Physics Challenge by JMR's — Solutions

Take C as origin. Let r = distance of boat from C and x = its distance from C measured along the river (downstream).

Boat velocity = V towards C + V along river ⇒ drdt = −V + Vxr,   dxdt = −Vxr + V

Adding: d(r + x)dt = 0 ⇒ r + x = constant.

At A: r = AC = bcos 37° = 5b4, x = CB = b tan 37° = 3b4 ⇒ r + x = 2b.

At D (on the bank): r = x = CD ⇒ 2·CD = 2b ⇒ CD = b ⇒ BD = b − 3b4 = b4

Answer (3) b/4

Plank is light ⇒ net force on it is zero ⇒ friction by A on plank = friction by B on plank = f.

A can supply at most μmg ⇒ f ≤ μmg. So friction between B and plank never exceeds μmg — (1) true — and never reaches B’s limit 2μmg, so B never slides — (2) true.

(3) F = 3μmg2: if no slipping, a = 2F − F3m = μg2; friction needed on A = F + ma = 2μmg > μmg ⇒ A slips, f = μmg.
B + plank: 2F − μmg = 2m·a ⇒ a = μg towards right — true.

(4) F = μmg4: common a = F3m = μg12; friction needed on A = F + ma = μmg3 < μmg ⇒ no slipping ⇒ a_A = μg12 ≠ 0 — wrong.

Answer (4)

Charge enclosed = charge on segment from (1, 2, 3) to (3, 4, 5).

Points on it: (1 + 2t, 2 + 2t, 3 + 2t), 0 ≤ t ≤ 1; dl = 2√3 dt; λ = (1 + 2t) − (2 + 2t) + 2(3 + 2t) = 5 + 4t

q = ∫₀¹ (5 + 4t)(2√3) dt = 2√3 (5 + 2) = 14√3 C

φ = qε₀ = 14√3ε₀ = 2x√3ε₀ ⇒ x = 7

Answer 7

Net force on each star must pass through the centre of mass G. For m₃ (sides a = m₃m₁, b = m₃m₂; θ₁, θ₂ = angles of these sides with m₃G):

Force components ⊥ to m₃G cancel: m₁a² sin θ₁ = m₂b² sin θ₂  …(1)

G lies on this line, so moments of m₁, m₂ about it balance: m₁a sin θ₁ = m₂b sin θ₂  …(2)

(1) ÷ (2) ⇒ a³ = b³ ⇒ a = b. Same argument at m₁, m₂ ⇒ a = b = c: triangle must be equilateral; no condition arises on the masses.

Answer (b), (c)

Feed current I in at node 1 (out at infinity): by symmetry it divides equally among 6 resistors ⇒ I6 flows 1 → 2.

Take current I out at node 2 (in from infinity): again I6 flows 1 → 2.

Superposing: current in 1–2 = I3 ⇒ V₁₂ = IR3 ⇒ R_eq = R3

Answer (4)