Thursday, October 22, 2015

Physics 9702 Doubts | Help Page 215

  • Physics 9702 Doubts | Help Page 215

Question 1028: [Work, Energy and Power]
A trolley runs from P to Q along a track. At Q its potential energy is 50 kJ less than at P.

At P, the kinetic energy of the trolley is 5 kJ. Between P and Q, the work the trolley does against friction is 10 kJ.
What is the kinetic energy of the trolley at Q?
A 35 kJ                        B 45 kJ                        C 55 kJ                        D 65 kJ

Reference: Past Exam Paper – June 2002 Paper 1 Q18 & November 2011 Paper 11 Q18 & Paper 13 Q20

Solution 1028:
Answer: B.
Let potential energy at P = PP and let potential energy at Q = PQ.
From the question, PP – PQ = 50kJ

Let kinetic energy at Q = KQ
So from conservation of energy,
Sum of energies at P = Sum of energies at Q + Work against friction
PP + KP = PQ + KQ + 10
PP + 5 = (PQ + KQ) + 10
KQ = (Pp – PQ) + 5 – 10 = 50 + 5 – 10 = 45kJ 

Question 1029: [Radioactivity]
(a) Explain what is meant by the binding energy of a nucleus.

(b) Fig. 7.1 shows the variation with nucleon number (mass number) A of the binding energy per nucleon EB of nuclei.

One particular fission reaction may be represented by the nuclear equation
23592U   +          10n               14156Ba             +          9236Kr   +          310n
 (i) On Fig. 7.1, label the approximate positions of
1. the uranium (23592U) nucleus with the symbol U,
2. the barium (14156Ba) nucleus with the symbol Ba,
3. the krypton (9236Kr) nucleus with the symbol Kr.

(ii) The neutron that is absorbed by the uranium nucleus has very little kinetic energy.
Explain why this fission reaction is energetically possible.

(c) Barium-141 has a half-life of 18 minutes. The half-life of Krypton-92 is 3.0 s.
In the fission reaction of a mass of Uranium-235, equal numbers of barium and krypton nuclei are produced.
Estimate the time taken after the fission of the sample of uranium for the ratio
number of Barium-141 nuclei / number of Krypton-92 nuclei
to be approximately equal to 8.

Reference: Past Exam Paper – November 2007 Paper 4 Q7

Solution 1029:
(a) Binding energy of a nucleus is the energy required to (completely) separate the nucleons (in a nucleus).

1. U must be labeled near the right-hand end of the line
2 & 3 Ba and Kr should be in approximately correct positions
{Iron-56 is among those with the highest binding energy per nucleon. Iron-56 corresponds to nucleon number = 56 and so, the highest value of EB on the curve would be at about A = 56. This can be used as a reference to plot the others.}

The binding energy is (A × EB).
EITHER The binding energy of U < binding energy of (Ba + Kr) OR EB of U < EB of (Ba + Kr)

Krypton-92 is reduced to {[1/2]3 =} 1/8 in 9s {3 × half-life}.
In 9s, very little decay of Barium-141 occurs. 
So, time taken is approximately 9s.   

(λ = 0.693 / t½)
λKr = 0.231      or         λBa = 6.42×10-4
8 = exp(-λBat) / exp(-λKrt)
Time taken, t = 9.0s

Question 1030: [Electric field]
(a) Define electric field strength.
(b) A sphere S has radius 1.2 × 10–6 m and density 930 kg m−3.
Show that the weight of S is 6.6 × 10−14 N.

(c) Two horizontal metal plates are 14 mm apart in a vacuum. A potential difference (p.d.) of 1.9 kV is applied across the plates, as shown in Fig. 3.1.

A uniform electric field is produced between the plates.
The sphere S in (b) is charged and is held stationary between the plates by the electric field.
(i) Calculate the electric field strength between the plates.
(ii) Calculate the magnitude of the charge on S.
(iii) The magnitude of the p.d. applied to the plates is increased.
Explain why S accelerates towards the top plate.

Reference: Past Exam Paper – November 2014 Paper 22 Q3

Solution 1030:
(a) Electric field strength is defined as the electric force per unit positive charge
mass = volume × density
{Volume = 4/3 × π × r3}
mass = 4/3 × π × (1.2 × 10–6)3 × 930 (= 6.73 × 10–15)
{Weight = mg}
weight = 4/3 × π × (1.2 × 10–6)3 × 930 × 9.81 = 6.6 × 10–14 N

{Electric field strength E = V / d}
E = 1.9 × 103 / 14 × 10–3
E = 1.4 (1.36) × 105 V m–1

Electric force, F = QE
{As the sphere is stationary, the electric force is equal to the weight.}
Q = 6.6 × 10–14 / 1.36 × 105   
Q = 4.9 (4.86) × 10–19 C

(iii) The electric force increases / is greater (than weight) and the charge (on S) is negative to give resultant / net / sum / total force up
{F = EQ = (V/d) Q = VQ / d. S should be negative in order to be attracted to the upper plate which has a positive potential.}

Question 1031: [Applications > Operational Amplifiers]
An operational amplifier (op-amp) is used in the comparator circuit of Fig. 10.1.

(a) (i) Show that the potential at the inverting input of the op-amp is +1.0 V.
(ii) Explain why the potential difference across resistor R is + 5 V when VIN is greater than 1.0 V and is zero when VIN is less than 1.0 V.

(b) The variation with time t of the input voltage VIN is shown in Fig. 10.2.

(i) On the axes of Fig. 10.2, draw the variation with time t of the output potential VOUT.
(ii) Suggest a use for this type of circuit.

Reference: Past Exam Paper – June 2015 Paper 41 & 43 Q10

Solution 1031:
(a) (i) Potential = 1.2 / (1.2 + 4.2) × 4.5 = +1.0 V
{The inverting potential is V-. A potential divider is formed with the total p.d. being 4.5V. We need to find the p.d. across the 1.2 kΩ.}

{V has been calculate to be +1.0V above}
for VIN > 1.0 V, V+ > V
The output (of op-amp) is +5 V          or positive
and the diode conducts giving +5 V across R            or Vout is +5 V

{The diode only conducts in one direction.}
for VIN < 1.0 V, output of op-amp is –5 V / negative so the diode does not conduct, giving Vout = 0          or 0 V across R

(i) The graph is a square wave with maximum value +5 V and minimum value 0 with vertical sides in correct positions and correct phase

(ii) For re-shaping (digital) signals/regenerator (amplifier)


  1. J15 p 13 Ques 29 ,37
    thank u in advance..

    1. For J15 p 13 Ques 29, check solution 1037 at

  2. Really appreciated your efforts for these wonderful resources and thumbs up


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