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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,028

The force on the charged particle due to electric and magnetic fields is given by

.Suppose E is along the x-axis and B along the y-axis,In what direction and with what minimum speed V should a positively charged particle be sent so that the net force on it is zero?

P.S. the capital letters in the equation have vector sign over them.

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What does the equation even mean?

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,528

If a force has two components, acting at right angles to each other, and you want the resultant to be zero then each component must be zero.

Perhaps the question needs re-phrasing?

[Do you mean V to be capitalised ie another vector ? Perhaps there should be a vector (cross) product sign in there too ?]

LATER EDIT: Silly me; of course you want this to be a vector product. It's somebody's Right or is it Left hand rule. Bit rusty on the details. The x isn't showing in my browser.

So you want

As

i, j, and k are unit vectors along the three axes. In your equation the k term comes first and order matters for the cross product so the minus will automatically come from

so you need

That deals with the speed; and the direction is the positive z axis.

Bob

ps. Don't know why it says 'minimum' speed. Only one speed will work.

*Last edited by bob bundy (2014-04-29 20:55:36)*

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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,028

got it now.

thanks

........................

I had one more question related to carrom board.

just coming back and posting it.

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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,028

A carrom board (4 ft × 4 ft square) has the queen at the centre .The queen, hit by the striker moves to the front edge, rebounds and goes in the hole behind the striking line. Find the magnitude of displacement of the queen

a) from the centre of the front edge

b) from the front edge to the hole.

I am not able to figure out the movement of striker on the board.

pls. help.

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That is not a physics question. To answer it, you need to know what line is the striking line which I do not know.

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,528

hi Niharika,

I've not heard of this game before so I've looked it up. In the rules I read, the holes are at the four corners. So I think my diagram shows it correctly. Q travels to point B, bounces back and goes in the corner hole at C. Now there will be friction as the queen slides and loss of energy at B but there is no information that would allow you to take account of these factors so I've had to assume that the slide is frictionless and the 'bounce' is perfectly elastic meaning the angle of incidence equals the angle of reflection.

On that basis you can imagine point A, the reflection of Q in the wall. A straight line from A through B to C represents the displacement which you can calculate using Pythagoras. The squares in my diagram are 6 inches.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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Seriously bob, did you never hear of it?

Another way to find the distance is to take a large printout of your diagram

'And fun? If maths is fun, then getting a tooth extraction is fun. A viral infection is fun. Rabies shots are fun.'

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,528

Seriously bob, did you never hear of it?

According to Wiki it is more popular in your part of the world.

Another way to find the distance is to take a large printout of your diagram

Tut tut. In most of my posts I try to encourage the OP to do some of the work.

I suppose you could also do it experimentally.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

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Well, printing it is some considerable amount of work.

They say carrom is very addictive. Even more than the internet. Unfortunately, I am a very poor player.

'God exists because Mathematics is consistent, and the devil exists because we cannot prove it'

'You have made another human being happy. There is no greater accomplishment.' -bobbym

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,528

Well, printing it is some considerable amount of work.

I see we have some confusion about what I meant by 'work'.

I am a teacher, so you must expect that I don't want my students to dodge doing the problem for themselves. So I try to give just enough help so that the student can finish the problem, making a contribution to the effort of getting the answer. Just shoveling some coal would involve work; but it would not improve the student's ability with the maths.

That's why some of my posts are brief. It is a compliment to the poster that I think their ability is sufficient to fill in the gaps.

Bob

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**ShivamS****Member**- Registered: 2011-02-07
- Posts: 3,629

niharika_kumar wrote:

The force on the charged particle due to electric and magnetic fields is given by

.

Suppose E is along the x-axis and B along the y-axis,In what direction and with what minimum speed V should a positively charged particle be sent so that the net force on it is zero?P.S. the capital letters in the equation have vector sign over them.

The velocity is the magnitudes of E/B in the direction of the z-axis.

niharika_kumar wrote:

A carrom board (4 ft × 4 ft square) has the queen at the centre .The queen, hit by the striker moves to the front edge, rebounds and goes in the hole behind the striking line. Find the magnitude of displacement of the queen

a) from the centre of the front edge

b) from the front edge to the hole.I am not able to figure out the movement of striker on the board.

pls. help.

a) 2 feet

b) sqrt(4^2 + 2^2) = sqrt(20) = 4.47 feet

Are these, by any chance, from HC Verma?

*Last edited by ShivamS (2014-05-01 08:12:26)*

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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,028

ShivamS wrote:

a) 2 feet

b) sqrt(4^2 + 2^2) = sqrt(20) = 4.47 feetAre these, by any chance, from HC Verma?

yes, these are from H C Verma, but the answers given are

a)

b)

bob,

I think I need some more help with it.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,528

hi Niharika,

Use Pythagoras to get AC.

B is one third along AC so

Bob

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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,028

oh, yes.

I didn't notice that we could take the whole AC and calculate.

thanks again

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**gourish****Member**- Registered: 2013-05-28
- Posts: 143

hey which chapter is it form... i.e. from HC VERMA?

"The man was just too bored so he invented maths for fun"

-some wise guy

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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,028

2nd chapter of class 11.

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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,028

here I have one more question.

Q.If a particle of mass m is moving with constant velocity v parallel to x-axis in x-y plane , calculate its angular momentum with respect to origin at any time t.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,528

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**niharika_kumar****Member**- From: Numeraland
- Registered: 2013-02-12
- Posts: 1,028

thanks.

I figured it out.

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