## Wednesday, October 20, 2010

### Multiple Choice Questions on Elastic and Inelastic Collisions

Today we will discuss two questions on collisions. The first question pertains to elastic collision in one dimension and the second question pertains to inelastic collision in two dimensions:

(1) A particle A of mass m moving along the positive x-direction with kinetic energy K suffers an elastic head-on collision with a stationary particle B of mass M. After the collision the particle A moves along the negative x-direction with kinetic energy K/9. What is the mass of particle B?

(a) 9 m

(b) 6 m

(c) 3 m

(d) 2 m

(e) m/3

Since the kinetic energy of A after collision is one-nineth of its initial kinetic energy, the momentum of A after collision is one-third of its initial momentum.

Since the momentum is to be conserved, we have

p = p p/3 where p is initial momentum of A and p’ is the momentum of B after the collision.

[The final momentum of particle A is negative since its direction is reversed].

Therefore, p’ = 4p/3

The kinetic energy gained by particle B due to the collision is p2/2M where M is the mass of particle B.

The kinetic energy lost by particle A due to the collision is (8/9)×p2/2m.

[Note that the initial kinetic energy of particle A is p2/2m and its final kinetic energy is (1/9) p2/2m].

Since the kinetic energy too is conserved in elastic collisions, the kinetic energy gained by particle B is equal to the kinetic energy lost by particle A. Therefore, we have

p2/2M = (8/9) p2/2m

Substituting for p’ = 4p/3, we have

(16/9) (p2/2M) = (8/9) p2/2m from which M = 2 m.

(2) A particle of mass m moving westwards with speed v collides with an identical particle moving northwards with speed v. If they stick together after the collision, what is their common speed after the collision?

(a) 4v

(b) 2√2 v

(d) √2 v

(e) v/√2

The total momentum before collision is the vector sum of the initial momenta mv and mv at right angles (since one is westwards and the other is northwards) and is equal to √2 mv acting along the north-west direction (Fig.).

The combined mass is 2m. If V is the common speed, the total momentum after collision is 2m V, acting along the north-west direction.

Since the momentum is conserved, we have

2m V = √2 mv so that V = v/√2

## Friday, October 15, 2010

### AIEEE 2010 Questions (MCQ) from Nuclear Physics

“I am a friend of Plato, I am a friend of Aristotle, but truth is my greater friend”

– Sir Isaac Newton

Multiple choice questions based on a given paragraph are seen in many entrance exam question papers. Here are two such questions on nuclear physics (which appeared in AIEEE 2010 question paper):

Directions : Questions (i) and (ii) are based on the following paragraph:

A nucleus of mass M + Δm is at rest and decays into two daughter nuclei of equal mass M/2 each. Speed of light is c.

(i) The speed of daughter nucleus is

(1) c√[Δm/(M + Δm)]

(2) c [Δm/(M + Δm)]

(3) c√(2Δm/M)

(4) c√(Δm/M)

The energy released in the process is Δmc2 in accordance with Einstein’s mass-energy relation. This energy is carried by the daughter nuclei (as kinetic energy).

Therefore we have

½ (M/2)v2 + ½ (M/2)v2 = Δmc2 where v is the speed of each daughter nucleus.

Or, Mv2/2 = Δmc2 from which v = c√(2Δm/M), as given in option (3).

[Note that the momentum is to be conserved in the process. Therefore, the daughter nuclei have to travel in opposite directions with the same speed since they have the same mass. If you were asked to give the velocities of the daughters, the answer would be v and – v]

(ii) The binding energy per nucleon for the parent nucleus is E1 and that for the daughter nuclei is E2. Then

(1) E1 = 2 E2

(2) E2 = 2 E1

(3) E1 > E2

(4) E2 > E1

The binding energy per nucleon for the daughter nuclei is greater than that for the parent nucleus since energy is released in the process[Option (4)].

