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!!

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## Comments

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TopNewestThis 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

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Thanks for mentioning the link...

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Thank you Shashwat for mentioning that. :)

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You're welcome :)

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

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

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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!

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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 :

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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?

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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 \(v_{0}\)

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Imgur

BTW Where did you make the image (Software)

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

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

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