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Mechanics Competition by shubham dhull

Welcome all to the Brilliant Mechanics Contest. Like the Brilliant Integration Contest, the aim of the Mechanics Contest is to improve skills and techniques often used in Olympiad/JEE-style Mechanics problems. But above all, the main reason is so we can have fun.

Anyone is allowed to participate, as long as they adhere to the following rules.

I will post the first problem. If someone solves it, he or she can post a solution and then must post a new problem.

A solution must be posted below the thread of the problem. Then, the solver must post a new problem as a separate thread.

Please make a substantial comment.

Make sure you know how to solve your own problem before posting it, in case no one else is able to solve it within 48 hours. Then, you must post the solution and you have the right to post a new problem.

If the one who solves the last problem does not post a new problem in 24 hours, the creator of the previous problem has the right to post another problem.

The scope of the problems is (IPhO) Olympiad Mechanics.

Keep yourself restricted to Newtonian Mechanics. Do not use Lagrangian mechanics

You are allowed to apply a slight dressing of Electricity And Magnetism to your Mechanics Problem.

Upload your solution along with the answer * Compulsory. *

It is NOT compulsory to post original problems. But make sure it has not been posted on brilliant.

Try to post the* simplest solution* possible. For example, if someone posted a solution using Newton's Laws, when there is a solution using only Work Energy theorem, the latter is preferred.

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

The problem creator should not give the answer to his/her question with the question.

Post only the problem and it's solution here. Extremely relevant comments are allowed. All discussions should strictly not be done here.

Note by Shubham Dhull
5 months, 3 weeks ago

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Hey guyz, @Aniket Sanghi @Prakhar Bindal @shubham dhull

Post any "good" mechanics or physics question you come across your preparation, I'll also post if I come across any.

Thanks. Harsh Shrivastava · 5 months, 2 weeks ago

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@Shubham Dhull I'll try to post a good mechanics question whenever I get any.

By good I mean a problem that is not a standard 'DC-Pandey/coaching material' problem. :) Harsh Shrivastava · 5 months, 2 weeks ago

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@Harsh Shrivastava

the green arrow shows the motion of block, find the least mass required to move the big block if coff. of friction is mew under the constraint that the system should remain stable @Aniket Sanghi its my original one ! l>h and rod is massless! Shubham Dhull · 5 months, 2 weeks ago

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Problem 6 SET I

A Bob of mass m is hanging by an inextensible thread. A similar of Bob of mass m and same radius is going to strike the Bob held with string as shown in the figure. It strikes the the Bob with a speed of \(v_{0} \).

Find the velocities of both the bobs after the collison. Harsh Shrivastava · 5 months, 2 weeks ago

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@Harsh Shrivastava Correct?? Aniket Sanghi · 5 months, 2 weeks ago

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@Harsh Shrivastava Lower bob's horizontal velocity \( v = 2\sqrt{3}/5 \)

Upper bob's after collision velocity \( v2 = \sqrt{13}/5 \) Aniket Sanghi · 5 months, 2 weeks ago

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@Aniket Sanghi bro, solution? rule here Shubham Dhull · 5 months, 2 weeks ago

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@Shubham Dhull You only told , no rules from now , and only we 3 are there , is their a need for solution?? Aniket Sanghi · 5 months, 2 weeks ago

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@Aniket Sanghi Correct!

Anybody want to post a question? Harsh Shrivastava · 5 months, 2 weeks ago

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@Harsh Shrivastava Somebody solvey optics q , it had got more than 180 views , but 0 solvers ! :P Aniket Sanghi · 5 months, 2 weeks ago

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@Aniket Sanghi i will try after phase test :P and after net also gets unlimited Prakhar Bindal · 5 months, 2 weeks ago

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@Aniket Sanghi I can't solve it, havent studied it yet. Harsh Shrivastava · 5 months, 2 weeks ago

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@Aniket Sanghi i'll try Shubham Dhull · 5 months, 2 weeks ago

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@Harsh Shrivastava i'll post Shubham Dhull · 5 months, 2 weeks ago

