Recent content by rockinwhiz

That's damn cool!

I really would like to know the process of your preparation... Like how and when you started preparing, what growth you saw in your problem solving capacity and so on. Kindly share about the beautiful journey. If possible, give me some tips/resources as I'm also an aspiring olympian.

Actually, the book this is from has generally ambiguous questions, so nothing is really specified as such. But I might look into that case too. However, for now, getting the answer is enough and I gained a lot of insights too. Thanks to all of you.

Oh yeah I missed that.

Oh yeah...! Since, ##A=dv/dt=-(g+ \frac {kv} m)##, therefore, ##\dot {K}_E## becomes: $$ \dot {K}_E = -mv \cdot (g+ \frac {kv} m)$$ So, again, differentiating this with respect to ##v## gives ##-mg -2kv=0##, giving: $$ |v|= |- \frac {mg} {2k} | \approx 12.2 m/s$$ Thank you guys so much!

I'll try this in a while and reply. Thanks!

So, if I'm not wrong, the kinetic energy keeps changing during upward motion. However, during downward motion, eventually the ball will eventually hit the terminal speed, so the kinetic energy will stop changing I guess?

A derivative with respect to ##v## here?

Rate of change of kinetic energy would be: $$ \frac {d K_E} {dt} \equiv \frac {d(mv^2/2) } {dt} \equiv mv \cdot \frac {dv} {dt}$$ So, I need to maximize this, most probably. So, should I take take another time derivative?

Because, ##F=ma=kv##, therefore, ##a=kv/m##. Clearly, the net acceleration ##A=-(g+a)##. Also, ##A=dv/dt=-(g+ \frac {kv} m )##, so cross multiplying and integrating LHS with respect to ##v## and RHS with respect to ##t## gives me: $$ v= e^{ \frac {-tk} m } * (u + \frac {gm} k) - \frac {gm} k $$...

Okay, I think I see something here. So, the tension in the string would also act on the big block, to drive it in the opposite direction as the top block, I guess? I think that works. That gets me the answer too. But if there are any nuances, then kindly point them out as well. Thank you!

I think I might've read something like this somewhere. Centre of mass at the start would be somewhere between the three masses, but closer to the big block. At the end, if the big block didn't move, it probably would shift somewhere away from that point. But no external force acts on the system...

Clearly, in the picture I can see that on the small block on top, tension and gravitational force act. Gravity gets balance by the normal force, so tension is the only force causing acceleration. On the block at the side gravity and tension result in vertical acceleration. However, I do not...

Oh yeah. I guess I'll check it out later. Thanks for everything man.

A big thank you to all of you helping me out.