The following question (MCQ) also was included in the AIEEE 2010 question paper:

A radioactive nucleus (initial mass number A and atomic number Z) emits 3 α-particles and 2 positrons. The ratio of number of neutrons to that of protons in the final nucleus will be

(1) (A Z 4)/(Z 2)

(2) (A Z 8)/(Z – 4)

(3) (A Z 4)/(Z 8)

(4) (A Z 12)/(Z – 4)

When an α-particle is emitted, the nucleus loses two protons and two neutrons. When a positron is emitted by the nucleus, a proton in the nucleus gets converted into a neutron. Since the number of neutrons in the original neucleus is A Z, the number of neutrons in the final nucleus will be (A Z) – (3×2) + 2 = A Z 4.

The protons in the final nucleus will be Z – (3×2) 2 = Z 8

Therefore, the ratio of number of neutrons to that of protons in the final nucleus will be(AZ – 4)/(Z – 8), as given in option (3).

## Saturday, October 02, 2010

### EAMCET (Engineering Entrance) 2010 Questions on Capacitors

“Non-violence is the greatest force at the disposal of mankind. It is mightier than the mightiest weapon of destruction devised by the ingenuity of man”

– Mahatma Gandhi

Today we will discuss three questions which appeared in EAMCET (Engineering Entrance) 2010 question paper. Questions (1) and (2) are from electrostatics while question no.3 is from transient phenomena.

(1) Two capacitors of capacities 1 μF and C μF are connected in series and the combination is charged to a potential difference of 120 V. If the charge on the combination is 80 μC, the energy stored in the capacitor of capacity C in microjoule is

(1) 1800

(2) 1600

(3) 14400

(4) 7200

The effective capacitance of the series combination is 1×C/(1+C) μF = C/(1+C) μF.

[Remember that when capacitors C1, C2, C3,….etc are connected in series, the effective capacitance Ceff is given by the reciprocal relation, 1/Ceff = 1/C1 + 1/C2 + 1/C3 + …. etc.]

We retain the unit of capacitance as μF for convenience.

Since the charge carried by the combination is 80 μC, we have (from Q = CV with usual notations)

80 = [C/(1+C)]×120

[We have substituted the charge in microcoulomb itself since the capacitance is in microfarad].

This gives 80 C + 80 = 120 C from which C = 2 μF

When capacitors are in series, all capactors carry the same charge. Therefore, the charge on the 2 μF capacitor is 80 microcoulomb.

The energy U of the 2 μF capacitor is given by

U = Q2/2C = 802/(2×2) = 1600 microjoule.

[Since Q is in microfarad and C is in microcoulomb, the energy is in microjoule].

(2) The potential difference between two parallel plates is 104 volts. If the plates are separated by 0.5 cm, the force on an electron between the plates is

(1) 32×10–13 N

(2) 0.32×10–13 N

(3) 0.032×10–13 N

(4) 3.2×10–13 N

This is a very simple question. The electric field (E) between the plates is given by

E = V/d where V is the potential difference and d is the separation between the plates

The force F on the electron (of caharge e = 1.6×10–19 coulomb) is given by

F = Ee = Ve/d = (104×1.6×10–19) /(0.5×10–2) = 3.2×10–13 N

(3) A capacitor of capacity 0.1 μF connected to a resistor of 10 MΩ is charged to a certain potential and then made to discharge through the resistor. The time in which the potential will take to fall to half its original value is (Given log10 2 = 0.3010)

(1) 2 sec

(2) 0.693 sec

(3) 0.5 sec

(4) 1 sec

If V0 is the initial voltage across the capacitor of capacitance C, the voltage V at time t during the discharge through the resistance R is given by

V = V0 et/RC where e is the base of natural logarithms.

Therefore V/V0 = et/RC

Since V/V0 = ½ in the question, we have

½ = et/RC

Or, et/RC = 2

Taking logarithm, t/RC = ln 2

Therefore, t = RC ln 2 = 10×106×0.1×10–6×ln 2

If you remember that ln 2 = 0.693, you can write the answer immediately as 0.693 sec.

[You should definitely know that the relation between natural logarithm (with base e) and common logarithm (base 10) of quantity x is

lne x = 2.303 log10 x

So ln 2 = 2.303×0.3010 = 0.693]