Physics 150  Fall 2003

Solution to HW 7

1.   Question B, Ch. 11 of March:

(1) An automobile battery is charged – Mass increases (because it gains energy)

(2)  A hot steel bar cools – Mass decreases (because it looses energy)

(3) Rubber band is stretched  Mass increases (because it gains energy)

(4) Reaction in sealed, insulated container – Mass unchanged (nothing escapes or enters)

(5) Reaction but heat escapes – Mass decreases (because it looses energy)

(6) Molecule pulled apart - Mass increases (because it gains energy – energy wa added to pull molecule apart)

2.  Question B, Ch. 12 of March: “action at a distance” means instantaneous action at a distance which violates the principle that nothing travels faster than light.

3.  Question C, Ch. 12 of March: Describe in your words, but the major point is that the twin on the rocket accelerated.  This can be understood from general relativity during

the period when the rocket accelerates to turn around. The twin on earth ages more and both twins agree that the earth twin has clocks that run faster than the rocket twin – i.e., the earth twin ages faster.  (This is hard question.)

Description made in pervious semester by Joseph Jun, teaching assistant:

For the twin on Earth, he observes the ship leave with speed v, travel a distance d, turnaround, and travel a distance d again at speed v.  The total time for this trip is roughly 2*d/v (plus time to turnaround and accelerate).  The Earthbound twin believes that the moving twin’s clock runs slow and therefore believes that his twin will be young when he steps off his ship.

The rocket-bound twin observes himself at rest and sees the Earth move away from him at speed v.  The Earth travels a distance d and then turns around and travels the same distance back again.  This twin thinks that Earth clocks are running slow because it is moving and he is not.  However, during the time that the Earth turns around, it is the rocket that experiences acceleration, the twin is placed in an equivalent gravitational field where he is near the bottom and the Earth is at the top.  If the acceleration is very large, which it would have to be in order to change directions from a relativistic speed within a reasonable time, then Earth clock’s will run much faster than his own and therefore when he returns to Earth he will be much younger than the twin at rest.

4.  Exercise 1, Ch. 11 of March:

For v = 0.8c, gamma = 1.67 and the mass appears to 3Kg x gamma = 5 Kg.

For v = 0.6c, gamma = 1.25 and the mass appears to 3Kg x gamma = 3.75 Kg.

Exercise 4, Ch. 11 of March:

60 km/s = 6 x 104 m/s, so v/c = 2 x 10-4. For this small value it is easiest to use the

approximate form gamma = 1 + ˝ (v/c) 2 and m = m0 gamma = m0 + ˝ m0 (v/c) 2 .   Thus the CHANGE in mass is:

˝ m0 (v/c) 2 = ˝ 1000Kg  x (2 x 10-4)2 = 2.0 x 10-5 Kg = 0.000002 Kg.

Exercise 6, Ch. 11 of March:

For v = 0.99999946 c, the mass increases by the factor gamma = 962

Exercise 8, Ch. 11 of March:

900 kilotons = 36 x 1014J.  This corresponds to mass = 36 x 1014J/ c2 = 4 x 10-2 Kg = 40g.

Exercise 2, Ch. 12 of March:

The problem is jut like finding the distance a ball would drop if it is throw at a given speed.  In this case it is light that moves at speed c.  The time to travel 30 km = 30,000 m is t = 3 x 104 m/3x 104 m/s = 1.0 x 10-4 s or 1/10,000th of a second.

If an object at rest dropped at freefall for that amount of time it would travel a distance:

d = 0.5*g*t2 = 0.5*(10 m/s2)*(1 x 10-4 s) 2 =5.0 x 10-8 m

Exercise 4, Ch. 12 of March:

The effect of a change in altitude by a height h = 3 x 105 m is given by: a*h/c2 where a = g is the acceleration of gravity on earth.  So a*h/c2 = 10*3 x 104 /9 x 1016  = 3.3 x 10-11.  Therefore, in 100,000 s the clock will be faster by an amount 100,000*3.3 x 10-11 = 3.3 x 10-6 s or 3.3 microseconds than clocks on Earth.

The last is an exercise we also did in class.  Try a triangle that includes the equator on two lines that go from the equator to the north pole.