A Special Relativity Kinematics Problem : \"The Rocket and Hangar \" ANSWER AS M

Question

A Special Relativity Kinematics Problem : "The Rocket and Hangar " ANSWER AS MANY PARTS AS POSSIBLE AND I WILL RATE THE BEST ANSWER. PRIORITY IS ON ANSWERING Q'S 4-10 One of the most surprising consequences of special relativity is that sequences of events may be seen very differently from different frames of reference. Imagine a car that is 50 meters long driving through a garage that is 100 m from front to back. The car approaches the closed front of the garage and someone opens the front door. The car passes through the garage, the rear door is opened and the car passes out of the garage. This sequence could be described as follows: Now image a 100 m rocket traveling at 0.6c toward a hanger that is 90 m from front to back. Two people -- an observer (H) at rest with respect to the hangar and the captain (R) at rest in the rocket – will both observe the rocket pass through the hangar. The observer H tells the captain R that (due to relativistic length contraction) he can close the front and rear doors of the hangar and the rocket will be briefly enclosed within the hangar. He will open the rear door just as the nose reaches it, and the rocket will pass through the hanger. Now consider this from the standpoint of the captain of the rocket ship. Will he see the rocket crash through either the front door of the rear door of the hangar? Since the hangar is moving toward him at 0.6c its length will be contracted to become even shorter than its 90m rest length. Recall that an event is described by its position and time coordinates. For our problem, we will need only one space dimension for the position and one time as the coordinates: xH, tH in the hangar frame, and xR, tR in the rocket frame. The origin xH = 0 is at the front door in the hangar frame, and the other origin xR = 0 is at the nose in the rocket frame. The clocks are synchronized, tH = 0 = tR just as the nose of the rocket ship enters the front door of the hangar (the two space origins are coincident at that instant). As you answer the following questions you should be able to explain the strange situation in which a long rocket is enclosed within the shorter hangar. 1. What is the formula for relativistic length contraction? Explain each of the symbols used. I GOT L=L_0/y 2. How long is the moving rocket according to H? Would H say that the rocket could be enclosed within the hangar (i.e. both front and rear doors closed simultaneously)? I GOT 80 m, H WOULD SAY THE ROCKET COULD NOT BE ENCLOSED IN THE HANGAR 3. How long is the moving hangar according to R? I GOT 72 m Remember that the distance traveled at constant speed is given by deltax = vdeltat, as long as the quantities deltax, deltat, and v are all measured with respect to the same coordinate system. Clearly show your calculations 4. Find the time H says the tail of the rocket arrives at the front door. 5. Find the time of this same event in R's frame of reference. (Hint: how far must the front door travel in R's frame from the nose of the rocket to the tail; how long does this take?) 6. Find the time in H's frame for the nose of the rocket to reach the rear door, i.e. find the time coordinate tH for event E3. (Hint: how far is it from the front to the rear door in the hanger frame; how long will it take the nose of the rocket to travel this distance? 7. Find the time in R's frame for event E3. (Hint: How far was the rear door of the hangar from the origin in (nose of) the rocket at tR = 0; how long does it take the rear door to travel this distance?) 8. List the events E1, E2, E3, in the sequence (time order) that they occurred in the hangar frame. Then list the same events in the sequence they occurred in the rocket frame. 9. Write one or two complete sentences describing the sequence of events as seen by H. 10. Write one or two complete sentences describing the sequence of events as seen by R. This should clearly explain how the rocket ship passed through the hangar, although it is too long in this frame of reference to fit inside the hangar.

Explanation / Answer

We will try to solve this problem with its very long sentence

Part 1)

The expression for the length contraction

L = Lp/

= 1/ sqrt ( 1- (v/c)2 )

L = Lp ( 1- (v/c)2 ) ½

where

L is the length measured rocket from the hangar (hangar)

Lp is the length measured rocket from the fixed reference system in the rocket (rocket)

v rocket speed

Part 2)

data

Lp = 100 m

v =0.6 c

Lr = 100 ( 1 – (0.6c/c)2 )1/2

Lr = 100 0.8

Lr = 80 m

The observer in the hangar measures a length of 80 m for the rocket

3)

In this case the captain sees the hangar moving toward him

Vh =-0.6c

Lph = 90 m

Lh = Lph (1 -(Vh/c)2 )1/2

Lh = 90 ( 1 -(-0.6c/c) ½

Lh = 90 0.8

Lh = 72 m

Part 4)

The man in the hangar sees the length Lr

Lr =80 m

V =0.6c

V= d/t

t = d/v

t = 80/ 0.6c

t = 133.33 c-1 = 44.44 10-8 s

Part 5

In the framework of the rocket it has the same time

t = tp

tp = t/

tp = t (1 -(Vh/c)2 )1/2

tp = 44 10-8 (1-(0.6c/c)2)1/2

tp =44 10-8 ( 0.8)

tp = 35.2 10-8 s

Part 6

reference hangar H

L =90 m

V = 0.6c

V=L/t

t =L/V

t = 90 / 0.6c

t =150 c-1 = 50 10-8 s

Part 7

Reference R

L =72 m

t = 72/ 0.6c

t =120 c-1

t = 40 10-8 s

Part 8

I am not clear what the E1 and E2 events, but what happens is the following

Under the hangar is the rocket reaches the front door is open, enters the hangar and its length is less than this, when entered the front door is closed and the rocket is momentarily waxed. When the nose sore to the back door is open and the rocket leaves the hangar

Part 9

The observer sees the hangar both doors are closed when the rocket is inside of him, Lc = 80m <Lh = 90m) passes a period of time 10-8 s 50 until the tip reaches the door ket tracera in this when this door opens and the rocket leaves

Part 10)

The rocket reaches the front door of the hangar that is shorter than the rocket, the rocket progresses and as the rocket moves opens the door trace 40 10-8 s the front door is closed and the rocket keeps coming out of the hangar

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