Meddling with Mechanics problems!

Greetings Mechanics Lovers!

Since many people are starting their preparations for entering into most prestigious institute of India,(IITs),lets start a Mechanics Marathon.This Marathon will be topic-oriented,i.e. we will try to post good questions of certain topics for first three weeks.This note will help us to improve our problem solving strategy.

Rules are simple as other contest/marathons.

•I'll post two problems.If anybody solves any one of the two problems,s/he should post next problem within 3 hours.

•There will be max 2 unsolved problems at a time.

•Make sure you know how to solve your own problem before posting it, in case no one else is able to solve it within 24 hours. Then, you must post the solution and you have the right to post a new problem.

•It is NOT compulsory to post original problems. But make sure it has not been posted on brilliant.Also please don't post HC Verma questions or DC Pandey questions.

•There should be only one question. No sub questions allowed.No parts. Only one answer should be there.

•The scope of the problems is Jee-Advance and NSEP.(mechanics only).

Topics for first three weeks:

• Vectors

• Kinematics

• Laws of motion

• Work Energy and Power

• Circular Motion

• Center of mass.

Post only the problem and it's solution here. Extremely relevant comments are allowed.You are allowed to discuss solutions and problems here.

Format your proof as follows:

Solution to Problem (Insert Problem no here)

[Post your solution here]

And put your Problem in a new thread following the format :

Problem (Insert Problem no here)

[Post your problem here]

Remember to reshare this note so it goes to everyone out there. And above all else, have fun!

Note by Harsh Shrivastava
3 years, 6 months ago

No vote yet
1 vote

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Problem 5

A chain of length L is placed on a smooth spherical surface of radius R with one of its ends fixed at the top of the sphere . What will be the acceleration of each element of the chain when its upper end is released

It is assumed that the length of chain is less than quarter of the sphere

Prakhar Bindal - 3 years, 6 months ago

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Is the answer a = (gR/L)(1 - cos (L/R))

Aniket Sanghi - 3 years, 6 months ago

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Correct! . Post next question and solution

Prakhar Bindal - 3 years, 6 months ago

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Solution to problem 5 :

Take a small element dx making angle dθd\theta with centre ( at an angle θ\theta with horizontal.)

Check for the tangential forces acting on it you will find that only one force act on it tangentially and that is

dmgcosθ dm g cos\theta (dm=MRdθL ( dm = \frac{MRd\theta}{L} ( where M is assumed as mass of chain which will cancel at end))

[

Edited : Sorry guys , I left one thing { Credit to Deeparaj Bhat }

There is one more tangential force , that is dT , but since tension is an internal force the integration of dT vanish .

]

This is the force on a single element , integrate it to get the total force and then divide it by mass to get acceleration , which you will get as

a=gRL(1cos(LR)) a = \frac{gR}{L} ( 1 - cos(\frac{L}{R}))

Calculations I left for you !

Aniket Sanghi - 3 years, 6 months ago

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@Harsh Shrivastava if you can remove the barrier of topics on posting questions , I will like post my next q that will be original! And will include a combination of topics.

Aniket Sanghi - 3 years, 6 months ago

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Problem 1:

Two particles, 1 and 2, move with constant velocities v1 and v2. At the initial moment their radius vectors are equal to r1 and r2. How must these four vectors be interrelated for the particles to collide?

Harsh Shrivastava - 3 years, 6 months ago

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Let the particles collide at point PP, whose position vector is r3{ r }_{ 3 }. Let the time taken be tt. By the vector laws of addition, we get

r3=r1+v1tr3=r2+v2t{ r }_{ 3 }={ { r }_{ 1 } }+{ v }_{ 1 }t\\ { r }_{ 3 }={ r }_{ 2 }+{ v }_{ 2 }t

\therefore. r1r2=(v2v1)t{r_1}- {r_2} = ({v_2} -{v_1})t

t\large t= (r1r2)(v2v1)\large\frac{({r_1} - {r_2})}{({v_2}-{v_1})}
From the system of equations,we get (r1r2)(r1r2)=(v2v1)(v2v1)\large \frac{({r_1} - {r_2})}{|({r_1}-{r_2})|} = \frac{({v_2} -{v_1})}{(|{v_2}-{v_1}|)}

By intuition, we can tell that the particles 11 and 22 collide when velocity of the second is directed towards the first. In that case, the velocity vectors turn parallel to the position vectors, thus making their unit vectors equal.

Swapnil Das - 3 years, 6 months ago

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Nice solution (+1) :).

Don't you think the velocities, in addition of being parallel, must be not be equal to each other in case r1r2r_{1} ≠ r_{2} ?

