View Full Version : Conservation of momentum
dave_wight
Jan 4, 2011, 02:29 PM
A proton of mass 1.67 X 10(to the power of -27) travelling at a speed of 6.8 X 10(to the power of 7) m/s hits a stationary unknown particle. The proton bounces back at a speed of 4.2 X 10(to the power of 6) m/s and the particle is projected forward at a speed of 2.6 X 10(to the power of 6) m/s. Calculate the mass of the unknown particle.
Thanks!
Dr.Jim
Jan 4, 2011, 11:30 PM
This is a classic billiard ball collision. Newton's law: for every action there is an equal and opposite reaction. Force is the change in momentum with respect to time. The force of the proton hitting the particle is the negative of the force of the particle hitting the proton. Since we are assuming no relativistic effects ∆(mv) = m ∆v. Consequently, m(proton) ∆v(proton) = m(particle) ∆v (particle). You do the math. The only unknown in the equation is m(particle).
Dr.Jim
Dr.Jim
Jan 4, 2011, 11:36 PM
The program here is not very good. That squirrelly notation (it looks like an a and a dagger) before (mv) etc. is supposed to be a delta. The equation is m(proton) delta v (proton) = m(particle) delta v(particle).
Dr.Jim
ebaines
Jan 5, 2011, 09:52 AM
Dr. Jim - the best way to embed math symbols is buy using Latex - with it way you can write equations such as:
m_{Proton} \Delta v_{Proton} = m_{Particle} \Delta v_{Particle}
You can learn about Latex here: https://www.askmehelpdesk.com/math-sciences/how-technical-scientific-documentation-formulas-50415.html