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@Shubham Dhull COOL!!! Harsh Shrivastava · 5 months, 2 weeks ago

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@Aniket Sanghi ok, no i dob't need the sol. , i can do it thnx Shubham Dhull · 5 months, 2 weeks ago

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@Harsh Shrivastava yes tell wheter elastic or not Shubham Dhull · 5 months, 2 weeks ago

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@Harsh Shrivastava Is e given?? Aniket Sanghi · 5 months, 2 weeks ago

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@Aniket Sanghi Sorry, forgot to mention that e=1:P Harsh Shrivastava · 5 months, 2 weeks ago

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@Harsh Shrivastava I won't try , otherwise I would be forced to post the next q :P ( And I lack time )

Prakhar also won't be online for 2 weeks . Furthermore this competition lacks participation , it is not arousing interest in me ... Aniket Sanghi · 5 months, 2 weeks ago

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@Aniket Sanghi Will you post questions?

@Aniket Sanghi Harsh Shrivastava · 5 months, 2 weeks ago

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@Harsh Shrivastava Whenever I will get time :)

By that time you can check my Brilliant questions ( I posted) if you want :) Aniket Sanghi · 5 months, 2 weeks ago

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@Aniket Sanghi K, we have just started rotational mechanics in Fiitjee, so after that, I'll become eligible to try mechanics questions on brilliant! Harsh Shrivastava · 5 months, 2 weeks ago

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@Shubham Dhull I can't understand why are you taking it on you , You are just the starter of the contest , I am blaming contest and that too participation , not you and even I was not blaming , I was just sharing my views! . I had never seen a starter of contest being so possessive about his contest !

A contest is started by one but then it becomes public , people take it ahead and in this one , people arn't coming and I think in that respect Harsh's thinking is right!

Sorry! If my lines hurt you . But I will again say I was saying Truth ( what I really felt) and the lines were for the 'Participation' (Not) 'You' as participation is not in your hand!

Even Harsh got me right! Sad to see ,You got me wrong! Aniket Sanghi · 5 months, 2 weeks ago

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@Shubham Dhull Its OK bro! :) Aniket Sanghi · 5 months, 2 weeks ago

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@Aniket Sanghi OK, what answer are you getting? Harsh Shrivastava · 5 months, 2 weeks ago

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@Aniket Sanghi No problem.

The new brilliant interface has made it difficult to discover "new" stuffs.

@Calvin Lin please do something, the community is going to a hibernation due to new boring interface, the new interface is not at all logical. Harsh Shrivastava · 5 months, 2 weeks ago

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Problem 4. SET I

Aniket Sanghi · 5 months, 2 weeks ago

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@Aniket Sanghi hey @Aniket Sanghi is the answer 8 ? Shubham Dhull · 5 months, 2 weeks ago

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@Shubham Dhull if my answer is correct, let me share the solution ! let the mass of the rods be m1 , m2 ( it would help if u assume some mass per unit length ) the B will move downward with some velocity v , the line bb will only have a horizontal velocity by using geometry u can fid the relations in their velocities and then using conservation of energy , we will get the velocity as 8 m/sec. easy enough !@Aniket Sanghi @Prakhar Bindal bhai ! Shubham Dhull · 5 months, 2 weeks ago

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@Shubham Dhull Answer is correct ! 👍 Post the next! Aniket Sanghi · 5 months, 2 weeks ago

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@Aniket Sanghi ok here is it \(set-1\) \(q-5\) An easy one A thin uniform equilateral triangular plate rests in a vertical plane with one of its ends (B) on a rough horizontal

floor and the other end (C) on a smooth vertical wall. The least angle its base (BC) can make with horizontal will be: Shubham Dhull · 5 months, 2 weeks ago

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@Shubham Dhull Am I correct ? Harsh Shrivastava · 5 months, 2 weeks ago

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@Harsh Shrivastava YES YOU'RE RIGHT ! also i liked it that u uploaded the question without uch delay , good work ! Shubham Dhull · 5 months, 2 weeks ago

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@Shubham Dhull

Harsh Shrivastava · 5 months, 2 weeks ago

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Why is the contest stubbed? Harsh Shrivastava · 5 months, 3 weeks ago

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@Shubham Dhull OK let's wait for 3 hrs. Harsh Shrivastava · 5 months, 3 weeks ago

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@Shubham Dhull I can't delete my comments , there's a bug.