Aditya Sky - 3 years, 6 months ago

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Thanks For The Invite.Is the answer to this question v2 vector-v1vector/mod(v2 vector -v1 vector)=r2vector-r1vector/mod(r2vector-r1 vector)

Kaustubh Miglani - 3 years, 6 months ago

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Problem 2:

5 equal masses hang as shown in the figure.Find acceleration of each body.Assume pulleys to be frictionless,light and string is inextensible.

Harsh Shrivastava - 3 years, 6 months ago

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Its an elementary problem on virtual work method

By Symmetry Of forces acceleration of left most and right most will be same and acceleration of middle 3 will be same.

Apply Virtual work method

Call acceleration of extremum's to be a1 and middle ones to be a2

You get

2Ta1 = 6Ta2

a1 = 3a2

Apply Newton's second for two blocks

mg-T =ma1

2T-mg = ma2

Rest is algebra which is left as an exercise for reader

On Solving a1 = 3g/7 and a2 = g/7

Prakhar Bindal - 3 years, 6 months ago

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@Prakhar Bindal Hello Prakhar. Could you please tell me where I could learn about this Virtual Work method, especially on how to apply it for JEE problems? I had never heard of it before and it seems quite interesting! Thanks in advance!

Ishan Dasgupta Samarendra - 3 years, 6 months ago

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@Ishan Dasgupta Samarendra Hey hello @Ishan Dasgupta Samarendra . Long time no see . how are you?

About your question actually i was taught this by my teacher in Laws Of Motion in FIITJEE .

I Am giving you some links for learning.

https://books.google.co.in/books?id=imp4CgAAQBAJ&pg=SA6-PA34&lpg=SA6-PA34&dq=virtual+work+method+JEE&source=bl&ots=ka5Dc74yLI&sig=JJ7L4LQfIFhgeWR0dOg_D95gJrw&hl=en&sa=X&ved=0ahUKEwiwjMHy-pnNAhWEk5QKHeykDjwQ6AEIGzAA#v=onepage&q=virtual%20work%20method%20JEE&f=false

https://www.youtube.com/watch?v=lsHXK8nlLBs&feature=youtu.be

Prakhar Bindal - 3 years, 6 months ago

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@Prakhar Bindal Thank you very much for the links. I'll go through them right away! I'm fine, thanks! Just buried under a huge amount of work, and hence sadly I've got no spare time for Brilliant. What about you? However I do see most of the problems you post on Chemistry; I really like them. Aachchaa, which books do you study from for P, C and M? As in theory as well as practice problems.Lastly, which center are you at FIITJEE? I seem to recall seeing your name once somewhere...

Ishan Dasgupta Samarendra - 3 years, 6 months ago

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@Ishan Dasgupta Samarendra Thanks! . I Am bad at chemistry :P.

You too know the amount of work fiitjee people give :P . so i am left with very less amount of time to complete other books but still i have some

Maths---> I refer Amit Mohan Agarwal's Skills in mathematics Series by Arihant

Physics----> I Have HC Verma for theory and DC Pandey for problems . i too have irodov but i hardly pick it up

Chemistry----> Physical - NCERT For Theory and RC Mukherjee for problems

Organic- Solomon's for theory and Himanshu Pandey for problems

Inorganic---> :P . i have JD Lee but i haven't opened it from last 6 months!

I Am from FIITJEE East Delhi centre (Laxmi Nagar) . Where are you from ? . where did you saw my name?

Prakhar Bindal - 3 years, 6 months ago

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@Prakhar Bindal Fun fact: Himanshu Pandey sir is my chemistry teacher(for inorganic and organic).

Harsh Shrivastava - 3 years, 6 months ago

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@Harsh Shrivastava Cool! . he must be a master in organic chemistry

Prakhar Bindal - 3 years, 6 months ago

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@Prakhar Bindal Oh, I too have many of the books you use! Anyway thanks very much for telling me ! I'm from the South Delhi Center. I must have seen your name in some AIITS result or somewhere else. i really can't remember.

Ishan Dasgupta Samarendra - 3 years, 6 months ago

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@Ishan Dasgupta Samarendra Might be. !

Prakhar Bindal - 3 years, 6 months ago

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@Ishan Dasgupta Samarendra @Prakhar Bindal @Ishaan Dasgupta Samarendra

How did you guyz managed your time in class 11?

How many hours did you studied?

Please guide me.Thanks!

