Antoha1's latest activity

My solution is: ##\frac{1}{2}mv^2=\frac{1}{2}kx^{2} \Longrightarrow x=v\sqrt{\frac{m}{k}}## x=0,354m block gains negative acceleration...

Here is my recent solve checking if I understand it now: Proton's total energy now is kinetic plus rest so E+mc^2, then conservation of...

do I understand correctly: Relativistic mass is just total energy so the energy ratio is 3Ep=Ea and I can manage everything with 2 first...

Applying conservation of energy law: Both particles got energy ##A## from eletric fiel: ##A_{p}=eU## and ##A_{\alpha}=2eU##. Lets say...

my other question is: if particle's speed has been increased lets say from 0 to around c, the mass of it increases, lets say rest mass...

ok, so adding rest energy of lithium, but what about rest energy of proton, because I am thinking that only its kinetic E is given...

So basically, mass of lithium given is useless in this problem?

I apply energy conservation law: $$E+(m_{li}+m)c^2=2m_{\alpha}(c^2+\frac{1}{2}v_{\alpha}^2)~~~~~~~~(1)$$ where ##v_{\alpha}## is speed...