(Don't reply as it will add one more comment.) Harsh Shrivastava · 5 months, 3 weeks ago

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An uniform rope of mass m is resting on a sphere such that it covers a circle of radius r on the sphere.

The radius is sphere is R.

Find the tension in the rope. Harsh Shrivastava · 5 months, 3 weeks ago

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@Harsh Shrivastava Take a small element in the rope dx (mass = dm =\( \lamda R (2d\theta)\) )

Check the normal reaction and divide it in components in horizontal and vertical direction and we get

\( N cos\theta = dm g \)

\( N sin\theta = 2 T sin (d\theta) \)

On simplifying we get

\( \boxed{ T = \frac {mgR}{2\pi \sqrt {R^2 - r^2}}} \) Aniket Sanghi · 5 months, 3 weeks ago

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@Aniket Sanghi Correct!

Post the next question. Harsh Shrivastava · 5 months, 3 weeks ago

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@Aniket Sanghi hey bro whats the situation can you explain or you @Harsh Shrivastava . if possible please attach a diagram Prakhar Bindal · 5 months, 3 weeks ago

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@Prakhar Bindal A rope is kept surroundimg a sphere of radius R, the circumference of rope is less than 2piR. Harsh Shrivastava · 5 months, 3 weeks ago

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Set(1) Problem 2

A Shell flying with velocity v= 500 m/sec bursts into three identical fragments so that kinetic energy of system increases 1.5 times. Maximise the velocity that one of the fragments can obtain.

this is an integer type problem (0 to 9)

give your answer in km/sec Prakhar Bindal · 5 months, 3 weeks ago

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@Prakhar Bindal Answer is 1.

It is more a maths problem than physics.

Let the velocities of the fragment be x,y,z after the explosion.

Conserving momentum along initial direction of motion, 500m = m/3 (x+y+z).

Also, using the KE data, x^2 + y^2 + z^2 = 1125000.

Using these two equations, we can say :

\[\dfrac{(1500-x)^2}{2} \leq 1125000-x^2\]

Thus we get -1000m/s <= x<= 0m/s Harsh Shrivastava · 5 months, 3 weeks ago

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@Harsh Shrivastava hey, post the next question ! Shubham Dhull · 5 months, 3 weeks ago

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@Harsh Shrivastava Right! . post the next.

Did you had your first AIITS? . What was ur rank? Prakhar Bindal · 5 months, 3 weeks ago

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@Prakhar Bindal Very bad, 126 :( Harsh Shrivastava · 5 months, 3 weeks ago

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@Harsh Shrivastava plz move such discussions to slack or other notes and delete ur comment here ! it's only for asking questions not talking !! Shubham Dhull · 5 months, 3 weeks ago

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@Shubham Dhull Ohk sorry Prakhar Bindal · 5 months, 3 weeks ago

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Some more Invitations

@Gauri shankar Mishra @Archit Agrawal @aryan goyat @Samarth Agarwal Aniket Sanghi · 5 months, 3 weeks ago

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I am starting a second set of question series (Advance Level / IPho Level) , as when 1st serious take a break the contest must go on! Both series will go independently . I hope @shubham dhull has no problem

* Question 1* [Set II ]

A heavy particle of mass m is placed symmetrically on the top of a smooth hemisphere also of mass m which is placed on a smooth horizontal plane . The system is released from rest (a slight puch given to the particle) . Find the angle \( \theta\) with the vertical where the particle will loose contact with the hemisphere . Give your answer as \( cos\theta = ?\)

[ Not Original] Aniket Sanghi · 5 months, 3 weeks ago

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@Aniket Sanghi no, not at all bro Shubham Dhull · 5 months, 3 weeks ago

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@Shubham Dhull but bro, set 1 is also for advance not mains ! Shubham Dhull · 5 months, 3 weeks ago