Harsh Shrivastava - 3 years, 6 months ago

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@Harsh Shrivastava I Think i am not the right person to be asked . you must contact the people who have given JEE This year

@Mayank Singh @Deeparaj Bhat @Abhineet Nayyar

Prakhar Bindal - 3 years, 6 months ago

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@Harsh Shrivastava I too am not the right person to ask. However one of the major mistakes I made in class 11 was that I never planned by which dates to finish which chapters. One should always have a countdown so that one knows exactly how far behind/ahead one is. For example, say you practise questions for Physics from sources A,B and C. Then you must have days scheduled by when you'll finish A, B and C respectively. That way you'll know whether you have been working the way you had to, or not.

Ishan Dasgupta Samarendra - 3 years, 6 months ago

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Let's start with an easy one!

Problem 3: The velocity of a particle moving in the positive direction of the xx axis varies as v=αxv=\alpha\sqrt{x}, where α\alpha is a positive constant. Assuming that at the moment t=0t=0 the particle was located at the point x=0x=0, find the time dependence of the velocity and acceleration of the particle.

Swapnil Das - 3 years, 6 months ago

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v=α2t/2v=\alpha^2t/2 and a=α2/2a=\alpha^2/2.

We know,

v=dxdt=αx1/2v=\frac{dx}{dt}=\alpha x^{1/2}

x1/2dx=αdtx^{-1/2}dx=\alpha dt

0xx1/2dx=0tαdt\displaystyle \int_{0}^{x}x^{-1/2}dx=\displaystyle \int_{0}^{t} \alpha dt

2x1/2=αt2x^{1/2}=\alpha t

x=α2t2/4x=\alpha^2 t^2/4

By differentiation,

v=α2t/2v=\alpha^2 t/2

a=α2/2a=\alpha^2/2

Akshay Yadav - 3 years, 6 months ago

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Correct.You perhaps can post the next question.

Swapnil Das - 3 years, 6 months ago

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Problem 4:

The trajectory for motion of a particle in xyx-y plane can be expressed as given below,

ex+ey=exey,x,yRe^x+e^y=e^x e^y, x,y \in \mathbb{R}

It is given that acceleration of this particle is zero and its initial velocity in xx and yy direction are 1m/s1 \text{m/s} and 2m/s2 \text{m/s} respectively. Also if its initial coordinates can be expressed in form integers, then find the initial coordinates (x,y)(x,y) of the particale.

If no such coordinates exist then enter your answer as 2016.

P. S. : Original problem.

Akshay Yadav - 3 years, 6 months ago

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Given equation:

1=ex+ey    0=ex+2ey(differentiating wrt time dxdt=1dydt=2)    1=eyNo real solutionQ.E.D.1=e^{-x}+e^{-y} \\\implies 0=e^{-x}+2e^{-y} \\ (\text{differentiating wrt time } \frac{dx}{dt}=1\: \frac{dy}{dt}=2) \\\implies -1=e^{-y} \\\text{No real solution}\\\large Q. E. D.

Deeparaj Bhat - 3 years, 6 months ago

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Correct! post the next question!

Akshay Yadav - 3 years, 6 months ago

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Can you please explain third line of you solution ?

Aditya Sky - 3 years, 6 months ago

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@Aditya Sky I meant that the second line is obtained after differentiating the given equation and using the given facts.

Deeparaj Bhat - 3 years, 6 months ago

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Problem 7: A lift of total mass M kg is raised by cables from rest to rest through a height h.The greatest tension which the cables can safely bear is n Mg Newtons(n>1).Find the shortest interval of time in which the ascent can be made,in terms of n,h,g.

Adarsh Kumar - 3 years, 6 months ago

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now this can have multiple ans depending on if cables are also attached at bottom for retardation or not i assume that the cable bring only acceleration and retardation is brought up by g only in that case i hope t(min)=root(2h/{n(n-1)g})(considering at a ht h v=0)

aryan goyat - 3 years, 6 months ago

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The latter case is correct!Well done!Post the next question please!

Adarsh Kumar - 3 years, 6 months ago

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Problem 8:

A cylinder of mass mm and radius RR rest on two supports in the same height .One support is stationery while the other slides from the cylinder at velocity vv. Determine the force of normal pressure NN by cylinder at stationery support when x=R2x=R\sqrt{2}.

aryan goyat - 3 years, 6 months ago

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Solution to Problem 8:

The answer is n=mg2mv22Rn=\frac{mg}{\sqrt{2}}-\frac{mv^2}{2R}. Due to time constraint I can't post the complete solution. But it can be found out by balancing forces. The angle made with horizontal can be found out by basic geometry. The velocity of the cylinder is v2\frac{v}{\sqrt{2}}. This can be found out by equating components.

