What is the kinetic energy of a proton moving at (a) 0.100c; (b) 0.500c; (c) 0.900c? How much wo

step 1 This is chapter 37 problem number 40. We're asked to calculate the kinetic energy of a proton if

What is the kinetic energy of a proton moving at (a) 0.100c;...

Key Concepts

Relativistic Kinetic Energy

This concept refers to the kinetic energy of an object moving at speeds close to the speed of light, where classical kinetic energy formulas no longer provide accurate results. It is calculated using the expression K = (? - 1)mc², where ? (the Lorentz factor) accounts for relativistic effects introduced by high velocities.

Lorentz Factor

The Lorentz factor, denoted by ?, is a multiplicative factor defined as 1/?(1 - v²/c²) that arises in special relativity. It quantifies the effects of time dilation, length contraction, and the increase in energy as an object approaches the speed of light, and is essential for converting between nonrelativistic and relativistic formulas.

Nonrelativistic Kinetic Energy

In the nonrelativistic limit, where the speed of the object is much less than the speed of light, kinetic energy is approximated by the classical formula K = ½mv². This approximation breaks down at high velocities and must be replaced by the relativistic kinetic energy expression to accurately describe energy changes.

Work-Energy Theorem in Relativistic Context

The work-energy theorem states that the work done on an object is equal to its change in kinetic energy. In a relativistic context, this theorem is applied using the relativistic kinetic energy formula, leading to results that significantly differ from nonrelativistic expectations as the speeds involved approach the speed of light.

Comparative Analysis of Relativistic and Nonrelativistic Limits

This concept involves comparing results obtained using relativistic equations with those from classical (nonrelativistic) equations. As speeds increase and approach substantial fractions of the speed of light, discrepancies emerge between the two approaches, highlighting the necessity of using relativistic mechanics at high velocities.

Transcript

00:01 This is chapter 37 problem number 40.

00:03 We're asked to calculate the kinetic energy of a proton if it has the speed of 0 .1 times the speed of light.

00:12 So straightforward enough, right? we need to remember the kinetic energy formula, the relativistic one.

00:21 It's going to be our gamma factor minus 1 times nc square.

00:25 So the gamma factor, as you know, is 1 over square root of 1.

00:30 Minus v squared over c squared and must be multiplied it's by m c squared where done the answer answering part a so um here the v is point one c if you plug that in um m is going to be 1 .67 right this is the proton 1 .67 times center of a negative 27 kilograms um speed of light is 3 times centered over 8 but please square this meters per second so you're going to have this here in b.

01:04 0 .1c so you're going to write 0 .1c squared here for v squared.

01:10 As you know, another c squared here in denominator.

01:16 When you do this algebra, you're going to find the kinetic energy to reach 7 .5, 7 times 10 to 0 .13 joules.

01:23 Now, how would our answer to 5b change if the velocity now, the speed now is 0 .5.

01:31 Times the speed of light, then the only difference we would have here is just to plug in 0 .5 times c, right? and the rest would be the same.

01:40 So if you do just that, the kinetic energy is going to be found out to be 2 .3, 2 times 10 to 0 .1912 joules.

01:49 Again, we're calculating everything in terms of joules.

01:53 So for part c, we are asked, what would the kinetic energy be if the velocity is? is now 0 .9 times c.

02:05 Again, now this would point 1c change that to 0 .9c.

02:11 The rest is going to remain the same.

02:13 So the kinetic energy after you do the algebra here is going to be 1 .94 times 10 % percent of negative 10 joules.

02:20 Now, as far as part d is concerned, we are asked to calculate the work done if the kinetic energy is changing.

02:33 From the kinetic energy when we have 0 .5 times c speed of light versus 0 .1 times c speed of light.

02:44 So we already calculated these two values.

02:47 All we need to do is to plug in the values, right? so we know the kinetic energy when the speed is 0 .5 times c is 2 .3 2 times centigrade of negative 11 joules.

03:01 This is the answer to part b, right? minus the answer to part a, basically, 7 .57 times 10 % from negative 13 joules.

03:11 This would give us 2 .24 times 10 to 10 to the power of negative 11 joules.

03:16 Now, in part e, we are asked to calculate the work done, if the speed now is changed from 0 .9c to 0 .5c, again, this is the answer to part c, this is the answer to part b.

03:32 So the answer to part c minus the answer to part b would be 1 .94 times 10 % 0 .10 joules minus 2 .3 .2 times 10 per negative 11 joules.

03:44 This is going to give us 1 .71 times 10 to 0 .10 joules.

03:49 Now in part f, we're asked how would our answer change two parts d and e if we're approaching this problem in a non -year? a relativistic way...

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