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:
`gChapter 12 Problems 42, 46,
48, 52, 53 //
`uUniversity Physics:
Ch. 21 Problems 23, 24, 26, 27, 30
Assignment 15:
`gCh. 23 Problems 4, 11, 14, 17, 20//
`uUniversity Physics: Ch. 34,
Problems 25, 28, 35, 40, 42//
Assignment 16:
`gCh. 23 Problems 23, 34, 37, 43, 52//
`uUniversity Physics: Ch. 35,
Problems 45, 52, 59, 62, 68, 75//
`gCh. 23 Problems 58, 61; Ch. 24 Problems
4, 7, 10 Submit Ch. 24 #10//
`uUniversity Physics: Ch. 37,
Problems 37, 40, 46, 49, 54 Submit Ch. 37 #44//
Assignment 18:
`gCh. 24 Problems 15, 23, 28, 31,
34 //
`uUniversity Physics: Ch. 38,
Problems 40, 45, 46, 49, 51//
Assignment 19:
`gCh. 24 Problems 40, 43, 46, 51, 57,
60 Submit Ch. 24 #43//
Assignment 25:
`gText Chapter 16, Problems 8, 11, 14, 18
`uUniversity Physics: Chapter 22
Problems 52, 54, 57, 60, 63, 66, 67, 74, 79, 83, 84
Assignment 26:
`gText Chapter 16, Problems 26, 29, 32, 35
`uUniversity Physics: Chapter 23
Problems 20, 21, 25, 27, 33, 34, 40, 41
Assignment 27:
`gText Chapter 17, Problems 9, 12, 16, 19,
22
`uUniversity Physics: Chapter 24
Problems 49, 50, 55, 57, 58, 61, 65, 72, 75
Assignment 28:
`gText Chapter 17, Problems 34, 37, 39, 45,
48
`uUniversity Physics: Chapter 25
Problems 36, 37, 42, 45, 46, 49, 52
Assignment 29:
`gText Chapter 18, Problems 6, 7, 9, 12,
29, 36
`uUniversity Physics: Chapter 26
Problems 44, 47, 50, 55, 60, 61
Assignment 30:
`gText Chapter 20, Problems 11, 13, 17, 21,
32 Submit #32//
`uUniversity Physics: Chapter 28 Problems 41, 42, 46, 52,
57, 65, 66; Chapter 29 Problems 43, 46, 51, 56
Assignment 31:
`gText Chapter 21, Problems 10, 12, 15,
22
`uUniversity Physics: Chapter 30
Problems 31, 36, 41, 46