Aditya Kumar - 3 years, 6 months ago

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I can't find a good problem now, can anyone else post it?

Aditya Kumar - 3 years, 6 months ago

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Solution to problem 8:

As the cylinder remains in contact with the stationary support it does circular motion about that point. As the cylinder is in contact with the block so the velocity of block on the normal of point of contact (which passes through center of cylinder) should be equal to that of cylinder.

vc=v2\Rightarrow v_c = \dfrac{v}{\sqrt{2}}

As the block does circular motion the forces perpendicular to the point of contact provide the required centripetal force.

mgcos45N=mvc2Rmg \cos {45^{\circ}} - N = \dfrac{mv_c ^2}{R}

N=mg2mv22RN = \dfrac{mg}{\sqrt{2}} - \dfrac{mv^2}{2R}

neelesh vij - 3 years, 6 months ago

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go and post the next q

aryan goyat - 3 years, 6 months ago

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@Aryan Goyat How to post image. ? I need image to illustrate the question.

neelesh vij - 3 years, 6 months ago

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@Neelesh Vij you just get the link by by copying it from publish place where you post q.

aryan goyat - 3 years, 6 months ago

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@Neelesh Vij when will you post q

aryan goyat - 3 years, 6 months ago

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Problem 10

A meteorite of mass m collides with a satellite which was oribiting around a planet of mass MM in a circular orbit of radius R . Due to collision, the meteorite sticks to satellite and satellite is seen to have entered an orbit in which its minimum distance from the planet is R2\frac{R}{2}. If mass of the satellite is is 10 times that of the meteorite , What was the velocity of the meteorite before collision ? (edit : Assume velocity of meteorite was initially perpendicular to that of the satellite)

Sumanth R Hegde - 2 years, 11 months ago

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Is the answer sqrt(288GM/R)?

Harsh Shrivastava - 2 years, 11 months ago

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no

Sumanth R Hegde - 2 years, 11 months ago

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Its a simple energy , momentum and angular momentum conservation problem

on solving v turns out to be sqrt (58GM/R)

Prakhar Bindal - 2 years, 11 months ago

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Yes :P.......Plz post the next problem

Sumanth R Hegde - 2 years, 11 months ago

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I might I have done a serious calculation mistake coz I also did the same.

Harsh Shrivastava - 2 years, 11 months ago

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Post the next problem!

Harsh Shrivastava - 2 years, 11 months ago

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PROBLEM 11

Moment of inertia of a uniform hexagonal plate about an axis LL' is I . The moment of inertia (about XX') of an equilateral uniform triangular plate of thickness double that of hexagonal plate is P . (specific gravity of triangular plate is 3 times that of hexagonal plate)

Find I/P

Prakhar Bindal - 2 years, 11 months ago

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file:///C:/Users/user/Downloads/Untitled.webp

visit this page for image

Prakhar Bindal - 2 years, 11 months ago

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ans is 5

aryan goyat - 2 years, 11 months ago

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@Aryan Goyat Yup correct!

Prakhar Bindal - 2 years, 11 months ago

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@Aryan Goyat Post the next problem , a tough one!

Harsh Shrivastava - 2 years, 11 months ago

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@Harsh Shrivastava wasn't this one tough? :P

Prakhar Bindal - 2 years, 11 months ago

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@Prakhar Bindal It was tough indeed, though I have only tried it once.

Harsh Shrivastava - 2 years, 11 months ago

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@Harsh Shrivastava This was very similar to a problem which appeared in our AITS 1

Prakhar Bindal - 2 years, 11 months ago

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http://tinypic.com/view.php?pic=2ludvn5&s=9#.WGoB81N96M8

Prakhar Bindal - 2 years, 11 months ago

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https://s28.postimg.org/vhjxmwknx/Untitled.png

Prakhar Bindal - 2 years, 11 months ago

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I am inviting few people to join the contest, sorry for mass taging.

@Akshay Yadav @Aniket Sanghi @Adarsh Kumar @Chinmay Sangawadekar @Vaibhav Prasad @Kalash Verma @Hrishik Mukherjee. @Nihar Mahajan @Prakhar Bindal @Swapnil Das @Rajdeep Dhingra @Shreya R @milind prabhu @Kaustubh Miglani @Vighnesh shenoy

Harsh Shrivastava - 3 years, 6 months ago

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Thanks for inviting :)

Swapnil Das - 3 years, 6 months ago

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Reshare the note so that it may reach out to others.Thanks!

Harsh Shrivastava - 3 years, 6 months ago

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@Aniket Sanghi @Deeparaj Bhat please post the next question!

