(a) A particle of mass m kg having an initial velocity V₀ is subjected to a retarding force proportional to its instantaneous velocity. Obtain the expression for the velocity and position of the particle as a function of time. 10 marks

(b) Show that

Question

(a) A particle of mass m kg having an initial velocity V₀ is subjected to a retarding force proportional to its instantaneous velocity. Obtain the expression for the velocity and position of the particle as a function of time. 10 marks

(b) Show that the kinetic energy of a system of n particles is given by T = ½ MV²_cm + ½ Σⁿᵢ₌₁ mᵢV'²ᵢ where M is the total mass, V_cm is the velocity of the centre of mass, V'ᵢ is the velocity of the particles about the centre of mass and mᵢ is the mass of the ith particle. 10 marks

(c) A charged π-meson with rest mass of 273mₑ at rest decays into a neutrino and a μ-meson of rest mass 207mₑ. Find the kinetic energy of the μ-meson and the energy of the neutrino. (mₑ is the rest mass of the electron) 10 marks

(d) The intensity at the central maximum observed on a screen in a double-slit experiment is 2×10⁻³ W/m². If the path difference between interfering waves reaching a point on the screen is λ/6, where λ is the wavelength of the light used in the experiment, determine the intensity at that point. 10 marks

(e) A telescope has an objective lens of diameter 10 cm. Determine whether this telescope can resolve two stars having an angular separation of 2·4 seconds of arc. (Assume the wavelength of starlight as 550 nm) 10 marks

UPSC Answer Check · Accepted Answer

(a) 
Let the retarding force be F = −k v, where k > 0 is a constant of proportionality. This is the standard linear drag model. Newton’s second law for the particle of mass m gives
m dv/dt = −k v.
Separating the variables,
dv/v = −(k/m) dt.
Integrating from t = 0, when the velocity is V₀, to time t, when the velocity is v:
∫_V₀^v dv′/v′ = −(k/m) ∫_0^t dt′.
This gives
ln(v/V₀) = −(k/m)t.
Taking exponentials,
v(t) = V₀ exp(−kt/m).
The velocity decays exponentially with time. As t → ∞, v → 0, which is consistent with a purely retarding force. The constant k has units kg/s, so kt/m is dimensionless. If V₀ were negative, the same expression holds with the direction reversed, because the force always opposes the instantaneous velocity.

For the position, use v = dx/dt:
dx/dt = V₀ exp(−kt/m).
Integrating from the initial position x₀ at t = 0 to position x at time t:
x(t) − x₀ = ∫_0^t V₀ exp(−kt′/m) dt′
= V₀ [ −(m/k) exp(−kt′/m) ]_0^t
= (m V₀/k) [1 − exp(−kt/m)].
Therefore,
x(t) = x₀ + (m V₀/k) [1 − exp(−kt/m)].
At large time, the particle approaches the limiting displacement x₀ + m V₀/k. This solution assumes the force remains linear in velocity throughout the motion and that no other forces act.

(b) 
This is König’s theorem for kinetic energy. Let the system consist of n particles with masses mᵢ, positions rᵢ, and velocities vᵢ. The total mass is
M = Σᵢ₌₁ⁿ mᵢ.
The centre of mass is at
R = (1/M) Σᵢ₌₁ⁿ mᵢ rᵢ.
Differentiating with respect to time gives the centre-of-mass velocity:
V_cm = Ṙ = (1/M) Σᵢ₌₁ⁿ mᵢ vᵢ.
Thus,
Σᵢ₌₁ⁿ mᵢ vᵢ = M V_cm.
Now define the velocity of each particle relative to the centre of mass:
v′ᵢ = vᵢ − V_cm.
Then,
Σᵢ₌₁ⁿ mᵢ v′ᵢ = Σᵢ₌₁ⁿ mᵢ vᵢ − V_cm Σᵢ₌₁ⁿ mᵢ
= M V_cm − M V_cm = 0.
The total kinetic energy is
T = ½ Σᵢ₌₁ⁿ mᵢ vᵢ²
= ½ Σᵢ₌₁ⁿ mᵢ (V_cm + v′ᵢ)·(V_cm + v′ᵢ).
Expanding,
T = ½ Σᵢ₌₁ⁿ mᵢ V_cm² + V_cm · Σᵢ₌₁ⁿ mᵢ v′ᵢ + ½ Σᵢ₌₁ⁿ mᵢ v′ᵢ².
The cross term vanishes because Σᵢ₌₁ⁿ mᵢ v′ᵢ = 0. Also,
½ Σᵢ₌₁ⁿ mᵢ V_cm² = ½ (Σᵢ₌₁ⁿ mᵢ) V_cm² = ½ M V_cm².
Hence,
T = ½ M V_cm² + ½ Σᵢ₌₁ⁿ mᵢ v′ᵢ².
This decomposition is valid in any inertial frame. The first term is the kinetic energy associated with the translational motion of the centre of mass, and the second term is the internal kinetic energy of the particles relative to the centre of mass. The theorem is useful because it separates the overall motion of the system from its internal motion.

