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

Consider a one-dimensional collision at relativistic speeds between two particles with masses $m_{1}$ and $m_{2}$. Particle 1 is initially moving with a speed of $0.700 c$ and collides with particle $2,$ which is initially at rest. After the collision, particle 1 recoils with a speed of $0.500 c$, while particle 2 starts moving with a speed of $0.200 c$. What is the ratio $m_{2} / m_{1} ?$

Key Concepts

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.

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.

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.

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.

Transcript

00:01 Consider a one -dimensional collision at relativistic speeds between two particles with masses m1 and m2.

00:08 Particle 1 is initially moving at 0 .700c.

00:12 It collides with particle 2, which is initially at rest.

00:15 After collision, particle 1 recoils with the speed of 0 .5 .00c.

00:21 Sure i got that right.

00:22 And particles 2 starts moving with 0 .2000c.

00:26 And what is the ratio of m2 to m1? so we're going to use the concept of conservation and momentum.

00:34 So let's begin with writing p1 plus p2 equals p1 plus p2.

00:49 Okay, both particles before and both particles after.

00:56 Massive particle at rest is m.

01:02 It has a velocity of v.

01:05 So particle one at rest will be m times v over the square root of 1 minus v squared over c squared and c equals 3 .0 times 10 to the eighth so let's go ahead and put p1 i'm going to do this on the next page p1 equal m1 v1 over the square root of 1 minus v squared over c squared and p2 equal m2 v2 times the square of 1 minus v2 minus v.

02:06 This should be v1, and this will be v2 squared over c squared.

02:14 The second particles at rest, so v equals 0, so p2 equals 0 initially.

02:23 So let's find the momentum of the first particle after the collision by using p11 equals m1 v1 over 1.

02:42 Minus v1 squared over c squared.

02:50 And the second particle after its collision will be p2 equals m2 v2 over the square root of 1 minus v2 squared over c squared.

03:08 Now we can start plugging some expressions in so we know that m1 v1, 1, m1 v1 over the square root of 1 minus v1 1 squared over c squared plus 0 will equal m1 v prime 1 over the square minus v 1 this should be 1 over c 1 squared and squared plus m 2 v2 2 over the square root of 1 minus v2 over c is squared and squared.

04:00 Ok, so let's see what we can do here.

04:12 We can cross out this term.

04:14 That's fine.

04:16 Let's subtract m1v1 from each side.

04:41 And this will get me.

04:45 I'm going to go to the next page.

05:17 And then that one's going to disappear.

05:19 So i'll have negative, i'll have m2v2...