR_1 +IR_2\) (and cancelling I gives) \(R_T =R_1 +R_2\)
or
V / I =\(V_1 / I +V_2 / I\) (and substituting R gives) \(R_T =R_1 +R_2\)
(d)(i) the (total) resistance (of the circuit) has decreased (and e.m.f. is unchanged)
(the current (in the cell) will) increase
(ii) (as the current is greater and so there is a) larger p.d. across the internal resistance
(terminal p.d. will) decrease
Questions 7
Topic – 11.2
Nuclei of an isotope of copper (Cu) each have 29 protons and 37 neutrons. This isotope is a β– emitter.
(a) State the nuclide notation in the form \(_{Z}^{A}\textrm{X}\) for this nucleus of copper.
(b) The energy spectrum of the β– radiation emitted by a sample of this isotope is shown in Fig. 7.1.
(i) Use Fig. 7.1 to explain why other particles apart from the β- particles must be emitted during this decay
(ii) State the name of the other particle emitted during the decay of this isotope.
(iii) The copper isotope decays to an isotope of zinc (Zn). Give the radioactive decay equation for this decay. Include the nucleon and proton numbers of all the particles involved.
▶️Answer/Explanation
Ans
(a) \(_{29}^{66}\textrm{Cu}\)
(b)(i) the energy of the decay is fixed / constant
the energies of the beta particles have a (continuous) range of values / varies / not constant
another particle / an (anti)neutrino must possess the extra / remaining energy (difference between energy of the decay and the \(\beta\) kinetic energy)
(ii) (electron) antineutrino
(iii) \( _{29}^{66}\textrm{Cu}\to _{30}^{66}\textrm{Zn}+_{-1}^{0}\rm{\beta}+_{0}^{0}\rm{\overline{\nu _e}}\)
values for Cu and Zn correct with no other extra particles on either side of the equation
second term correct (\(_{-1}^{0}\rm{\beta}\))
third term correct (\(_{0}^{0}\rm{\overline{\nu _e}}\) )