Consider a one-dimensional collision at relativistic speeds between two particles with masses m1

Consider a one-dimensional collision at relativistic speeds...

Key Concepts

Relativistic Momentum

In special relativity, momentum is defined as p = ?mv, where m is the rest mass, v is the velocity of the particle, and ? (the Lorentz factor) accounts for the effects of high speeds near that of light. This modification of the classical momentum formula ensures that the momentum calculation remains valid even when speeds become a significant fraction of the speed of light, making it a crucial concept in analyzing high-speed collisions.

Lorentz Factor

The Lorentz factor, denoted as ? = 1/?(1 - v²/c²), is a key component in special relativity that quantifies how time, length, and relativistic mass change for an object moving at a significant fraction of the speed of light. It is essential for adjusting classical physics equations (like those for momentum and energy) to account for the relativistic effects observed at high velocities.

Conservation of Momentum

The law of conservation of momentum holds true in relativistic collisions when momentum is correctly defined using the Lorentz factor. In analyzing collisions at speeds comparable to the speed of light, one must use the relativistic momentum (?mv) to ensure that the total momentum before and after the collision remains the same, providing a fundamental tool for solving collision problems in special relativity.

Conservation of Energy

In relativistic dynamics, energy conservation includes both the rest energy (mc²) and the kinetic energy modified by the Lorentz factor, resulting in the total energy being given by E = ?mc². This principle is critical in dealing with collisions and interactions at high speeds, as it ensures that the sum of the rest and kinetic energies is conserved in the process, alongside momentum conservation.

Transcript

00:01 In this problem, we have a head -on collision, one -dimensional head -on collision, and we are told that m2 is initially at rest, and then at the end, m1 recoils, which means it goes back the opposite direction that it came in, and m2, now you might say, well, they didn't tell me that m2 moved to the right.

00:23 How do i know that? well, let's talk about that.

00:26 We know there's going to be a force of interaction.

00:28 M1's trying to go through m2.

00:31 So it means m2 applies a force to left.

00:34 So on m1, there are three possibilities.

00:37 It's slowed but still moving to the right.

00:39 It's brought to rest, or slowed, brought to rest, and sent back to the left.

00:45 And that's obviously what happens.

00:47 M2, with the third law, that force had applied to the left on m1, m1 applies that same magnitude force to the right on m2.

00:55 M2 can only move to the right from rest, from rest.

01:01 Now, remember something also about momentum.

01:03 Doesn't matter if it's classical momentum, relativistic momentum, it is a vector.

01:08 So these are not speeds.

01:11 These are, if you like, x components of the velocity vector.

01:16 Or just remember, if you don't want to put the x subscript to indicate x component, just remember that the vs are one -dimensional vectors.

01:23 You've got to worry about pluses and minuses, as i've done here.

01:26 Plus direction i will use to the right, so minus on the recoil of m1, everything else positive.

01:32 We're going to need the gamma factor, one over the script, root in a minute, so i just wrote that off to the side to remind us.

01:38 So, is momentum conserved here? are the forces of interaction internal? the answer is yes.

01:44 There's f -net external is zero.

01:48 So we're fine with momentum conservation.

01:50 Same rules apply here as would in a classical situation.

01:55 Nothing changes.

01:57 So let's write our momentum conservation.

02:00 M1, gamma -1, v -1.

02:03 That's the relativistic momentum now.

02:05 Got that gamma factor in there plus m2 gamma 2 v2 is equal to m1 gamma 1 prime v1 prime plus m2 gamma 2 prime v2 prime i use prime to indicate after the collision common common so you don't have to do heavy subscripting it's enough that's going on here okay so you write this out you normally you don't want to even when you did classical problems you don't want you don't want to put things in before you've written your basic expression doesn't matter if it's energy momentum conservation angle momentum conservation then in the next step you could say oh okay m2 is not moving so this is zero so this term is gone anything else is zero you take knock out here but that's that's usually what you want to do because it's so symmetric here you don't you don't have a chance of losing anything.

03:07 Okay.

03:08 Now, the goal of the problem is, actually, you might say, well, it is the goal of problem.

03:13 You don't have any of the ms.

03:14 They're looking for, they're looking for the ratio of m2 over m1.

03:20 That's the goal of the problem.

03:21 So let's divide both sides by m1.

03:23 It's going to give me gamma 1, gamma 1 v1 is equal to gamma 1 prime v1 v1 plus m2 over m1 gamma 2 prime v2 prime so we have now in one equation and if you think of m2 over m1 is your 1 -0 you got one equation 1 on 9 okay let's do some out let's just do some algebra now let's bring all the other terms in the m2 ratio to the left gamma 1 v1 minus gamma 1 prime, v1 prime, is equal to m2 over m1, gamma 2 prime, v2 prime.

04:13 So of this i can solve for m2 over m1 is equal to 1 over gamma 2 prime, v2 prime, gamma 1 v1, minus gamma 1 prime, v1 prime.

04:37 So that's what we have.

04:39 At this point, we can just put it in our values.

04:44 M2 or m1, just moving over here to give myself a little more room...

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