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**.

## Comments

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TopNewestHey 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 · 3 months, 1 week ago

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– Shubham Dhull · 3 months, 1 week ago

sure i will bro. i'll try your problem too as soon as i get out of this season-time(diwali-house cleaning) haha !!Log in to reply

By good I mean a problem that is not a standard 'DC-Pandey/coaching material' problem. :) – Harsh Shrivastava · 3 months, 1 week ago

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its my original one !l>h and rod is massless! – Shubham Dhull · 3 months, 1 week agoLog in to reply

@Harsh Shrivastava @Aniket Sanghi @Prakhar Bindal – Shubham Dhull · 2 months, 3 weeks ago

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@Prakhar Bindal – Shubham Dhull · 3 months ago

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– Shubham Dhull · 3 months, 1 week ago

i know- i know what u mean !Log in to reply

Problem 6 SET IA 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 · 3 months, 1 week ago

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– Aniket Sanghi · 3 months, 1 week ago

Correct??Log in to reply

Upper bob's after collision velocity \( v2 = \sqrt{13}/5 \) – Aniket Sanghi · 3 months, 1 week ago

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– Shubham Dhull · 3 months, 1 week ago

bro, solution? rule hereLog in to reply

– Aniket Sanghi · 3 months, 1 week ago

You only told , no rules from now , and only we 3 are there , is their a need for solution??Log in to reply

Anybody want to post a question? – Harsh Shrivastava · 3 months, 1 week ago

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– Aniket Sanghi · 3 months, 1 week ago

Somebody solvey optics q , it had got more than 180 views , but 0 solvers ! :PLog in to reply

– Prakhar Bindal · 3 months, 1 week ago

i will try after phase test :P and after net also gets unlimitedLog in to reply

– Harsh Shrivastava · 3 months, 1 week ago

I can't solve it, havent studied it yet.Log in to reply

– Shubham Dhull · 3 months, 1 week ago

i'll tryLog in to reply

– Shubham Dhull · 3 months, 1 week ago

i'll postLog in to reply

– Harsh Shrivastava · 3 months, 1 week ago

COOL!!!Log in to reply

– Shubham Dhull · 3 months, 1 week ago

ok, no i dob't need the sol. , i can do it thnxLog in to reply

– Shubham Dhull · 3 months, 1 week ago

yes tell wheter elastic or notLog in to reply

– Aniket Sanghi · 3 months, 1 week ago

Is e given??Log in to reply

– Harsh Shrivastava · 3 months, 1 week ago

Sorry, forgot to mention that e=1:PLog in to reply

@Aniket Sanghi @Prakhar Bindal @shubham dhull – Harsh Shrivastava · 3 months, 1 week ago

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Prakhar also won't be online for 2 weeks . Furthermore this competition lacks participation , it is not arousing interest in me ... – Aniket Sanghi · 3 months, 1 week ago

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@Aniket Sanghi – Harsh Shrivastava · 3 months, 1 week ago

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By that time you can check my Brilliant questions ( I posted) if you want :) – Aniket Sanghi · 3 months, 1 week ago

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– Harsh Shrivastava · 3 months, 1 week ago

K, we have just started rotational mechanics in Fiitjee, so after that, I'll become eligible to try mechanics questions on brilliant!Log in to reply

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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 · 3 months, 1 week ago

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– Aniket Sanghi · 3 months, 1 week ago

Its OK bro! :)Log in to reply

– Harsh Shrivastava · 3 months, 1 week ago

OK, what answer are you getting?Log in to reply

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 · 3 months, 1 week ago

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Problem 4.

– Aniket Sanghi · 3 months, 1 week agoSET ILog in to reply

@Aniket Sanghi is the answer 8 ? – Shubham Dhull · 3 months, 1 week ago

heyLog in to reply

– Shubham Dhull · 3 months, 1 week ago

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 !Log in to reply

– Aniket Sanghi · 3 months, 1 week ago

Answer is correct ! 👍 Post the next!Log in to reply

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– Harsh Shrivastava · 3 months, 1 week ago

Am I correct ?Log in to reply

– Shubham Dhull · 3 months, 1 week ago

YES YOU'RE RIGHT ! also i liked it that u uploaded the question without uch delay , good work !Log in to reply

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@Prakhar Bindal @Harsh Shrivastava @Aniket Sanghi – Shubham Dhull · 3 months, 1 week ago

Don't let the contest go in a cold hault guys buck up !Log in to reply

@Prakhar Bindal @Harsh Shrivastava @shubham dhull try it guys! – Aniket Sanghi · 3 months, 1 week ago

