PHYS1015 MOTION AND RELATIVITY JAN 2015 EXAM ANSWERS

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1 PHYS1015 MOTION AND RELATIVITY JAN 2015 EXAM ANSWERS Section A A1. (Based on previously seen problem) Displacement as function of time: x(t) = A sin ωt Frequency f = ω/2π. Velocity of mass is v(t) = dx dt = Aω cos ωt Maximal velocity is therefore Aω and is obtained when total energy E is equal to kinetic energy, E = 1 2 mv2 = 1 2 ma2 ω 2. [ 2 ] Alternatively for the 2 marks: recall that E = 1 2 ka2 for total energy and that k = mω 2 for oscillating spring. or Hence A 2 = 2E mω 2 = A = 1 2π f 2E m(2π f ) 2 2E m.

2 Putting in values we find A = 0.16 m.

3 A2. (Similar to example given in workshops) Weight of skydiver is mg. Forces balance at terminal velocity, since there is no acceleration. Hence drag force also has magnitude mg and is opposite to direction of travel. Work done by drag force over distance d is therefore mgd. Power is rate of doing the work hence P = mgv. Putting in numbers we find P = 37 kw.

4 A3. (Unseen problem) Relativistic energy E = γmc 2, where γ = 1 1 v 2 /c 2. Rest energy is mc 2 and kinetic energy is E mc 2 = 0.99 MeV. We have γ = E/(mc 2 ) so and finally 1 = E2 1 v2 m 2 c 4 c 2 v c = hence 1 m2 c 4 E 2 = 0.94 v 2 c = 1 m2 c 4 2 E 2

5 A4. (Bookwork) Magnitude of force on object at radius r is GMm r 2. Work done by gravity moving mass from position r 0 to position r is r [ ] W = dr GMm r 1 = GMm (r ) 2 r r 0 r 0. and potential energy is defined to be U(r) U(r 0 ) = W. Taking reference point r 0 to be at infinite separation and declaring U( ) = 0 we have U(r) = GMm r.

6 A5. (Based on previously seen problem in lectures) Relativistic Doppler formula: ν = ν 0 1 ve /c 1 + v e /c where ν 0 and ν are emitted and observed frequencies. Hence for wavelengths: λ λ 0 = 1 + ve /c 1 v e /c Hence λ 2 0(1 + v e /c) = (λ ) 2 (1 v e /c) So that (v e /c)(λ (λ ) 2 ) = (λ ) 2 λ 2 0 or (v e /c) = (λ ) 2 λ 2 0 λ (λ )

7 Section B B1 (a) (bookwork) In an elastic collision the total kinetic energy is conserved (Or: relative velocity after collision is opposite to that before). In an inelastic collision the total kinetic energy is not conserved. After a totally inelastic collision the two bodies move with common velocity. (b) (similar to problem in lectures) (i) Conservation of momentum: m 1 u = m 1 v 1 + m 2 v 2, where v 1 and v 2 are the velocities of the two balls in the positive x direction. Conservation of kinetic energy: 1 2 m 1u 2 = 1 2 m 1v m 2v 2 2. Rearranging KE equation: m 1 (u 2 v 2 1 ) = m 2v 2 2, hence m 1 (u v 1 )(u + v 1 ) = m 2 v 2 2. Rearranging momentum equation: m 1 (u v 1 ) = m 2 v 2. Divide to get u + v 1 = v 2. Put back in to momentum equation: m 1 (u v 1 ) = m 2 (u + v 1 ). Solve for v 1 : i.e. v 1 (m 1 + m 2 ) = u(m 1 m 2 ) ( so v 1 = u. ) m 1 m 2 m 1 +m 2 Moves in positive x direction if m 1 > m 2 and in negative x direction if m 1 < m 2. (ii) We had v 2 = u + v 1. But we know v 1 in terms of the initial velocity and the masses. So substitute in: ( We have: v 2 = u ) 1 + m 1 m 2 m 1 +m 2 = 2u m 1 m 1 +m 2 Kinetic energy of Ball 2 after collision is 1 2 m 2v 2 2 = 2 u2 m 2 1 m 2 (m 1 +m 2 ) 2.

8 (iii) Final height up incline z reached when KE is converted to potential energy U = m 2 gz. So we need m 2 gz = 2 u2 m 2 1 m 2 (m 1 +m 2 ) 2. Hence z = 1 v g = 2u2 m 2 1 g(m 1 +m 2 ). 2 (c) (Unseen problem) Totally inelastic means the balls move with common velocity after collision. Momentum conservation: m 1 u = (m 1 + m 2 )v for common velocity v afterwards. Hence v = um 1 m 1 +m 2. Potential energy of combined ball at top of incline is (m 1 + m 2 )gh. We need kinetic energy just after collision equal to (or greater than) this. i.e. 1 2 (m 1 + m 2 )v 2 = (m 1 + m 2 )gh, i.e. v 2 = 2gh i.e u2 m 2 1 (m 1 +m 2 ) 2 = 2gh i.e. u = (1 + m 2 /m 1 ) 2gh.

