Mechanics II - Assignment II - Answers
Winter 2012
1
Resnick - Problem 2.10
a) AD and AC are the two events that can be used to measure the rate of the clock carried at A, although, BD and
BC can practically serve the same purpose for the clock at B, which runs at the same rate as the one at A.
b) The observer on the cart at A, measures the proper time interval, as it uses the same clock at the same place
in his frame. This will be shorter than the corresponding time interval measured by any non-proper observer. So, if
the ground observer, with her own clocks
q measures the interval between events AD and AC to be ∆t, the clock carried
at A will measure it to be
∆t
γ
= ∆t ×
1−
v2
c2 .
c) Recall that the ground observer measures a contracted length for the cart. So, for the ground observer to observe BC and AD happening simultaneously, the event at the front end of the cart (i.e. BC) should happen first in
the cart frame.
2
Resnick - Problem 2.12
a) Suppose there exists another frame S ′ , moving in the A→B direction, in which events A and B happen simultaneously. In that frame one will have t′B − t′A = 0. This time interval is related to the tB − tA interval in frame S,
according to the lorentz transformation
t′B − t′A =
tB −tA − cv2 .(xB −xA )
q
2
1− vc2
or, substituting the corresponding values,
0=
10−6 − cv2 .(600)
q
2
1− vc2
c
Solving this for v, one finds v = 1.5 × 108 m
s = 2 . The positive sign of the obtained answer confirms our choice of
direction for v, which should be physically understandable. The S ′ observer will see the length AB contracted to:
q
√
2
ABS ′ = ABS × 1 − vc2 = 600 × 1 − 0.25 = 519.6m
b) Repeating the same calculations, this time with ABS = 100m, we get:
v = c2 ×
∆tAB
∆xAB
= 9.00 × 108 ×
10−6
100
= 9 × 108 m
s ,
which exceeds c and is not acceptable. Here, the two events are not separated far enough from each other to be
simoltaneous in any other frame.
3
Resnick - Problem 2.28
The Earth observer measures a longer time interval than the proper time measured in the particle’s rest frame:
∆tEarth = ∆tπ+ × γ =
2.5×10−8
√
1−0.992
= 17.7 × 10−8 s
And so,
hf inal = hinitial − ∆tEarth × 0.99c = 2 × 105 − 17.7 × 10−8 × 0.99 × 3.00 × 108 = 199947m
1
4
Resnick - Problem 2.32
y
A
v
θ
θ
x
A
B
a) Looking at points A nd B, the time it takes for the point of touch of the rod with the x-axis to slide from A to B, is
AB. sin θ
v
∆tAB =
The displacement of this point of contact along the x-axis during this time interval is
∆xAB = AB. cos θ
⇒u=
∆xAB
∆tAB
=
AB. cos θ
AB. sin θ
v
= v cot θ
b) (Correction should be made to the given value of θ from 60◦ to 15◦ .)
v = 3c
⇒u=
c
3
× cot 15◦ = 1.24c
This is not the physical speed of the center of mass of any object, but merely the relative speed of moving points along
one object. This should not be interpreted as a physical velocity and only has geometrical meaning. Special relativity
puts limit on the actual physical velocities, not on velocity differences, and so there is no contradiction with special
relativity here.
5
Resnick - Problem 2.33
N
0.6c
0.8c
s
Take the frame attached to the particle approaching from North (with v = 0.8c) to be the S frame and the frame
attached to the earth as the S ′ frame. According to the S observer, Earth has the relative speed v = −0.8c along
the North-South direction. According to the Earth (S ′ ) observer, the second particle has speed u′ = −0.6c. Relative
speed of the two particles as measured from the point of view of the S observer, would therefore be:
u=
u′ +v
′
1+ uc2v
=
−(0.8+0.6)c
1+0.8×0.6
2
= −0.946c