Tuesday, December 22, 2015

Physics 9702 Doubts | Help Page 229

  • Physics 9702 Doubts | Help Page 229



Question 1082: [Waves > Double slit]
(a) State one difference and one similarity between longitudinal and transverse waves.

(b) A laser is placed in front of two slits as shown in Fig. 6.1.

The laser emits light of wavelength 6.3 × 10–7 m.
The distance from the slits to the screen is 2.5 m. The separation of the slits is 0.35 mm.
An interference pattern of maxima and minima is observed on the screen.
(i) Explain why an interference pattern is observed on the screen.
(ii) Calculate the distance between adjacent maxima.

(c) State and explain the effect, if any, on the distance between adjacent maxima when the laser is replaced by another laser emitting ultra-violet radiation.

Reference: Past Exam Paper – November 2014 Paper 22 Q6



Solution 1082:
(a)
Difference:
The vibration / oscillation (of particles) / displacement of the particles is parallel to the energy transfer / wavefronts in longitudinal waves and perpendicular for transverse waves
OR Transverse waves can be polarised, longitudinal waves cannot be polarized

Similarity: both transfer / propagate energy

(b)
(i) Waves from the slits are coherent / have a constant phase relationship
The waves overlap (at screen) with a phase difference or have a path difference

EITHER Maxima are obtained where the phase difference is an integer ×360° (or ×2π rad) OR the path difference is an integer ×λ
OR equivalent explanation of minima e.g. (n+½)×360°

(ii)
{λ = wavelength, D = distance from slits to screen, a = separation of slits}
Maxima spacing = λD / a
Maxima spacing = (6.3×10–7 × 2.5) / 0.35 × 10–3
Maxima spacing = 4.5 × 10–3 m

(c) The (ultra-violet radiation has) a shorter wavelength, hence there will be a smaller separation / distance











Question 1083: [Current of Electricity > Capacitors]
Some capacitors are marked ‘48 μF, safe working voltage 25 V’.
Show how a number of these capacitors may be connected to provide a capacitor of capacitance
(a) 48 μF, safe working voltage 50 V,
(b) 72 μF, safe working voltage 25 V.

Reference: Past Exam Paper – November 2002 Paper 4 Q5



Solution 1083:
(a)
Two capacitors in series          OR any circuit such that V ≤ 25V across any C
In parallel with second series pair       OR any correct combination
{Facts:
For capacitors in series: 1/C = 1/C1 + 1/C2 + …
For capacitors in parallel: C = C1 + C2 + …
From Kirchhoff’s laws, the sum of p.d. across any branch in a parallel combination is the same for any all branches.

To have a voltage of 50V from these capacitors, we can connect 2 capacitors in series. The sum of p.d. is then (25 + 25 =) 50 V. However, this would produce an overall capacitance of 24 μF. To increase the capacitance, we connect the 2 capacitors in series above in parallel with another pair of capacitors in series. Since this is a parallel connection, the voltage of 50 V across the branch is still the same, but the overall capacitance is now 48 μF }

(b) Two capacitors in series in parallel with a single capacitor           OR other correct combination
{The overall capacitance needs to be 72 μF. This can be obtained by connecting ‘2 capacitance in series’ with another capacitor in parallel.
For the 2 capacitors in series 1/C = 1/48 + 1/48 giving C = 24 μF
Overall capacitance of parallel connection = 24 + 48 = 72 μF
One of the branches contain a single capacitor with a safe working voltage of 25V. Since the sum of p.d. should be the same in all branches, the p.d. across EACH capacitor in the other branch would be (25 / 2 =) 12.5 V. The safe working voltage is 25V: this represents the maximum voltage that can be used. Any lower voltage can also be used though.}










Question 1084: [Kinematics > Projectile motion]
Water emerges horizontally from a hose pipe with a velocity of 4.0 ms-1 as shown in the figure. The pipe is pointed at P on a vertical surface 2.0 m from the pipe. If the water strikes at S, calculate PS.


Reference: ???



Solution 1084:
This is a projectile motion problem. The motion can be broken down into a horizontal and a vertical motion.

The vertical distance PS needs to be found. Time, t is the quantity which is common for both the vertical and horizontal motion. [For the amount of time the water would have travelled a distance of 2.0m horizontally, it would also have dropped a distance PS vertically]

For the horizontal motion,
Speed, u = 4.0 ms-1
Distance to travel, s = 2.0 m
No acceleration (The horizontal motion is not affected by the acceleration of free fall. Thus, the horizontal motion occurs at a constant speed.)
Speed u = s / t             giving time t = s / u = 2.0 / 4.0 = 0.5 s

For the vertical motion,
Distance dropped, s = PS = ???
Acceleration (due to gravity), a = 9.81 ms-2
Initial speed = 0 ms-1 (since the water emerges horizontally, it has no vertical component initially)
Time, t = 0.5s
Equation for uniformly accelerated motion:   s = ut + ½at2
 PS = 0 + 0.5(9.81)(0.5)2 = 1.72625 = 1.73 m


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