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Question 715: [Dynamics > Momentum]

A body of mass m, moving at velocity v, collides with stationary body of the same mass and sticks to it.

Which row describes momentum and kinetic energy of the two bodies after the collision?

momentum                  kinetic energy

A         mv                               ¼ mv²

B         mv                               1/8 mv²

C         2mv                             ½ mv²

D         2mv                             mv²

Reference: Past Exam Paper – June 2011 Paper 11 Q9

Solution 715:

Answer: A.

In any collision in a closed system (such as this one), momentum should be conserved. That is, the sum of momentum before collision should be equal to the sum of momentum after collision.

Sum of momentum before collision = mv + m(0) = mv

After collision, mass of body = 2m. Let the velocity after collision be u.

Sum of momentum after collision = (2m)u = 2mu

From the conservation of momentum,

2mu = mv

Velocity u = v / 2

Momentum after collision = 2mu = 2m (v/2) = mv

(The above momentum could be deduced directly from the conservation of momentum, but the final velocity would be unknown.)

Kinetic energy after collision = ½ (2m) (v/2)² = ¼ mv²

Question 716: [Nuclear Physics]

Nuclear reaction between two helium nuclei produces a second isotope of helium, two protons and 13.8 MeV of energy. Reaction is represented by the following equation.

³₂He     +          ³₂He    - - - >   ^........._.........He     +          2 ^........._.........p     +          13.8 MeV

(a) Complete the nuclear reaction

(b) By reference to this reaction, explain meaning of the term isotope.

(c) State quantities that are conserved in this nuclear reaction

(d) Radiation is produced in this nuclear reaction.

State

(i) possible type of radiation that may be produced

(ii) why energy of this radiation less than the 13.8MeV given

(e) Calculate minimum number of these reactions needed per second to produce power of 60W

Reference: Past Exam Paper – November 2012 Paper 22 Q7

Solution 716:

(a) {A = upper value, Z = lower value}

He:      A = 4               Z = 2

p:         A = 1               Z = 1

(b) The Helium produced and those reacting are isotopes since the nuclei of both have 2 protons (same number of protons). The reacting Helium nuclei have 1 neutron while the one produced has 2 neutrons.

(c)

Proton number and neutron number

Energy – mass

Momentum

(d)

(i) γ radiation  

{γ radiation is an EM radiation. It is neutral (uncharged) and do not has mass. An alpha particle is a helium nucleus. It has a charge (+2) and some mass. As stated in part (c), some quantities must be conserved in the nuclear reaction. Since these are already conserved, we cannot add more mass, charge, … The same applies for beta particles.}

(ii) The product(s) must have some kinetic energy.

(e)

Energy in Joules:

13.8MeV = 13.8 × 1.6×10^-19 × 10⁶ = 2.208×10^-12 J

Power:

Let the number of reactions per second be n

60 = n × 2.208×10^-12

n = 2.7(2) × 10¹³ reactions s^-1

Question 717: [Radioactivity]

Scientists investigating count rate from a radioactive source observed that the count rate fluctuates.

What do these fluctuations imply about the nature of radioactive decay?

A It involves atomic nuclei.

B It is predictable.

C It is random.

D It is spontaneous.

Reference: Past Exam Paper – June 2013 Paper 11 Q38

Solution 717:

Answer: C.

Fluctuations in the count rate of a radioactive source indicates the radioactive decay is random. It cannot be predicted.

Question 718: [Transformer]

A transformer is illustrated in Fig.1.

(a)

(i) Explain why coils are wound on a core made of iron.

(ii) Suggest why thermal energy is generated in the core.

(b)

(i) State Faraday’s law of electromagnetic induction.

(ii) Use Faraday’s law to explain why potential difference across the load and the e.m.f. of the supply are not in phase.

(c) Electrical energy is usually transmitted using alternating current. Suggest why transmission is achieved using

(i) high voltages,

(ii) alternating current.

Reference: Past Exam Paper – June 2011 Paper 41 Q6

Solution 718:

(a)

(i) To concentrate the (magnetic) flux / reduce flux losses

(ii) The changing flux (in the core) induces current in the core. The currents in core give rise to a heating effect.

(b)

(i) Faraday’s law of electromagnetic induction states that the e.m.f. induced is proportional to the rate of change of (magnetic) flux (linkage)

(ii) The magnetic flux is in phase with / proportional to e.m.f. / current in primary coil. The e.m.f. / p.d. across the secondary coil is proportional to rate of change of flux. So the e.m.f. of the supply is not in phase with the p.d. across the secondary coil.

{For the primary coil, the magnetic flux is proportional to the current while for the secondary, the e.m.f. induced is proportional to the rate of CHANGE of flux. So, if the flux is constant for the secondary, no e.m.f. is induced – it should be changing. This is not the case for the primary. As long as there is an e.m.f. in the primary, the magnetic flux is proportional to it – it does not need to be changing.}

(c)

(i) For the same power (transmission) {P = VI}, a high voltage is transmitted with low current. With low current, there is less energy losses in transmission cables {P = I²R}.

(ii) The voltage is easily / efficiently changed.

{With a changing current, the voltage can be easily changed.}

Question 719: [Kinematics]

Diagram shows a velocity-time graph for a vehicle.

The vehicle, moving at 4.0 m s^–1, begins to accelerate at time = 0.

What is vehicle’s acceleration at time = 3.0 s?

A 0.67 m s^–2                B 1.0 m s^–2                  C 1.3 m s^–2                  D 2.0 m s^–2

Reference: Past Exam Paper – June 2010 Paper 11 Q8 & Paper 12 Q9 & Paper 13 Q14

Solution 719:

Answer: C.

From a velocity-time graph, the acceleration at time t is given by the gradient at time t. This gives the instantaneous acceleration at that time.

In this graph, the gradient and hence the acceleration is changing.

Note that the gradient at time t = 3s is obtained by drawing a tangent at that point and calculating its gradient. We cannot just take the coordinate at that point (3, 6) to obtain acceleration (as done for choice D).

A tangent at time t = 3s would pass almost at/close to points (0, 2) and (6, 10).

Gradient = (10 – 2) / (6 – 0) = 8/6 = 1.33 ms^-1  

Question 720: [Kinematics]

The dotted line shows path of a competitor in a ski-jumping competition.

Ignoring air resistance, which graph best represents variation of his speed v with the horizontal distance x covered from the start of his jump at P before landing at Q?

Reference: Past Exam Paper – November 2012 Paper 12 Q10

Solution 720:

Answer: D.

At the maximum height, speed v is minimum (due to conservation of energy – gravitational potential energy is maximum and kinetic energy is minimum) [A is incorrect] but the speed of the jumper is not zero at the top of the flight as he still has horizontal velocity. [C is incorrect]

Just before landing (at maximum value of x in graph), the speed is not zero. Since P is at point higher than Q, the person has both potential energy and kinetic energy at P. From the conservation of energy, kinetic energy is higher at Q (since Q is at a lower height – the potential energy is converted into kinetic energy), and so, speed v at Q must be higher than at P.

Due to acceleration due to gravity, the change in speed v with x (i.e. the gradient) is not uniform. [B is incorrect]