AP Kaity Problem solns

Kaity Vermillion is skating at .8c. How long does it take her in her frame to reach the end of rink 100m away in the rink’s frame?

Her time is t’=gamma (t-vx/c^2) x also =vt. so

t’ = gamma(t-v^2t/c^2) = gamma t (1/gamma^2) = t/gamma where gamma =1.67 Another, easier way is is t=gamma(t’+vx’/c^2) but Kaity is at the origin in her frame so x’ =0. and t=gamma t’

How long does she think the rink is?

This requires finding out how far from her in her frame is a colleague moving at.8c with respect to the rink who is at the end of the rink when Kaity is the beginning. Colleagues x’ is given by x’=gamma(x-vt) but what is t at this time? Try a simpler way.

x=gamma(x’+vt’), but t’ is 0 when Kaity is at beginning so x= gamma x’ and x’=x/gamma or 60 m.

If she starts skating at noon on her watch and the rink’s clock at the back end of the rink, when she thinks she the rink has gone -25m, where is she in the rink and what time does the rink clock at that point say?

Kaity’s time, t’, will be 25 m/.8c, x =gamma(x’+vt’) x’ still =0 for Kaity

so we have x = gamma(0+.8c x 25m/.8c )= 1.67 x25 m. Rink time is 1.67x25m/.8c

She is shining a green 6x10¹⁴ Hz laser over her shoulder ( pointing backwards). What frequency light do you, standing at the back of the rink, see?

f = f’(sqrt((c-.8c)/(c+.8c)) =f’/3 = 2x10^14 Hz

Her coach passes her at a speed Kaity sees as .2c. How fast do you think the coach is going?

ucoach in rink = (.8c+.2c)/ (1+.2c*.8c/c^2) =c/1.16

Given that momentum is gmv and KE is (g-1)mc², and Kaity’s mass is 58 kg, what are her momentum and KE.