(c) 
Let mπ = 273 mₑ, mμ = 207 mₑ, and mν ≈ 0. The pion is initially at rest, so its total energy is
E_initial = mπ c² = 273 mₑ c².
After decay, let the magnitude of the momentum of each final particle be p. Momentum conservation requires the muon and neutrino to have equal and opposite momenta. For the massless neutrino,
Eν = pν c = p c.
For the muon, the relativistic energy-momentum relation gives
Eμ = √(p² c² + mμ² c⁴).
Energy conservation gives
mπ c² = Eμ + Eν.
Let x = Eν = p c. Then,
√(x² + mμ² c⁴) + x = mπ c².
Write a = mμ c² and b = mπ c². Then,
√(x² + a²) = b − x.
Squaring both sides,
x² + a² = b² − 2 b x + x²,
so
2 b x = b² − a²,
and therefore,
x = (b² − a²)/(2 b)
= (mπ² − mμ²) c² / (2 mπ).
Substituting the masses:
Eν = [(273² − 207²) / (2 × 273)] mₑ c²
= (31680 / 546) mₑ c²
= 5280/91 mₑ c² ≈ 58.02 mₑ c² ≈ 29.65 MeV.
The muon total energy is
Eμ = mπ c² − Eν
= (mπ² + mμ²) c² / (2 mπ).
Its kinetic energy is
Kμ = Eμ − mμ c²
= [(mπ − mμ)² / (2 mπ)] c².
Substituting,
Kμ = [(273 − 207)² / (2 × 273)] mₑ c²
= (66² / 546) mₑ c²
= (4356 / 546) mₑ c²
= 726/91 mₑ c² ≈ 7.98 mₑ c² ≈ 4.08 MeV.
As a check, Kμ + Eν = (726/91 + 5280/91) mₑ c² = (6006/91) mₑ c² = 66 mₑ c², and adding the muon rest energy 207 mₑ c² gives 273 mₑ c², the original pion rest energy. Here the neutrino is treated as massless, which is an excellent approximation for this decay.

(d) 
In a double-slit experiment, the intensity at a point is given by
I = I₀ cos²(δ/2),
where I₀ is the intensity at the central maximum and δ is the phase difference between the two interfering waves. The phase difference is related to the path difference Δ by
δ = (2π/λ) Δ.
Given Δ = λ/6,
δ = (2π/λ)(λ/6) = π/3.
Therefore,
I = I₀ cos²(π/6)
= I₀ (√3/2)²
= I₀ × 3/4.
With I₀ = 2 × 10⁻³ W/m²,
I = (3/4) × 2 × 10⁻³ W/m²
= 1.5 × 10⁻³ W/m².
This assumes the two slits have equal intensities and that the diffraction envelope and background light are negligible.

(e) 
By Rayleigh’s criterion, the minimum angular separation that a telescope can resolve is
θ_min = 1.22 λ / D,
where λ is the wavelength of light and D is the diameter of the objective lens. Here,
D = 10 cm = 0.10 m,
λ = 550 nm = 5.50 × 10⁻⁷ m.
Thus,
θ_min = (1.22 × 5.50 × 10⁻⁷ m) / 0.10 m
= 6.71 × 10⁻⁶ rad.
To convert radians to seconds of arc, use
1 rad = (180/π) × 3600 arcsec ≈ 206265 arcsec.
Hence,
θ_min ≈ 6.71 × 10⁻⁶ × 206265 arcsec
≈ 1.38 arcsec.
The given angular separation of the two stars is 2.4 arcsec, which is greater than the minimum resolvable angle 1.38 arcsec. Therefore,
the telescope can resolve the two stars.
This conclusion assumes diffraction-limited performance; atmospheric turbulence and optical aberrations are ignored.

(a) A particle of mass m kg having an initial velocity V₀ is subjected to a retarding force proportional to its instantaneous…

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