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.

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2^{34}

2^{34}

a_{i-1}

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\sqrt{2}

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Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

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Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
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Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

Markdown	Appears as
`italics` or `_italics_`	italics
`bold` or `__bold__`	bold
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Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1 paragraph 2	paragraph 1 paragraph 2
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Math	Appears as
Remember to wrap math in `$` ... `$` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$

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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 . "

A Former Brilliant Member - 3 years, 6 months ago

2as rt s^2+r^2/s^2 I think

Kaustubh Miglani - 3 years, 6 months ago

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 - 3 years, 6 months ago

now u post the next question !

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

Aniket Sanghi - 3 years, 6 months ago

@Prakhar Bindal @Aniket Sanghi @Kaustubh Miglani @Swagat Panda

I am taking part in this contest

Kaustubh Miglani - 2 years, 9 months ago

@Swapnil Das @Chirayu Bhardwaj @neelesh vij @Sambhrant Sachan @Nishant Rai @Jyotisman Das and all others

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]

A slight push *

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!

no, not at all bro

but bro, set 1 is also for advance not mains !

@A Former Brilliant Member – What I mean to say , u will understand by solving this q !! :)

@Aniket Sanghi –

@A Former Brilliant Member – 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 – Correction : In the first set * we can go on with easy as well as hard q

@Aniket Sanghi – $Q-2\quad set-\quad II\\$ You asked for it , you got it, 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 "

@A Former Brilliant Member – 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 – Yaar maine alag alag line main likha tha , sab Ek saath Ek line main aa gaya! :|

@A Former Brilliant Member – Answer I/2m correct? ( I have to go to school , I am in hurry)

@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

@A Former Brilliant Member – Post the next problem.

Harsh Shrivastava - 3 years, 6 months ago

Some more Invitations

@Gauri shankar Mishra @Archit Agrawal @aryan goyat @Samarth Agarwal

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

@Aniket Sanghi @shubham dhull @Archit Agrawal @Prince Loomba @Harsh Shrivastava @Samarth Agarwal @neelesh vij

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

Right! . post the next.

Did you had your first AIITS? . What was ur rank?

hey, post the next question !

Momentum equation would be $\vec{v_1}+ \vec{v_2}+ \vec{v_3}=1500 \vec{i}.$

How can you say that the maximum condition would be when all velocities are parallel or antiparallel?

Harsh Poonia - 4 months, 3 weeks ago

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.

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

$\boxed{ T = \frac {mgR}{2\pi \sqrt {R^2 - r^2}}}$

hey bro whats the situation can you explain or you @Harsh Shrivastava . if possible please attach a diagram

@Prakhar Bindal – A rope is kept surroundimg a sphere of radius R, the circumference of rope is less than 2piR.

Correct!

Post the next question.

It should be r instead of R

Akash Yadav - 2 years, 9 months ago

Why is the contest stubbed?

Problem 4. SET I

Aniket Sanghi - 3 years, 5 months ago

@Prakhar Bindal @Harsh Shrivastava @shubham dhull try it guys!

hey @Aniket Sanghi is the answer 8 ?

A Former Brilliant Member - 3 years, 5 months 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 !

@A Former Brilliant Member – Answer is correct ! 👍 Post the next!

@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:

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 - 3 years, 5 months ago

Try it guyz!

@Aniket Sanghi @Prakhar Bindal @shubham dhull

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 – 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.

@Aniket Sanghi – OK, what answer are you getting?

@Aniket Sanghi – Will you post questions?

@Aniket Sanghi

@Harsh Shrivastava – Whenever I will get time :)

By that time you can check my Brilliant questions ( I posted) if you want :)

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

Is e given??

Sorry, forgot to mention that e=1:P

yes tell wheter elastic or not

Lower bob's horizontal velocity $v = 2\sqrt{3}/5$

Upper bob's after collision velocity $v2 = \sqrt{13}/5$

Correct??

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.

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$