Solve this...

A cubic dice is sliding on a frictionless table. The dice is a cube with edges of length 1 cm and a mass of 30 g. A kid reaches down and gives a horizontal flick to the dice, causing it to change which face is up. What is the minimum impulse in g~cm/s the kid must give to the dice in order to change the face?

Details and assumptions The acceleration of gravity is . You can model the dice as a perfect cube of uniform density. The dice never leave the table.

please help out with this!!

#Mechanics

Easy Math Editor

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Comments

@Deepanshu Gupta, @Mvs Saketh ,@Ronak Agarwal @Nathanael Case @Shashwat Shukla ,please help us!!!

satvik pandey - 6 years, 3 months ago

I can't see why a horizontal flick will result in toppling of block

Ronak Agarwal - 6 years, 3 months ago

This problem has already been posted on this website as a problem by Sir David Mattingly.

You can check out the problem here.

This is one of the most beautiful and challenging questions that I have ever solved and I don't think it would be fair to discuss it here. A full solution has been posted if you want to take a look.

Cheers.

@satvik pandey @manish bhargao

Shashwat Shukla - 6 years, 3 months ago

Thanks for mentioning the link...

manish bhargao - 6 years, 2 months ago

Thank you Shashwat for mentioning that. :)

satvik pandey - 6 years, 2 months ago

You're welcome :)

Shashwat Shukla - 6 years, 2 months ago

@Shashwat Shukla – As my class 10 boards are over so I am looking forward to prepare for KVPY. As you are a KVPY scholar, could you please tell me from where( book if any) did you prepare for it and how?

Congratulation for reaching the milestone of 100 followers!!! :D

@Ronak Agarwal --- I would like to hear your views too.

satvik pandey - 6 years, 2 months ago

@Satvik Pandey – I'm sure that you must have heard this a hundred times, but I will say it again: the NCERT textbook is very important! I say this because every single chemistry question in our paper came from NCERT and the ones that I couldn't answer were from the 12th standard NCERT textbook(which I hadn't studied properly)...I wouldn't recommend studying the entire 12th textbook just for KVPY because chapters like biomolecules and polymers are a waste...I used Resonance material for KVPY and if you can get that, then you can study 12th portions from there as they have some good questions too.

Every year, math is the hardest and physics always has some beautiful questions which are hard to get in just 2-3 minutes...So do study bio as it's just free marks if you remember tenth grade biology.

Best of luck :)

And I saw that you had 99 followers. I just made it a hundred :D ...Cheers to both of us!

Shashwat Shukla - 6 years, 2 months ago

@Shashwat Shukla – Thanks for guiding me and making my followers list 100. :D :)

satvik pandey - 6 years, 2 months ago

@Satvik Pandey – Could you help me in a question ?

Two small balls of equal mass are joined by a light rigid rod. If they are released from rest in the position shown and slide on the smooth track of diameter 0.5 m in the vertical plane. The speed of balls when A reaches B's position and B is at B' is :

Rajdeep Dhingra - 6 years, 2 months ago

@Rajdeep Dhingra – Sure. :)

First see which forces are acting on the system. There would be a normal force on A and on B. And it is well known that work done by these forces would be zero. Because force acting and displacement are perpendicular to each other. Also remember that work done is the dot product of the force acting and displacement of the point of application of force.

So we can easily use conservation of energy. Consider the zero potential level to be at the base of the wedge. Now initial potential energy of the system is equal to the sum of potential energies of ball A and ball B. Now the potential energy of ball B is zero and the potential energy of A is $0.5mg$ . And final energy of the system would be $0.5 mv^{2}+0.5mv^{2}$ (both the balls would have same velocities)

So $0.5mg =0.5 mv^{2}+0.5mv^{2}$

Solving this we get $v=\sqrt(0.5g)$ . If I am not wrong then this should be the answer. Is it correct?

satvik pandey - 6 years, 2 months ago

@Satvik Pandey – Yes thanks very much.

Rajdeep Dhingra - 6 years, 2 months ago

@Rajdeep Dhingra – You are welcome.

The situation would be more complicated if you were asked to find the velocity at any of A and B at any position when A is in contact withe the circular part of the wedge.

Like in this situation

picture.

Here the velocity of both the balls would not be same. Here you will need a constraint relation between $v$ and $v_{0}$

satvik pandey - 6 years, 2 months ago

@Satvik Pandey – I will try to solve this. Imgur Imgur.
BTW Where did you make the image (Software)

Rajdeep Dhingra - 6 years, 2 months ago

@Rajdeep Dhingra – I made it using paint.

satvik pandey - 6 years, 2 months ago

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}`	$\boxed{123}$