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Why is the contest stubbed? – Harsh Shrivastava · 3 months, 2 weeks ago

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– Harsh Shrivastava · 3 months, 2 weeks ago

OK let's wait for 3 hrs.Log in to reply

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(Don't reply as it will add one more comment.) – Harsh Shrivastava · 3 months, 2 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 · 3 months, 2 weeks ago

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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 · 3 months, 2 weeks ago

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Post the next question. – Harsh Shrivastava · 3 months, 2 weeks ago

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@Harsh Shrivastava . if possible please attach a diagram – Prakhar Bindal · 3 months, 2 weeks ago

hey bro whats the situation can you explain or youLog in to reply

– Harsh Shrivastava · 3 months, 2 weeks ago

A rope is kept surroundimg a sphere of radius R, the circumference of rope is less than 2piR.Log in to reply

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 · 3 months, 2 weeks ago

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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 · 3 months, 2 weeks ago

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– Shubham Dhull · 3 months, 2 weeks ago

hey, post the next question !Log in to reply

Did you had your first AIITS? . What was ur rank? – Prakhar Bindal · 3 months, 2 weeks ago

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– Harsh Shrivastava · 3 months, 2 weeks ago

Very bad, 126 :(Log in to reply

– Shubham Dhull · 3 months, 2 weeks ago

plz move such discussions to slack or other notes and delete ur comment here ! it's only for asking questions not talking !!Log in to reply

– Prakhar Bindal · 3 months, 2 weeks ago

Ohk sorryLog in to reply

@Aniket Sanghi @shubham dhull @Archit Agrawal @Prince Loomba @Harsh Shrivastava @Samarth Agarwal @neelesh vij – Prakhar Bindal · 3 months, 2 weeks ago

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Some more Invitations

@Gauri shankar Mishra @Archit Agrawal @aryan goyat @Samarth Agarwal – Aniket Sanghi · 3 months, 2 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 · 3 months, 2 weeks ago

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– Shubham Dhull · 3 months, 2 weeks ago

no, not at all broLog in to reply

– Shubham Dhull · 3 months, 2 weeks ago

but bro, set 1 is also for advance not mains !Log in to reply

– Aniket Sanghi · 3 months, 2 weeks ago

What I mean to say , u will understand by solving this q !! :)Log in to reply

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– Harsh Shrivastava · 3 months, 2 weeks ago

Post the next problem.Log in to reply

– Aniket Sanghi · 3 months, 2 weeks ago

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 :)Log in to reply

My Problem !

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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 · 3 months, 2 weeks ago

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– Aniket Sanghi · 3 months, 2 weeks ago

Yaar maine alag alag line main likha tha , sab Ek saath Ek line main aa gaya! :|Log in to reply

– Aniket Sanghi · 3 months, 2 weeks ago

Answer I/2m correct? ( I have to go to school , I am in hurry)Log in to reply

@Aniket Sanghi – Shubham Dhull · 3 months, 2 weeks ago

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 !Log in to reply

– Aniket Sanghi · 3 months, 2 weeks ago

Correction : In the first set * we can go on with easy as well as hard qLog in to reply

– Aniket Sanghi · 3 months, 2 weeks ago

A slight push *Log in to reply

Time provided here is 72 hours after the question is posted. Rest , the same rules go here! – Aniket Sanghi · 3 months, 2 weeks ago

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@Swapnil Das @Chirayu Bhardwaj @neelesh vij @Sambhrant Sachan @Nishant Rai @Jyotisman Das and all others – Shubham Dhull · 3 months, 2 weeks ago

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@Prakhar Bindal @Aniket Sanghi @Kaustubh Miglani @Swagat Panda – Shubham Dhull · 3 months, 2 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 · 3 months, 2 weeks ago

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Firstly 1/2

mv^2 = 2as^2/mAs 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 · 3 months, 2 weeks ago

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– Aniket Sanghi · 3 months, 2 weeks ago

Bro! I guess u mean to say v^2 = 2as^2/m ( correct the typo bro! ) :)Log in to reply

– Shubham Dhull · 3 months, 2 weeks ago

now u post the next question !Log in to reply

– Kaustubh Miglani · 3 months, 2 weeks ago

2as rt s^2+r^2/s^2 I thinkLog in to reply

– Shubham Dhull · 3 months, 2 weeks ago

you have to show how you did it, it's a rule bro !Log in to reply

– Kaustubh Miglani · 3 months, 2 weeks ago

How to post a pic?Log in to reply