Question 1:
1 A well has a depth of 36m from ground level to the surface of the water in the well, as shown in Fig. 1.1.
A student wishes to find the depth of the well. The student plans to drop a stone down the well and record the time taken from releasing the stone to hearing the splash made by the stone as it enters the water.
(a) Assume that air resistance is negligible and that the stone is released from rest.
Calculate the time taken for the stone to fall from ground level to the surface of the water.
time = ……………………………………………… s
(b) The time recorded by the student using a stop-watch is not equal to the time in (a).
Suggest three possible reasons, other than the effect of air resistance, for this difference.
1 ……………………………………………………….
2 ………………………………………………………
3 ………………………………………………………
(c) The student repeats the experiment three times and uses the results to calculate the depth of the well. The values are shown in Table 1.1.
The true depth of the well is 36.0m. Explain why these results may be described as precise but not accurate.
▶️Answer/Explanation
Ans :
1(a) t = √(2 s / g)
= √[(2 × 36) / 9.81]
= 2.7 s
1(b) $\bullet$ reaction time between hearing the splash and stopping the stop-watch
$\bullet$ the sound (of the splash) takes time to reach the student or the stone hits the water at a different time to the sound
being heard or the sound (of the splash) has to travel to the student
$\bullet$ the student might not let go of the stone from ground level
$\bullet$ the student might not let go of the stone and start the stop-watch at the same time
$\bullet$ stop-watch may not be properly calibrated / has a zero error
$\bullet$ (local value of) g is not (exactly) 9.81 (ms$^{2}$2)
$\bullet$ stone given initial velocity / initial velocity not zero
$\bullet$ stone does not fall (exactly) vertically / in a straight line
Any three points, 1 mark each
1(c) precise: results are close together / have little scatter
not accurate: the values are not close to / 50% different / (very) different from the true value
Question 2:
A sphere floats in equilibrium on the surface of sea water of density 1050kgm$^{-3}$, as shown in Fig. 2.1.
(a) 21% of the volume of the sphere is below the surface of the water. Calculate the density of the sphere.
density = ………………………………………. kgm$^{-3}$
(b) The sphere is now held so that its entire volume is below the surface of the water. The sphere is then released.
(i) Calculate the initial acceleration of the sphere.
acceleration = ………………………………………… ms$^{-2}$
(ii) The sphere accelerates upwards but remains entirely below the surface of the water. State and explain what happens to the acceleration of the sphere as its velocity begins to increase.
▶️Answer/Explanation
Ans :
2(a) F = 1050 × 9.81 × 0.21 V or W = $\rho$ × 9.81 × V
0.21 V × 1050 ( × 9.81) = V ( × 9.81) ×$\rho$
$\rho$ = 220 kg m $^{-3}$
2(b)(i) F = 1050 × 9.81 × V or W = 220 × 9.81 ×V
(V × 1050 × 9.81) – (V× 220 × 9.81) = (V × 220) × a
a = 37 ms $^{-2}$
2(b)(ii) the (downward) drag / viscous force increases (with speed)
resultant force decreases (as upthrust and weight remain the same)
acceleration decreases (as its velocity increases)
Question 3:
3 (a) State the principle of conservation of momentum.
(b) A firework is initially stationary. It explodes into three fragments A, B and C that move in a horizontal plane, as shown in the view from above in Fig. 3.1.
Fragment A has a mass of 3m and moves away from the explosion at a speed of $4.0ms^{–1}$. Fragment B has a mass of 2m and moves away from the explosion at a speed of $6.0ms^{−1}$ at right angles to the direction of A. Fragment C has a mass of m and moves away from the explosion at a speed v and at an angle θ as shown in Fig. 3.1.
Calculate:
(i) the angle θ
(ii) the speed v.
v = ………………………………………… ms$^{-1}$
(c) The firework in (b) contains a chemical that has mass 5.0g and has chemical energy per unit mass 700J kg$^{-1}$. When the firework explodes, all of the chemical energy is transferred to the kinetic energy of fragments A, B and C.
