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**!

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## Comments

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TopNewestProblem 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

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Solution to problem 5 :

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

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

\( dm g cos\theta \) \( ( dm = \frac{MRd\theta}{L}\) ( 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

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

Calculations I left for you !

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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.

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Now, note that tension is an internal force and hence, \(\int dT\) over the chain length vanishes.

We end up with the same answer finally. ;)

I know this is a very small point, but just thought that it's worth mentioning.

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And should I delete my solution and you will post yours?

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

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Correct! . Post next question and solution

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Its Not right to say acceleration of the chain As a whole as every element will be having an acceleration of different direction but their magnitude will be the same that i am asking to calculate

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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.

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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

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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!

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

Hey helloAbout 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

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@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!

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@Mayank Singh @Deeparaj Bhat @Abhineet Nayyar

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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?

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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?

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Let the particles collide at point \(P\), whose position vector is \({ r }_{ 3 }\). Let the time taken be \(t\). By the vector laws of addition, we get

\({ r }_{ 3 }={ { r }_{ 1 } }+{ v }_{ 1 }t\\ { r }_{ 3 }={ r }_{ 2 }+{ v }_{ 2 }t\)

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

\(\large t\)= \(\large\frac{({r_1} - {r_2})}{({v_2}-{v_1})}\)

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

By intuition, we can tell that the particles \(1\) and \(2\) 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.Log in to reply

Nice solution (+1) :).

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

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Nah.Anyone can participate.

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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)

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Problem 10A meteorite of mass m collides with a satellite which was oribiting around a planet of mass \(M\) 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 distancefrom the planet is \(\frac{R}{2}\). 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)Log in to reply

Its a simple energy , momentum and angular momentum conservation problem

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

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Post the next problem!

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

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

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Is the answer sqrt(288GM/R)?

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no

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Problem 8:A cylinder of mass \(m\) and radius \(R\) rest on two supports in the same height .One support is stationery while the other slides from the cylinder at velocity \(v\). Determine the force of normal pressure \(N\) by cylinder at stationery support when \(x=R\sqrt{2}\).

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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.

\(\Rightarrow v_c = \dfrac{v}{\sqrt{2}} \)

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

\(mg \cos {45^{\circ}} - N = \dfrac{mv_c ^2}{R} \)

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

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go and post the next q

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Solution to Problem 8:The answer is \(n=\frac{mg}{\sqrt{2}}-\frac{mv^2}{2R}\). 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 \(\frac{v}{\sqrt{2}}\). This can be found out by equating components.

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

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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.

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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)

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

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Problem 4:The trajectory for motion of a particle in \(x-y\) plane can be expressed as given below,

\[e^x+e^y=e^x e^y, x,y \in \mathbb{R}\]

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

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

P. S. : Original problem.

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Given equation:

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

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Can you please explain third line of you solution ?

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Correct! post the next question!

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Yeah! Write the solution to it too.

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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 \(x\) axis varies as \(v=\alpha\sqrt{x}\), where \(\alpha\) is a positive constant. Assuming that at the moment \(t=0\) the particle was located at the point \(x=0\), find the time dependence of the velocity and acceleration of the particle.Log in to reply

\(v=\alpha^2t/2\) and \(a=\alpha^2/2\).

We know,

\(v=\frac{dx}{dt}=\alpha x^{1/2}\)

\(x^{-1/2}dx=\alpha dt\)

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

\(2x^{1/2}=\alpha t\)

\(x=\alpha^2 t^2/4\)

By differentiation,

\(v=\alpha^2 t/2\)

\(a=\alpha^2/2\)

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Correct.You perhaps can post the next question.

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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

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

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

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

visit this page for image

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ans is 5

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Is this dead? Shall I participate

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

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

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Is the contest dead?

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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!

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

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Problem 9A metal rod is of length \(1m\) is clamped at points \(5cm \) and \(15 cm \) from either ends. Find the minimum frequency of natural longitudinal oscillations of rod.

Given :

Young's modulus of rod \( = 1.6 \times 10^{11} Pa\)

Mass per unit volume \( \rho = 2500 kgm^{-3} \)

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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

firstnode 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 \(40,000 Hz\)Log in to reply

Awesome! We got a new problem poster!

Post the next problem please!

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correct!

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Comment deleted Jan 05, 2017

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no

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pic

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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.

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

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Problem 6 : [Original]

Take a bomb of mass m and give it a velocity \( v = 10m/sec \) at an angle \( \theta = 53° \)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 \( 6 m/sec \) 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 \( e = 0.4 \)

The ground is smooth.

Take \( g = 10m/s^2 \)

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

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

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@aryan goyat for posting this!

Thanx!Log in to reply

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You may post the solution and the next problem.

@Aniket Sanghi

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Comment deleted May 19, 2016

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Got at same 14.78379 m .

@aryan goyat @Aniket Sanghi

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Nope , my answer is different

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Comment deleted May 20, 2016

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Nope

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@Prakhar Bindal we really need your help!

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

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

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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

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Thanks for inviting :)

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

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