Recent content by Eirik

That's what I thought. Looks like it's just a mistake in the textbook then, as they were referencing the total energy of the composite particle. Thanks again:)

@PeroK Is it okay if I ask another question on the same topic? If the other proton had been at rest, is it true that ##E=K+mc^2+mc^2=2mc^2(1+\frac{K}{2mc^2})## My physics book seems to think it's ##E=2mc^2*\sqrt{1+\frac{K}{2mc^2}}##:oldconfused:

I must've typed something wrong into my calculator the first time I tried to solve it haha! It's the same equation. Hope I don't make mistakes as dumb as this one during my exam tomorrow:blushing:. Thanks!

I suddenly realized I forgot to account for the fact that kinetic energy isn't conserved in inelastic collisions.. 🤦‍♂️ That probably has something to do with it doesn't it

Hi! Hope I'm posting this in the right place! I'm practicing for exams and came over this question: A proton with mass ##m_p## is accelerated to a relativistic velocity, with kinetic energy ##K##. It collides completely inelastic with another proton, which has the same kinetic energy, ##K##...

@TSny Yes, of course! 🤦‍♂️ I thought ##\ddot R## would be M's downward acceleration, but R will get smaller over time, so it has to be ##Mg-2S=-M\ddot R## That finally gives me ##\ddot R + \frac{M}{M+2m}g - \frac{\frac{2mR_0^4\omega_0^2}{M+2m}}{R^3}=0## with ##\alpha = \frac{M}{M+2m}## and...

Thank you so much @TSny ! That was really, really helpful! Then we have ##L_i=L## ##2mR_0^2\omega_0=2mR^2\omega## Solving for ##\omega## gives us ##\omega=\frac{R_0^2\omega_0}{R^2}## And if we choose the positive direction for the accelaration along -z, it should be ##S=m*(\omega^2R-\ddot R)##...

Here's a diagram of what the system looks like: So far I have figured out what the initial angular velocity is, if the system is balanced (no movement): ## \sum F_m = m*\frac{v^2}{R_0}-\frac{Mg}{2}=0 ## ##m \frac{v^2}{R_0}-\frac{Mg}{2}=0 ## divide both sides by m ##\omega_0 =...

To begin with, it would have velocity in the X direction, and Vy=0. Because of the velocity in the x direction, it's then going to be accelerated along the y-axis. This acceleration should be equal to a=q*v*B/m, right? Is it then correct to say that the acceleration is going to be...

The movement in the z-direction is easy to solve for, as it's only affected by the gravitational force. However, if there's a magnetic field pointing down along the z-axis, the particle is going to be accelerated along the y-axis (F=q*v *B). The force is always going to be perpendicular to the...

Thank you! This was extremely helpful! However, won't pounds not being an SI unit affect the results? Souldn't I measure it in Newtons (or kilograms)? Same question for using degrees instead of radians. I should also probably mention that I don't need this to be extremely precise by the way! :)...

Homework Statement So I have been building a mouse trap car at my school, and I need to hand in a report on it tomorrow. :biggrin: I don't have all of the measurments at the moment, but the only thing I want to know is how to calculate this. I want to see how much energy is lost to...