9 B2 (a) (bookwork) Newton s second Law: F = ma. Gravitational force on projectile is F = mgˆk with ˆk a vertical unit vector. Hence acceleration a = gˆk. Integrating once we find velocity v = v 0 gt ˆk. Integrating again we find position r = r 0 + v 0 t 1 2 gt2 ˆk. We can set initial position r 0 = 0. Projecting onto horizontal component x(t) = v 0 t cos θ. and vertical component z(t) = v 0 t sin θ 1 2 gt2. (b) (bookwork) Time of flight is time when height vanishes (but not t = 0), hence v 0 sin θ = 1 2 gt. Rearranging we have t = 2v 0 sin θ g. (c) (Unseen problem) (i) Maximum height obtained at half the total time of flight, i.e. at t = v 0 sin θ g. At this time the height is z max = v2 0 sin2 θ g 1 v 2 0 sin2 θ 2 g = 1 v 2 0 sin2 θ. 2 g Hence (using sin(π/6) = 1/2) we have v 2 0 = 8gz max or v 0 = 8gz max 100 m/s. Total distance travelled is value of x at total time of flight, i.e. x max = 2v2 0 sin θ cos θ g = v2 0 sin 2θ g. Hence we have x max = 8z max sin 2θ = 4 3z max = 60 3 m 901 m. (ii) Now elevation angle θ = 0. So we can use simplified equations for x and z from before:

10 x(t) = v 0 t. and z(t) = 1 2 gt2 We want the distance travelled when z(t) = z max with z max the same as before. When z(t) = z max we have t = 2zmax g. and hence horizontal distance travelled x = v 0 2zmax g. Putting in numbers: x = 520 m.

11 B3 (unseen problem) (a) (i) Let S be the laboratory reference frame and S the electron beam frame. S moves in the direction of the electrons at v = 0.995c. The muon velocity in frame S is u = 0.9c. For the muon in the laboratory reference frame we have: 1 γ = = u2 c 2 Thus its momentum in the S frame is p = γmu Hencep = (2.3)(105.7 MeV/c 2 )(0.9c) = 218 Mev/c (ii) To find the momentum in the electron-beam reference frame we use the velocity transformation equation to find the muon s velocity in frame S : u = u v 1 uv c 2 = 0.99c 0.995c 1 (0.9c)(0.995c)/c 2 = 0.91c [ 2 ] In the laboratory frame the faster electrons are overtaking the slower muon. Hence the muon s velocity in the electron-beam frame is negative. γ for the muon in frame S is: γ = = 2.41 The muon s momentum in the electron-beam reference frame is p = γ mu = (2.41)(105.7 MeV/c 2 )( 0.91c) = 231 MeV/c (b) (i) In the CM frame both incoming protons have equal & opposite velocities and therefore momenta i.e. the total momentum is zero. [ 2 ] Relativistic energy E of one of the protons is γm p c 2. Total energy is twice this. Energy conservation: 2E = 2γm p c 2 = 2m p c 2 + m π c 2 hence energy of each of the protons is E = m p c m πc MeV [ 2 ] Since E = γm p c 2, we have γ = 1 + m πc 2 2m p c 2 = = 1.072

12 and hence v = 0.36c [ 2 ] (ii) Now one of the protons is at rest. So the velocity of the moving proton is found by relativistically adding the velocity in the centre of mass frame (0.36c) to the relative velocity between frames (also 0.36c): u = u v 1 = 0.36c+0.36c uv c = 0.64c [ 2 ] 2 We have γ = 1/ = 1.3 [ 2 ] And for kinetic energy we have: K = (γ 1)m p c 2 = = 280 MeV [ 2 ] Alternatively for the 6 marks: In new frame collision products are moving with velocity v. Momentum p of moving proton before collision is constrained by momentum conservation p = γ(2m p + m π )v. Energy of incoming proton E constrained by invariance of E 2 p 2 c 2, i.e. we have E 2 p 2 c 2 = (E ) 2 (p ) 2 c 2 [ 2 ] and hence (E ) 2 = E 2 p 2 c 2 + (p ) 2 c 2 = E 2 (γm p v) 2 c 2 + (γ(2m p + m π )v) 2 c 2. which gives E = 1218 MeV. Kinetic energy K = E m p c 2 = = 280 MeV.

13 13 PHYS999 W1 B4 (a)(bookwork) Postulate 1. The laws of physics are the same in any inertial frame. Postulate 2. The speed of light is the same in any inertial frame. An inertial frame is one in which Newton s first law holds, i.e. a body continues in uniform motion unless acted upon by an external force. [ 2 ] (b)(bookwork) Lorentz transformation: Orthogonal directions: y = y and z = z t = γ(t vx/c 2 ) x = γ(x vt) 1 where γ = 1 v2 c 2 (c) (Unseen problem) [ 2 ] [ 2 ] Let the first event occur at x = t = 0 in S (and hence at x = t = 0 in S ). The second event occurs at x = x in S and at t = 0 in S. Using Lorentz transformation above we then have 0 = γ(t v x/c 2 ). Hence the two events are separated in time in S by time t = v x/c 2. (d) (Unseen problem) Signals are simultaneous in S and separated in S, as in part (c): Earlier event is A at origin x = t = 0. Later event is B at x = x and time t = v x/c 2 in S. Hence v = tc 2 /( x) = 90 m/s (= c). S was moving in positive x direction hence lander is flying in direction from A towards B. Yes - a pilot travelling faster in the same direction would see B occur first. [ 2 ] Copyright 9999 c University of Southampton Page 13 of 13

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