Unformatted text preview: 54. (a) The strategy is to ﬁnd the γ factor from E = 14.24 × 10−9 J and mp c2 = 1.5033 × 10−10 J and
from that ﬁnd the contracted length. From the energy relation (Eq. 3845), we obtain
γ= E
= 94.73 .
mc2 Consequently, Eq. 3813 yields
L= L0
= 0.222 cm = 2.22 × 10−3 m .
γ (b) and (c) From the γ factor, we ﬁnd the speed:
v=c 1− 1
γ 2 = 0.99994c . Therefore, the trip (according to the proton) took ∆t0 = 2.22 × 10−3 /0.99994c = 7.40 × 10−12 s.
Finally, the time dilation formula (Eq. 387) leads to
∆t = γ ∆t0 = 7.01 × 10−10 s
which can be checked using ∆t = L0 /v in our frame of reference. ...
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Energy

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