Recent content by a.man

Yeah, it's the relative velocity of one end of the dashpot to the other (thus the relative velocity of one mass to another). So for the first mass it'll be derivate of x2 minus that of x1 with respect to time... and for the second mass it'll be negative of that? But the force opposes the...

"... where v is the relative velocity of its two ends. The force opposes the motion." So yeah, it should be the same for both cases – both magnitude and direction.

ma1=-kx1-k(x1-x2) - bv ma2=-kx2-k(x2-x1) - bv These were my first two equations of motion. I just added and subtracted them, using k = ω2m to get the equations for 2. Quick question: can I take relative velocity v as equal to the derivative of y2 with respect to time?

Sorry, this is the image:

Homework Statement Two particles, each of mass M, are hung between three identical springs. Each spring is massless and has spring constant k. Neglect gravity. The masses are connected as shown to a dashpot of negligible mass. The dashpot exerts a force of bv, where v is the relative...

Okay, I got the second answer as well, so never mind this. (Unless anyone else has the sam problem, I guess...?) You DID have to get an equation of the velocity of neutrons in terms of initial energy, find for which values of energy will lead to a REAL, POSITIVE value of the velocity in the...

Right, I figured how to do the first part: In center of mass system, the final particle's velocities will be zero respective to the center of mass (they will be the same speed), since this is threshold energy. So final energy is zero and initial energy is 2.8 MeV (The fact that system has the...

Homework Statement A thin target of lithium is bombarded by helium nuclei of energy E0. The lithium nuclei are initially at rest in the target but are essentially unbound. When a Helium nucleus enters a lithium nucleus, a nuclear reaction can occur in which the compound necleus splits apart...