Physics II Class 05/02


A spear, which you have previously observed to be 2 meters long while at rest, is chucked right by your face at .7 times the speed of light.  What is the length of the sphere as observed in your frame of reference?

The length of a moving object contracts in its direction of motion by factor sqrt(1 - v^2 / c^2).  In this case v = .7 c so

sqrt(1 - v^2 / c^2) = sqrt( 1 - (.7 c)^2  / c^2) = sqrt(1 - .7^2) = sqrt(1 - .49) = sqrt(.52) = .72.

The length as observed in your reference frame is therefore .72 * 2 meters = 1.44 meters.

The spear has rest mass 4 kg.  As it goes by you give it a little shove, applying a force of 10 N during a time interval of .01 second.  By how much does the speed of the spear increase?

We could apply the impulse-momentum theorem:  impulse is 10 N * .01 s = .1 N s = .1 kg m/s.

The mass of the spear is 4 kg * 1 / sqrt(1 - v^2 / c^2) = 4 kg / .72 = 5.6 kg. So the speed of the spear changes by

So now the spear has velocity about .7 c + .018 m/s.  How much work did you have to do to make that change?

Figuring the KE at .7c and at .7c + .018 m/s it is asserted that the KE has changed by 2.1 * 10^-7 J.

To confirm this let's figure force * distance.  A 10 N force applied for .01 s to an object moving at .7 c implies distance .7 c * .01 s and work

How much work would be required to increase the velocity of the spear from .7 c to .8 c?

At .7 c the KE is .5 m v^2, where m = 5.6 kg and v = .7 * 3 * 10^8 m/s.  We get KE = 1.23 * 10^17 J.

At .8 c the KE is .5 m v^2, where m = 4 kg / sqrt(1 - (.8 c)^2 / c^2) = 6.67 kg and v = .8 * 3 * 10^8 m/s.  We get KE = 1.92 * 10^17 J.

The change in KE is therefore 1.92 * 10^17 J - (1.23 * 10^17 J) = 6.9 * 10^16 J.

Suppose we exert a force F on the spear while it is moving at velocity v.  Then at what rate is the velocity changing?

The rate of change of velocity is a = F / m.  The mass m of the spear is m0 / sqrt(1-(v^2/c^2)), where m0 stands for the rest mass (4 kg in this case).

Since a = dv/dt this gives us the differential equation

dv/dt = F / [ m0 / sqrt(1 - (v^2 / c^2) ) ] = F/m0 * sqrt(1 - (v^2/c^2)).

Separating variables we get

which we solve to get

Solving for v we obtain

 

 

Assignment 14:

Assignment 15:

Assignment 16:

Assignment 18:

Assignment 19:

Assignment 25:

Assignment 26: 

Assignment 27: 

Assignment 28:    

Assignment 29: 

Assignment 30: 

Assignment 31: