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[Mentor's note: this post does not use the homework template because it was originally posted in a non-homework forum.]

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A ship (attached to reference frame S') is passing us (standing in reference frame S) with velocity v= 0.933

a) v=0.933c

u'=-0.959c

=> gamma = 3.52851

=> L'=214.255m

By time = distance/speed, temporal separation is 0.745microseconds

b)

v=0.933c

u'=-0.959c

Using velocity addition formula => u=-0.247024c

=> gamma = 1.03198

=> L'=712.571m

By time = distance/speed, temporal separation is 9.89microseconds

c) Same as a?

d) v=0.933c

u'=+0.959c

Using velocity addition formula => u=0.9985c

=> gamma = 26.2631

=> L'=28.7856m

By time = distance/speed, temporal separation is 0.136microseconds

Would love some hints on where I am going wrong, these are all incorrect.

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A ship (attached to reference frame S') is passing us (standing in reference frame S) with velocity v= 0.933

*c. A*proton is fired at speed 0.959*c*relative to the ship from the front of the ship to the rear. The proper length of the ship is 756 m. What is the temporal separation between the time the proton is fired and the time it hits the rear wall of the ship according to**(a)**a passenger in the ship and**(b)**us? Suppose that, instead, the proton is fired from the rear to the front. What then is the temporal separation between the time it is fired and the time it hits the front wall according to**(c)**the passenger and**(d)**us?**Attempt at a solution:**a) v=0.933c

u'=-0.959c

=> gamma = 3.52851

=> L'=214.255m

By time = distance/speed, temporal separation is 0.745microseconds

b)

v=0.933c

u'=-0.959c

Using velocity addition formula => u=-0.247024c

=> gamma = 1.03198

=> L'=712.571m

By time = distance/speed, temporal separation is 9.89microseconds

c) Same as a?

d) v=0.933c

u'=+0.959c

Using velocity addition formula => u=0.9985c

=> gamma = 26.2631

=> L'=28.7856m

By time = distance/speed, temporal separation is 0.136microseconds

Would love some hints on where I am going wrong, these are all incorrect.

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