(i) Show that the total chemical energy in the firework is 3.5J.
(ii) Calculate the mass m.
m = ……………………………………………. kg
▶️Answer/Explanation
Ans :
3(a) sum/total momentum before = sum / total momentum after
or
sum/total momentum (of a system of objects) is constant
if no (resultant) external force / for an isolated system
3(b)(i) 3m×4 = m× v sin$\theta$
(v sin$\theta$ = 12)
2 m × 6 = m×v cos $\theta$
(v cos $\theta$ = 12)
therefore sin $\theta$ = cos $\theta$ or tan$\theta$ = 1
$\theta$= 45°
3(b)(ii) mv × cos 45° = 12m
or
mv × sin 45° = 12m
or
(mv)$^{2}$ = (3m × 4)$^{2}$ + (2m × 6)$^{2}$
v = 17 m s$^{-1}$
3(c)(i) (chemical energy) = 0.0050 × 700 = 3.5 (J)
or
(chemical energy) = 5.0 × 0.700 = 3.5 (J)
3(c)(ii) E = 1⁄2mv$^{2}$ total
E = (0.5 × 3 m × 4$^{2}$) + (0.5 × 2 m × 6$^{2}$) + (0.5 × m × 17$^{2}$)
3.5 = 204m
m = 0.017 kg
Question 4:
4 (a) For a progressive wave, state what is meant by the frequency.
(b) A loudspeaker, microphone and cathode-ray oscilloscope (CRO) are arranged as shown in Fig. 4.1.
The loudspeaker is emitting a sound wave which is detected by the microphone and displayed on the screen of the CRO as shown in Fig. 4.2.
The time-base on the CRO is set to 0.50ms cm−1 and the y-gain is set to 0.20Vcm$^{-1}$.
Calculate:
(i) the frequency of the sound wave
frequency = ……………………………………………. Hz
(ii) the amplitude of the signal received by the CRO.
amplitude = ……………………………………………… V
(c) The intensity of the sound wave in (b) is reduced to a quarter of its original intensity without a change in frequency. Assume that the amplitude of the signal received by the CRO is proportional to the amplitude of the sound wave. On Fig. 4.2, sketch the trace that is now seen on the screen of the CRO.
(d) A metal sheet is now placed in front of the loudspeaker in (b), as shown in Fig. 4.3.
A stationary wave is formed between the loudspeaker and the metal sheet.
(i) State the principle of superposition.
(ii) The initial position of the microphone is such that the trace on the CRO has an amplitude minimum. It is now moved a distance of 1.05m away from the loudspeaker along the line joining the loudspeaker and metal sheet.
As the microphone moves, it passes through three positions where the trace has an amplitude maximum before ending at a position where the trace has an amplitude minimum.
Determine the wavelength of the sound wave.
wavelength = …………………………………………….. m
(iii) Use your answers in (b)(i) and (d)(ii) to determine the speed of the sound in the air.
speed = ………………………………………… ms$^{-1}$
▶️Answer/Explanation
Ans :
4(a) the number of wavefronts/crests/troughs passing a fixed point per unit time or
the number of oscillations per unit time (of source / point on wave / particle of medium)
4(b)(i) T = 4× 0.50× 10 $^{-3}$
( = 2.0× 10$^{-3}$ s)
f = 1 / 2.0 × 10$^{-3}$
= 500 Hz
4(b)(ii) amplitude = 2.8× 0.20
= 0.56 V
4(c) period same as original trace sinusoidal wave of constant amplitude less than 2.8 cm throughout amplitude 1.4 cm
4(d)(i) when (two or more) waves meet (at a point) (resultant) displacement is the sum of the individual displacements
4(d)(ii) node-to-node separation is $\lambda$/2
or
microphone moves through 3 node-to-node separations
or
d = 1.5 $\lambda$
$\lambda$ = 1.05 / 1.5
= 0.70 m
4(d)(iii) v = f ×$\lambda$
= 500 × 0.70
= 350 ms$^{-1}$
Question 5:
A student sets up a circuit with a battery, an ammeter, a heater and a light-dependent resistor (LDR) all in series.
