Recent content by Kishor Bhat

So be it. If I use F = qE, and acceleration a = F/m, I get 8:1 since the velocities are proportional to a^2. But if we had to take the inter-particular forces into consideration, I couldn't use that formula, since the repulsive force would vary with intervening distance.

But we don't know the magnitude of the electric field..

Ah. Well then, thank you. :)

I'm afraid I'm not getting it. Conservation of momentum gives 2:1, and for conserving energy, I don't know the distance between the charges to calculate potential energy.

Homework Statement Two charges of masses m and 2m, charges 2q and q respectively are placed in a uniform electric field and are allowed to move at exactly the same time. Find the ratio of their kinetic energies. Homework Equations Field = qE Force =(Kq1q2)/r^2 Kinetic energy=1/2mv^2...

Gracias.

The MI's are correct, and yes, they are all of same mass and radius. One question: can we find the acceleration of c.o.m and then use the equations of motion to find final velocity? I haven't tried that.

Right. Sorry.

Homework Statement Find the ratio of the translational kinetic energies of a ring, a coin, and a solid sphere at the bottom of an inclined plane. The bodies have been released from rest at the top. Assume pure rolling without any slipping. The Attempt at a Solution Well, I'm really not...

I'm assuming that was rhetorical. :P I think you should just find the resultant of the momentum vectors, and then arctan[(y-component)/(x-component)] (components of resultant) should give you the angle.

Well, the work done by the force is equal to (-k*x^2)/2, where x is the max. deformation of the spring. This can be found by integration. But I'm not sure how the setup looks and how you're applying the force.

Truer words have not been spoken. :P

Nope. All words were directly from the worksheet.

Mm I'll try 'em out. Thanks a bunch. Both of you. I think we can conclude: Case closed. :)

Oh yeah... whenever we add or subtract MI's it has to be about the same axis right? Epic fail on my part. Thanks again.