(5%) Problem 18: Two spaceships are moving along the line connecting them. One o

Question

(5%) Problem 18: Two spaceships are moving along the line connecting them. One of the spaceships shoots a missile towards the other. The missile leaves the attacking spaceship at 0.965c and approaches the other at 0.74c. Randomized Variables u=0.74 c u' = 0.965 c > A What is the relative velocity of the two spaceships? Express it as a ratio to the speed of light. Let the relative velocity of the spaceships be positive when they are approaching each other Grade Summary Deductions Potential 0% 100% cos) asn() atan() acotan) sinhO sinO Submissions tan acosO cotanh) Attempts remaining: 7 cotan % per attempt) detailed view END Degrees O Radians BACKSPACE DELCLEAR Submit Hint I give up! Hints: 1 % deduction per hint. Hints remaining: 3 Feedback: %deduction per feedback. - -

Explanation / Answer

When two objects are approaching,

relative velocity = (Vb - Va)/(1 - Vb*Va/c^2) where all velocities are in vector notation

0.965 c = V missile - V(ship)/(1 - Vm* Vship/c^2)

Vm (1+0.965* Vship/C) = 0.965c + V ship

Vm = (V + 0.965c)/(1 + 0.965*V/c)

Similarly from point of view of B

0.74c = V'+Vm/(1 + V'*Vm/c^2)

So Vm = (V' - 0.74c)/(1 - V'*0.74/c)

So equating

(V + 0.965c)/(1 + 0.965*V/c) = (V' - 0.74c)/(1 - V'*0.74/c)

V'(1 + 0.74/c* (V+0.965c)/(1 +0.965*V/c) = 0.74 c

V' = 0.74c * (c+0.965V)/(1.704V +1.714c)

Now Relative velocity of ships = V+V' /(1 + v*v'/c^2)

= We can assume one object be completely at rest

eg: Ship A which is firing

So V = 0

So V' = 0.74/1.714 *c

So relative velocity = 0.43 c

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