The battery has negligible internal resistance.
A voltmeter is connected across (in parallel with) the heater.
(a) On Fig. 5.1, complete the circuit diagram of this arrangement.
Fig. 5.1
(b) The heater is a wire made of metal of resistivity 1.1 × 10$^{-6}$Ωm. The wire has length 2.0m and cross-sectional area 3.8 × 10$^{-7}$m$^{2}$.
The reading on the voltmeter is 4.8V.
Calculate:
(i) the resistance of the heater
resistance = …………………………… Ω
(ii) the reading on the ammeter.
reading on ammeter = ……………………..A
(c) The heater is replaced by a new wire. The new wire is made of the same metal as the wire in (b) and has the same length but a larger diameter.
The resistance of the LDR remains constant.
(i) State and explain whether the new wire has a resistance that is greater than, less than or the same as that of the wire in (b).
(ii) State and explain whether the new reading on the voltmeter is greater than, less than or equal to 4.8V.
▶️Answer/Explanation
Ans :
5(a) correct symbol for the heater or for the LDR
all correct symbols in series (ignore voltmeter) and no extra symbols
correct symbol for voltmeter and in parallel with the heater
5(b)(i) R = $\rho$L/A
= (1.1 × 10−6 × 2.0) / 3.8 × 10$^{-7}$
= 5.8 Ω
5(b)(ii) I = 4.8 / 5.8 = 0.83
5(c)(i) A larger (for new wire)
or
A$\infty$ d$^{2}$ (and d larger for new wire)
or
R$\infty$ 1 / d$^{2}$ (and d larger for new wire)
so R is less (than that of first wire)
5(c)(ii) (heater / total resistance decreases so) current (in circuit) increases (so p.d. across LDR increases)
or
heater resistance decreases so it has a smaller share/proportion/fraction of the (total) voltage / e.m.f.
(so voltmeter) reading is less (than 4.8 V)
Question 6:
6 (a) Define the Young modulus.
(b) A uniform wire is suspended from a fixed support. Masses are added to the other end of the wire, as shown in Fig. 6.1.
The variation of the length l of the wire with the force F applied to the wire by the masses is shown in Fig. 6.2.
The cross-sectional area of the wire is 0.95mm.
(i) Determine the unstretched length of the wire.
unstretched length = …………………………………………….. m
(ii) For an applied force F of 30N, determine:
● the stress in the wire
stress = ………………………………………………… Pa
● the strain of the wire.
strain = ……………………………
▶️Answer/Explanation
Ans :
6(a) (Young modulus =) stress / strain
6(b)(i) unstretched length = 1.9980 m
6(b)(ii) stress = F / A
= 30 / 9.5× 10$^{-7}$
= 3.2 ×10$^{7}$ Pa
strain = 0.0050 / 1.9980
= 2.5 × 10$^{-3}$
Question 7:
7 (a) Table 7.1 shows incomplete data for three flavours (types) of quark. The elementary charge is e.
Complete Table 7.1 by inserting the missing charges.
(b) Using the symbols given in Table 7.1, state a possible quark combination for the following hadrons:
(i) a neutral baryon
(ii) a meson with a charge of +e.(c) Quarks are fundamental particles.
Electrons are in another group (class) of fundamental particle.
(i) State the name of this group.
(ii) State the name of another particle in this group.
▶️Answer/Explanation
Ans :
7(a) down charge = –1/3( e) and charm charge = (+)2/3(e)
all antiquarks have opposite sign and same (non-zero) magnitude of charge as the corresponding quarks
7(b)(i) udd or cdd
7(b)(ii) ud or cd
7(c)(i) lepton(s)
7(c)(ii) positron / neutrino / antineutrino