Adarsh Kumar - 3 years, 6 months ago

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Problem 6 : [Original]

Take a bomb of mass m and give it a velocity v=10m/sec v = 10m/sec at an angle θ=53° \theta = 53° with horizontal. The bomb bursts into two identical parts when it's velocity made an angle +45° with horizontal. Just after the breakdown, One of the part's velocity was found to be 6m/sec 6 m/sec in horizontal direction. (i. e. It retained it's horizontal component of velocity ) . Find the final displacement of one part with respect to other after they come to rest.

Details And Assumptions :

  • The coefficient of restitution for the ball ground collision is e=0.4 e = 0.4

  • The ground is smooth.

  • Take g=10m/s2 g = 10m/s^2

Aniket Sanghi - 3 years, 6 months ago

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You may post the solution and the next problem.

@Aniket Sanghi

Harsh Shrivastava - 3 years, 6 months ago

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aryan goyat - 3 years, 6 months ago

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this gives ans 18.1527 as mentioned by @Adarsh Kumar i would request to find a mistake in this?????

aryan goyat - 3 years, 6 months ago

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@Aryan Goyat Thanx!@aryan goyat for posting this!

Adarsh Kumar - 3 years, 6 months ago

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@Adarsh Kumar welcome

aryan goyat - 3 years, 6 months ago

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@Adarsh Kumar Go! Post the next question for now @AdarshKumar

Aniket Sanghi - 3 years, 6 months ago

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@Aniket Sanghi Ohk,gimme a minute I just woke up :P.

Adarsh Kumar - 3 years, 6 months ago

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aryan goyat - 3 years, 6 months ago

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this gives ans as 16.71 as suggested by @Aniket Sanghi i request to find a mistake in this.

aryan goyat - 3 years, 6 months ago

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aryan goyat - 3 years, 6 months ago

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this gives ans as 14.782 as suggested by @Prakhar Bindal i request to find a mistake in it??

aryan goyat - 3 years, 6 months ago

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Hey guyz,please delete unnecessary comments as the note is getting slow.Thanks.

Harsh Shrivastava - 3 years, 6 months ago

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Problem 9

A metal rod is of length 1m1m is clamped at points 5cm5cm and 15cm15 cm from either ends. Find the minimum frequency of natural longitudinal oscillations of rod.

Given :

  • Young's modulus of rod =1.6×1011Pa = 1.6 \times 10^{11} Pa

  • Mass per unit volume ρ=2500kgm3 \rho = 2500 kgm^{-3}

neelesh vij - 3 years, 6 months ago

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pic

aryan goyat - 3 years, 6 months ago

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Rod is clamped at a distance 5cm from one end and 15 cm from other end. Its easily understandable i guess.

neelesh vij - 3 years, 6 months ago

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@Neelesh Vij yep it is but i thought you were not able to do as guided by me.

aryan goyat - 3 years, 6 months ago

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@Aryan Goyat yes i didn't understand the method u said to post the pic.

neelesh vij - 3 years, 6 months ago

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@Neelesh Vij be on slack

aryan goyat - 3 years, 6 months ago

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@Aryan Goyat I need to go to coaching right now. so talk to you later

neelesh vij - 3 years, 6 months ago

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@Neelesh Vij ok bye today i have coaching too from 2-30 so better talk at 9-00

aryan goyat - 3 years, 6 months ago

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I did it like this. The particles at ends of the rod are the ones which will oscillate with max amplitude(antinodes). In case of minimum frequency , we are looking to maximize wavelength. Hence consider the case when antinode is at one end and first node is at the hinge 5cm from it.(This also takes care of a node being at the other point that is hinged) . hence max wavelength is 20cm. Thus min frequency is 40,000Hz40,000 Hz

Sumanth R Hegde - 2 years, 11 months ago

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correct!

neelesh vij - 2 years, 11 months ago

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Awesome! We got a new problem poster!

Post the next problem please!

Harsh Shrivastava - 2 years, 11 months ago

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Is the contest dead?

Harsh Shrivastava - 3 years, 6 months ago

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Yeah :P . No more activity as we all are quite busy with our class 12th!

Prakhar Bindal - 3 years, 6 months ago

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Oh, sad. But no worries! I'll be coming up with an Electromagnetism contest on Monday, be sure to take part!

Swapnil Das - 3 years, 6 months ago

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Is this dead? Shall I participate

Harry Jones - 2 years, 11 months ago

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If you want you may post problems here and I will participate here.So post the next problem !

Harsh Shrivastava - 2 years, 11 months ago

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Can u derive the equation of trajectory of each particle in the Irodov question no 1.12?

Harry Jones - 2 years, 11 months ago

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