Download 1-A swimmer wants to swim straight across a river with

1-A swimmer wants to swim straight across a river with current flowing at a speed of v1 =
0.27 km/h. If the swimmer swims in still water with speed v2 = 0.81 km/h, at what angle
should the swimmer point upstream from the shore and at what speed v will the
swimmer swim across the river?
=
Answer:
°v=
km/h
 = 180 26’ or 18.430,  = 0.8538 km/h
B
Starting from point A, if the swimmer (say
he) swims in the still water of the river at the
speed V2, then he reaches at point B,
swimming in a straight line as shown in the
figure behind.
Still river
A
B

v1
B1

v1

v2




  v 2  v1
Flowing river
B2

A
B
- v1

v1



v1
B1


v2
  v 2  v1

Flowing river
A
Starting from point A in the flowing river, if
the swimmer continues swimming in the
straight line, then he is pushed by the river
current and is swift towards point B. Thus
instead of along AB, he now swims with the

 
velocity   v 2  v1 along AB1 making
angle  with the actual required path AB.
To avoid reaching at point B’, the swimmer
must overcome the effect of flow. For this he
has to flow upstream at angle of  in the
direction of B2 with velocity


 
 
  v 2  - v1 i.e.   v 2 - v1
so that he is drifted by the river current and
reaches at point B, i.e. at the expected
destination.
 

 
The magnitude of this velocity   v 2 - v1 is given by parallelogram law of vector
addition (or one can judge from the right angle triangle as seen in the above figure),

 i.e. 
v 2  v 2  0.27 2  0.812  0.0729  0.6561  0.729  0.8538 km / h
1
2
As seen from the right angle triangle ABB’, the direction of this velocity i.e. the angle 
is given by
v
0.27 1
tan θ  1 

v 2 0.81 3
v 
1
 θ  tan 1  1   tan 1    18.430 i.e.180 26'
 3
 v2 
2-In the 1887 experiment by Michelson and Morley, the length of each arm was 11 m.
The experimental limit for the fringe shift was 0.005 fringes. If sodium light was used
with the interferometer ( = 589 nm), what upper limit did the null experiment place on
the speed of the Earth through the expected ether?
km/s
Answer:
3.471 km/s
[See a proper text book for the derivation of the fringe width. I am directly giving here
the formula for the fringe shift. ]
Fringe shift (say n) is given by
2Dv 2
n 2
c λ
Where
n – fringe shift.
D– arm length (i.e. the distance of the each mirror from the
semitransparent silver plate.
V – speed of the earth through the ether medium.
c – speed of light in vacuum/air , c = 3 x 108 m/s.
Data: Fringe shift n = 0.005, arm length D = 11 m,
 = 589 nm = 589 x 10-9 m = 5.89x10-7 m
c = 3 x 108 m/s.
Solve the equation for v.
nc 2 λ
2
 v 
2D
nc 2 λ
i.e. v  
2D

nc 2 λ
2D


2
0.005  3  10 8  5.89  10 7

2  11

0.005  9  1016  5.89  10 7
22

5  10 3  9  1016  5.89  10 7
22
5  10 3  9  1016  5.89  10 7

22
265.05  10 6
22
  12.05  10 3
v =3.471 x 103 m/s = 3.471 km/s

3-Astronomers discover a planet orbiting around a star similar to our sun that is 15 light
years away. How fast must a rocket ship go if the round trip is to take no longer than 30
years in time for the astronauts aboard?
c 1c
How long will the trip take as measured on Earth?
yr
30 year
Answer:
The rocket ship should travel at the speed of light i.e. c.
1 light year = distance traveled by a light in 1 year. =
The space ship must travel at a speed comparable to that of light and thus it’s a
relativistic problem.
Let us set the earth as a frame of reference S.
Then the rocket ship is an inertial frame S’ that must travel at some constant velocity v,
comparable to the speed of light c, to make a round trip within 30 years. Then in 15
years, it will travel a distance of 15 light years to reach the planet and in next 15 years, it
will come back, traveling again 15 light years.
Thus the rocket must travel a distance of 1 light year in 1 year.
But 1 light year is the distance traveled by the light in 1 year.
Hence the rocket must travel at a speed of light c to travel 1 light year distance in 1 year.
Thus v = c.
Star
S’
v
x’
S
v=0
x
Rocket
Ship
Earth
The round trip time 0 =30 years is the time of travel expected for the round trip of the
rocket. But when the ship travels this distance at speed of light, then it will measure the
time 0 =0 from the following equation.
Refer ‘time dialation’ in books on relativity.
The time interval  as measured in the frame of reference (i.e. stationary frame) i.e. on the
earth is given by
v2
τ 0  τ 1  2 , where 0 is the time measured by the space ship.
c
c2
 0 for any value of  (the time measured on the earth).
c2
This means the time literally stops for the object moving at speed of light. The time may
have 30 years
i.e.
τ0  τ 1 
With respect to space ship, no time is required for the journey.
If the duration of the journey is measured from the earth, then it is 30 years.
4-A spaceship is moving at a speed ot 0.7c away from an observer at rest. A boy in the
spaceship shoots a proton gun with protons having a speed of 0.8c.
(a) What is the speed of the protons measured by the observer at rest when the gun is shot
away from the observer?
c 0.9615 c
(b) What is the speed of the protons measured by the observer at rest when the gun is shot
toward the observer?
c
-0.2273 c
Answer:
The observer is in the stationary frame S. The space ship is moving with a speed of 0.7 c.
As this speed remains constant, we treat the space ship as an inertial frame S’.
When proton gun is fired away from the stationary observer (i.e. in the direction of
motion of the space ship):
Speed of protons as observed by the boy in space ship
v’x = 0.8 c
Speed of the space ship
v = 0.7 c
We take the velocity of the space ship as positive direction of X-axis.
(a) When proton gun fires protons in the direction of motion i.e. away from the
observer, along +x-axis:
v'  v
The velocity of the protons as observed by the stationary observer is v x  x '
v
1  v 2x
c


 0.8c  0.7c   1.5 c 
i.e. v x  

  0.9615 c
1  0.7c  0.8c   1  0.56 
c 2 

Thus a stationary observer on the earth observes the protons to be moving away from
him, in the direction of the space ship at speed 0.9615 c.
(b) When proton gun fires protons towards the earth, i.e. along -x-axis:
Replace v’x in the above expression by –v’x. Then we get
- v 'x  v
vx 
- v 'x
1 v 2
c


 - 0.8c  0.7c   - 0.1c 
i.e.
vx  

   0.2273 c
1  0.7c  - 0.8c   1  0.56 
c 2 

Thus a stationary observer on the earth observes the protons to be moving at a speed of
0.2273 c. The negative sign indicates that the direction of the protons is opposite to that
of the space ship, i.e. towards the earth.
Thus the protons travel towards the earth with velocity of -0.2273 c.
--------------------------
 