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@Shubham Dhull What I mean to say , u will understand by solving this q !! :) Aniket Sanghi · 5 months, 3 weeks ago

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@Aniket Sanghi

Shubham Dhull · 5 months, 3 weeks ago

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@Shubham Dhull Post the next problem. Harsh Shrivastava · 5 months, 3 weeks ago

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@Shubham Dhull That's what I wanted to say ! Q taking at least 4-5 mins This is an easy start and I hope lot more tougher q in this set ! In the second set we can go on with easy as well as hard q ! Btw yr answer is correct @shubhamdhull Post a hard q ! Hardness one step up to this :) Aniket Sanghi · 5 months, 3 weeks ago

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@Aniket Sanghi \(Q-2\quad set-\quad II\\ \) You asked for it , you got it,

 My Problem !

My Problem !

Two similar rods are hinged by a massless pivot between them and it was given impulse " I " at its end what it the velocity of end point of other rod initially , everything is smooth and ideal . express your answer in m,L, " I " Shubham Dhull · 5 months, 3 weeks ago

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@Shubham Dhull I told u , I was in hurry to go to school! , I don't think that must be some rule!!!!! Let the hinge applies I1 towards left on the upper rod and towards right on the lower rod.

Eq goes like this I - I1 = m v1 (I + I1)L/2 = mw^2l/12 I1 = m v2 I1l/2 = mw2^2l/12

And last one

v1 - wL/2 = v2 + w2L/2

And then the thing we have to find v2 - w2L/2

On solving , we get the answer as I/2m Aniket Sanghi · 5 months, 3 weeks ago

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@Aniket Sanghi Yaar maine alag alag line main likha tha , sab Ek saath Ek line main aa gaya! :| Aniket Sanghi · 5 months, 3 weeks ago

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@Shubham Dhull Answer I/2m correct? ( I have to go to school , I am in hurry) Aniket Sanghi · 5 months, 3 weeks ago

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@Aniket Sanghi ok , correct answer ,good job, now upload ur ques but keep in mind , it should be one(or two may be) level up in hardness ! a sooner reply would be appreciated and also upload a ques in set -1 as u r the solver here also ! @Aniket Sanghi Shubham Dhull · 5 months, 3 weeks ago

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@Aniket Sanghi Correction : In the first set * we can go on with easy as well as hard q Aniket Sanghi · 5 months, 3 weeks ago

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@Aniket Sanghi A slight push * Aniket Sanghi · 5 months, 3 weeks ago

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@Aniket Sanghi In this second set \( [Set II] \) All questions posted must be of Advance level / IPhO level .

Time provided here is 72 hours after the question is posted. Rest , the same rules go here! Aniket Sanghi · 5 months, 3 weeks ago

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\(Q-1\) \ i'll post the first problem so," If the kinetic energy of a particle moving on a circle varies as a function of distance travelled on the circumference of the circle as E=as^2, find the force on particle as a function of distance travelled on the circumference of the circle . " Shubham Dhull · 5 months, 3 weeks ago

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@Shubham Dhull Its quite easy.

Firstly 1/2mv^2 = 2as^2/m

As velocity vector is along the tangent to circle . differentiate this wrt s

you get

vdv/ds = 2as/m

which is nothing but a in tangential direction.

For normal accleration simply v^2/R

Now force is resultant of Normal force and tangential force :) Prakhar Bindal · 5 months, 3 weeks ago

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@Prakhar Bindal Bro! I guess u mean to say v^2 = 2as^2/m ( correct the typo bro! ) :) Aniket Sanghi · 5 months, 3 weeks ago

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@Prakhar Bindal now u post the next question ! Shubham Dhull · 5 months, 3 weeks ago

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@Shubham Dhull 2as rt s^2+r^2/s^2 I think Kaustubh Miglani · 5 months, 3 weeks ago

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@Kaustubh Miglani you have to show how you did it, it's a rule bro ! Shubham Dhull · 5 months, 3 weeks ago

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@Shubham Dhull How to post a pic? Kaustubh Miglani · 5 months, 3 weeks ago

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