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 \times 3
2^{34}
a_{i-1}
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Top NewestHey 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.
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Comment deleted Mar 04, 2017
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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. :)
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@Harsh Shrivastava @Aniket Sanghi @Prakhar Bindal
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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.
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Correct??
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Lower bob's horizontal velocity \( v = 2\sqrt{3}/5 \)
Upper bob's after collision velocity \( v2 = \sqrt{13}/5 \)
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bro, solution? rule here
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Anybody want to post a question?
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yes tell wheter elastic or not
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Is e given??
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Sorry, forgot to mention that e=1:P
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Try it guyz!
@Aniket Sanghi @Prakhar Bindal @shubham dhull
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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 ...
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@Aniket Sanghi
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By that time you can check my Brilliant questions ( I posted) if you want :)
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Comment deleted Oct 13, 2016
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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!
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Comment deleted Oct 22, 2016
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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.
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Problem 4. SET I
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hey @Aniket Sanghi is the answer 8 ?
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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 !
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@Prakhar Bindal @Harsh Shrivastava @shubham dhull try it guys!
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Why is the contest stubbed?
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Comment deleted Oct 10, 2016
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OK let's wait for 3 hrs.
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Comment deleted Oct 10, 2016
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(Don't reply as it will add one more comment.)
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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.
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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
\( \boxed{ T = \frac {mgR}{2\pi \sqrt {R^2 - r^2}}} \)
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It should be r instead of R
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Correct!
Post the next question.
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hey bro whats the situation can you explain or you @Harsh Shrivastava . if possible please attach a diagram
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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
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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
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hey, post the next question !
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Right! . post the next.
Did you had your first AIITS? . What was ur rank?
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@Aniket Sanghi @shubham dhull @Archit Agrawal @Prince Loomba @Harsh Shrivastava @Samarth Agarwal @neelesh vij
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Some more Invitations
@Gauri shankar Mishra @Archit Agrawal @aryan goyat @Samarth Agarwal
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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]
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no, not at all bro
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but bro, set 1 is also for advance not mains !
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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
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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 !Log in to reply
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A slight push *
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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!
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@Swapnil Das @Chirayu Bhardwaj @neelesh vij @Sambhrant Sachan @Nishant Rai @Jyotisman Das and all others
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@Prakhar Bindal @Aniket Sanghi @Kaustubh Miglani @Swagat Panda
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I am taking part in this contest
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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 . "
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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 :)
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Bro! I guess u mean to say v^2 = 2as^2/m ( correct the typo bro! ) :)
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now u post the next question !
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2as rt s^2+r^2/s^2 I think
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you have to show how you did it, it's a rule bro !
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It should be r instead of R
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