Mechanics 2, solutions

Problem 1:  (A)

Definition of the Lagrangian

L = T - U = ½m[(dr/dt)2 + r2(dθ/dt)2] - U(r) =  ½m[(dr/dt)2 + r2(dθ/dt)2] - kr2.

Problem 2:  (E)

Cyclic coordinates

The coordinate θ is cyclic, pθ =  ∂L/∂(dθ/dt) is constant.

Problem 3:  (E)

Canonical or conjugate momentum

pq -∂L/∂(dq/dt) = 4m(dq/dt)3.

Problem 4:  (D)

Constants of motion

The Lagrangian does not explicitly depend on time and the generalized coordinates do not explicitly depend on time, so H = T + U = E and the energy is a constant of motion.

E = (dq/dt)p - L = 3 m(dq/dt)4 + g(q)

Problem 5:  (D)

Relativistic energy and momentum

v/c = pc/E,  pc = (25 + 9 + 2)½ = 6 MeV,  E = 10 MeV,  v/c = 3/5.

Problem 6:  (A)

The rest energy mc2 is Lorentz invariant.

E2 - p2c2 = m2c4 =  64 MeV2.  We need E'2 - p'2c2 =  64 MeV2.

Problem 7:  (B)

Relativistic energy

E = γmc2,  γ = (1 - 0.64)-1/2 = 1/0.6 = 1.67.

Problem 8:  (D)

Length contraction

100/(γv) = t.  ct = 100*(1 - v2/c2)1/2/(v/c),  0.5 = (1 - v2/c2)1/2/(v/c),  (v/c)2 = 1/1.25 = 4/5.

Problem 9:  (D)

Relativistic energy and momentum

E = (m2c4 + p2c2)1/2 ≈ pc,  p ≈ E/c = 500 GeV/c.

Problem 10:  (B)

Energy and momentum conservation

M1c2 = (M22c4 + p22c2)1/2 + hf
|hf/c| = |p2|.
M12c4 + p22c2 - 2 M1c3 p2 = M22c4 + p22c2.
p2c = (M12c2 - M22c2)/2M1.
E22 = (M24c4 + M14c4 - 2M22 M12c4)/(4M12) + 4M12M22c4/(4M12) = (M12c2 + M22c2)/2M1.