Recent content by coldblood

Thank you friends, I have got the answer. I was doing a silly mistake. Special Thanks To NascentOxygen and ehild

in the equation, i = [E1 + E2 +------+En] - [E1 + E2 +------+E N-n] / R1 + R2 + ------- + RN here E = αR Hence, i = [αR1 + αR2 +-------+ αRn] - [αR1 + αR2 +-------+ αR N-n] / [αR1 + αR2 +-------+ αRN] typical to solve this, so If I assume all the resistances identical, then i =...

NascentOxygen: Please tell me how we'll find the current in this circuit?

Hi friend I am Stuck in a problem. Please help me in solving this. Thank you all in advance. The problem is as follows. https://fbcdn-sphotos-a-a.akamaihd.net/hphotos-ak-xpf1/t1.0-9/10270329_1576533655907072_4352617381833809587_n.jpg...

That means when I was conserving linear momentum, it was giving the answer as v'(hoop + bullet) = v/2, which was wrong but giving the correct answer(coincidence). I took conservation of linear momentum in wrong manner. This is the best approach. What do you say Tanya?

Yes, I got it. Thank you all for the help.

From now on I will always try to take a fixed point as the reference, then I watch over the quick approach. Thanks a lot for the help.

The post which I wrote, [FONT="Arial Black"]Because during pure rolling contact point does not slip. Hence friction is zero.. Is this statement correct?

I got it. That means Any of the choice could be there. Hence Can't be interpreted?

But this will give the answer ω = v/ 4r, and answer is v/ 3r

I got it haruspex, I will take care for all these. But If in any problem collision takes place like in Prob. 4 and porb 7, we take it as abrupt change or what should we think?

Last one when x = R/2, f = 0, Because during pure rolling contact point does not slip. Hence friction is zero.

I conserve A.m. about the center of mass of the system.

https://fbcdn-sphotos-f-a.akamaihd.net/hphotos-ak-prn2/1476572_1462877957272643_1028005643_n.jpg https://fbcdn-sphotos-g-a.akamaihd.net/hphotos-ak-prn1/q71/s720x720/936643_1462877960605976_1351577648_n.jpg

Now I got where I was doing mistake. conserving linear momentum, m(v) = 2m v' => v' = v/2 conserving angular momentum about centre of the loop, mv(R/2) = mR2 + m(R2)/4(loop) + {mR2/4(particle)}. ω + (m). (v/2).(R/2) - (m). (v/2).(R/2)(particle +loop com) it gives the answer, ω = v/3R