Question

A spaceship with rest mass m₀ is traveling with an x-velocity V_0x=+4/5 in the frame of the earth. It collides with a photon torpedo (an intense burst of light) moving in the -x direction relative to the earth. Assume that the ship's shield totally absorbs the photon torpedo.

a)The oncoming torpedo is measured by terrified observers on the ship to have an energy of 0.75m₀. What is the energy of the photon torpedo in the frame of earth?

b)Use convervation of four-momentum to determine the final x-velocity (in the earth frame) and mass (in terms of m₀) of the damaged ship after it absorbs the torpedo. Use an energy-momentum diagram to solve this problem graphically.

c) Find the ship's final mass and x-velocity quantitatively, using four-dimensional column vectors.

Answers: a) 1/4m₀ b) final speed= 13/23, final mass= 1.58m₀

Answer 1

A spaceship with rest mass m₀ is traveling with an x-velocity V_0x=+4/5 in the frame of the earth.

(a) The oncoming torpedo is measured by terrified observers on the ship to have an energy of 0.75m₀.

The energy of the photon torpedo in the frame of earth which is given as :

E_P1S = (0.75) m₀ = - P_xp1s

| E_P1E | = | r (E_P1S + P_xp1s | { eq.1 }

| P_xP1E | = | r (P_xP1S + _EP1S |

where, r = 1 / 1 - V_ox² 1 / 1 - (4 / 5)²

r = (5 / 3)

and     = (4 / 5)

inserting the values in above relation,

| E_P1E | = | (5/3) [0.75 m₀ + (4/5) (0.75 m₀)] |

| P_xP1E | = | (5/3) [-0.75 m₀ + (4/5) (0.75 m₀)] |

| E_P1E | = | (1/4) m₀ |

| P_xP1E | = | -(1/4) m₀ |

Tha answer is given by,     E_P1E = (1/4) m₀

(b) Use conservation of four-momentum, the final x-velocity (in the earth frame) of the damaged ship after it absorbs the torpedo which will be given as :

Using an equation, we have

E_s = m₀ / 1 - V_ox²                                                              { eq.2 }

inserting the value of V_ox in eq.2,

E_s = m₀ / 1 - (4/5)²

E_s = (5/3) m₀

And

P_s = m₀ V_ox / 1 - V_ox²                                                              { eq.3 }

inserting the value of V_ox in eq.3,

P_s = m₀ (4/5) / 1 - (4/5)²

P_s = (4/3) m₀

We have an equation, To determine the final velocity -

| M / 1 - V² | = | E_s + E_P1E |                                                      { eq.4 }

| MV / 1 - V² | = | P_s + P_xP1E |

solving eq.4, we get

| M / 1 - V² | = (23 / 12) m₀    { eq.5 }

| MV / 1 - V² | = (13 / 12) m₀                                                                     { eq.6 }

dividing eq.6 by eq.5, we get

V = (13 / 23)

(c) Use conservation of four-momentum, the final mass (in terms of m₀) of the damaged ship after it absorbs the torpedo which will be given as :

using eq.6, we have

MV / 1 - V² = (13 / 12) m₀        

inserting the value of 'V' in eq,6,

M (13/23) / 1 - (13/23)² = (13 / 12) m₀      

M (13/23) / (0.824) = (13 / 12) m₀    

M = [(23 / 12) (0.824)] m₀   

M = 1.58 m₀