Recent content by Kumar Singh

Sorry! question set up is right. A simpler version (only a(x)) has already been asked in IITJEE 2007. See here question number 3. http://www.kshitij-iitjee.com/IITJEE-Past-Year-Papers/IITJEE-2007-Paper-1 This question has been taken from a test paper of a reputed coaching institute for IITJEE.

Yes , you are right. But acceleration perpetually changes magnitude along x axis...and x component of acceleration / force will vanish ultimately. But what's the nature of acceleration and velocity along x axis..i really don't know. I feel calculus is involved here. Can you help?

Finally whole speed is in y-axis only, i hope.

Well...you see at the moment they meet, speed along x-axis will be zero and as I am taking energy concept in use I have to care only about initial and final positions. initially speed along x as well as y-axis was zero and finally speed is only along y axis.

Ok. This is the diagram I used to make equations.. On the other hand I used some other method to solve it... Let the point of application moves distance d before masses meet, worked done on system would be F.d By work energy theorem F.d = 2 (1/2 m v^2) v^2 = Fd/m again for motion along...

Ok. This is what I did. I have assumed whole case at some angle θ with vertical and then resolved the force http://filesystem:https://web.telegram.org/temporary/851516233_42736_5351388760177376947.jpg so, F = 2Tcosθ T = F/(2cosθ) Now, for an individual mass F(x) = T sinθ = (F tanθ)/2 a(x) =...

Ok. Got it :smile:

Certainly not homework. I am not somebody's teacher to give someone some homework. I thought this is a forum for Physics lovers to exchange the ideas and shared problem solving. :cry:

Ok. I am new here so bit unfamiliar with the prevailing norms.

Basically I worked over it by taking some angle at some moment and resolving the force then writing down velocity and acceleration equation for x and y-axis but I am getting entangled in calculus part.

[Mod Note: moved to homework forum, template missing but homework-type problem] There are two masses (m) tied at two ends of a string of length 'l'. Whole system is placed on a friction-less surface and force 'F' is applied at the centre of string in direction perpendicular to the string. Find...