Recent content by Ganesh

I have written all the equations I know. Is the application correct? P.S. I did not know you had typed out replies before I had finished editing (which I did immediately after my post came up)

Since l is infinite, the second side of bredth b never enters the field. Also, the force on each of the sides of infinite length is equal and opposite. So net force is the expression I typed out.

I did just that.

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I just edited myn post now

What happens when a rectangular loop is dropped into a field of uniform, magnetic field B extending to infinity? The loop has a mass m and its resistance is R. It has length L, breadth b, L tending to infinity. Gravity is present in downward direction. The loop is dropped such that emf is...

I agree with Shashank. Once the system is arranged and all the required data is given, only a single solution must exist in the field of classical physics.

I think I am pretty clear on the situation now. The COM of A and B has accelerted in the time of duration of the collision as its velocity has changed. As we are viewing from an accelerating frame, the acceleration of other bodies will be relative to this frame. So the rel. accn = (actual...

Hi Arildno I read your reply. Indeed angular momentum is conserved about the system's COM. Thinking further I realized, If you take the angular momentum in the frame of the COM of A and B only, it is an accelerated frame (as its velocity has changed and is changing.) So, pseudo forces are...

I am sorry that a lot of confusion has arisen due to some improper wording. Reilly, please note that masses connected at the ends are m/2 and m/2 while that of colliding mass is m. Gokul, to avoid confusion, consider masses attached to the ends as A and B and that of the colliding mass as...

Hi Tide and Gokul! What I am saying is : There are two equations for linear momentum and elastic collision. (The third mass won't move as the rod can exert forces only along its length.) So solving eqns. mv = mv'/2 + mv" and v - 0 = v' - v" we get v" = 4v/3 and v' = v/3...

Two masses of 'm/2' each are connected by a massless rigid rod of length 'l' and placed on a horizontal frictionless table. A third mass 'm' moving with velocity 'v' on the table strikes one of the masses in a head on elastic collision. Line of action is perpendicular to the length of the rod...

Two masses of 'm/2' each are connected by a massless rigid rod of length 'l' and placed on a horizontal frictionless table. A third mass 'm' moving with velocity 'v' on the table strikes one of the masses in a head on elastic collision. Line of action is perpendicular to the length of the rod...