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#2 Re: Help Me ! » Find the value » 2014-11-15 15:48:06

anyhow, manually.

Did you appear in KVPY @agnishom.

#5 Re: Science HQ » collision » 2014-10-18 22:42:27

bob bundy wrote:

Now for the horizontal motion.  At impact with the block, the velocity is still V and the block is moving at -V/4.  We aren't told the mass of the block so no momentum equation is possible.  But we can use the elasticity equation.  We'll have to assume e=1  again as we aren't told anything else.  So

You'll need to be careful with negatives here, but you should be able to get the velocity of the ball after the impact (assume block is still -v/4).


I am not sure if I could understand the way of using that equation properly.

Will you pls show how you placed all the terms in that.

#6 Science HQ » collision » 2014-10-18 03:10:40

Replies: 5

The block A begin to move with a velocity v/4.At the same time a ball is thrown from the wall of height h with a velocity v which traces the path  as shown before coming back to the original position.Find x.


#8 Re: Help Me ! » coordinate » 2014-10-03 16:23:26

here is one more:-

Show that all the chords of the curve

which subtend a right angle at the origin pass through a fixed point.Also find that point.

#9 Re: Help Me ! » coordinate » 2014-10-02 22:34:55

my thanks to him too. smile

Even I was unable to do that and hence my 2nd equation was not in latex form.

#10 Re: Help Me ! » coordinate » 2014-10-02 16:16:34

thanks bob. smile

I didn't consider  fact that y-mx is the factor of both the equations, but yes you used it, so
if we have that (y-mx) is the factor of both the given equations,
why don't we put y=mx in both of them, then eliminate the term x^2 and solve by cross multiplication.

that gives m^2 =(hA-Ha)/(bH-hB)={(aB-Ab)/2(bH-Bh)}^2

That has given a clear solution without m,n and p and the steps are less complicated.

#11 Re: Help Me ! » coordinate » 2014-10-01 19:11:03

Here is another question:-

Obtain the condition that one of the straight lines given by the equation

may coincide with one of those given by the equation

#14 Re: Science HQ » find the max. height. » 2014-09-27 16:12:04

h is the height to which the ball is raised up from its hanging position.

#16 Science HQ » find the max. height. » 2014-09-26 20:15:34

Replies: 4

In the given figure the ball of mass m is held above as shown.It is then allowed to swing freely.Find the maximum height attained by the block.


#17 Re: Science HQ » projectile » 2014-09-18 16:49:42

Yes you are.
Your explainations have always proved to be very beneficial. smile

#18 Re: Science HQ » projectile » 2014-09-18 00:30:46

hmm, so it was simple.
thanks again bob. smile

#19 Re: Science HQ » projectile » 2014-09-17 18:55:19

It's over now.

and I couldn't make this one:-
Q.A ball rolls off the top of a stairway with a horizontal velocity u.If the steps are h meter high and b meter wide, show that the ball will just hit the edge of nth step if


#20 Re: Science HQ » projectile » 2014-09-17 04:01:35

thank you so much for such  simplified answers.
It is really very helpful(because my brain is not working right now and I have my Physics exam tomorrow). smile

#21 Science HQ » projectile » 2014-09-16 19:36:58

Replies: 14

A particle is projected over a triangle from one end of a horizontal base and grazing the vertex falls on the other end of the base.If

be the base angle and
the angle of projection , prove that

#24 Re: Puzzles and Games » replace the question mark » 2014-09-11 01:29:43

no I couldn't solve it and so put it up here.

#25 Re: Help Me ! » coordinate » 2014-09-11 01:22:19

are you sure x=a and not -a.
I am getting x=-a.

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