Mechanics - JEE QUESTIONS AND DOUBTS IN KINEMATICS-2

341

Hari Shankar ·2010-10-03 03:53:31

Choosing the conventional coordinate system, we have
v_{1x}(t) = v_1, v_{1y} = -gt which gives \tan \theta_1(t) = -\frac{gt}{v_1}

Similarly, \tan \theta_2(t) = \frac{gt}{v_2}

Since the vectors are perpendicular, we get \tan \theta_1(t)\tan \theta_2(t) = -1 \Rightarrow t = \frac{\sqrt{v_1v_2}}{g}

The vertical displacement is same for the particles and hence the distance between them is the same as the horizontal distance between them which is obviously (v_1+v_2)t = (v_1+v_2)\frac{\sqrt{v_1v_2}}{g}

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scintillating dev ·2010-10-05 04:59:17

yeah u r right Qwerty. thank you.

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21

Shubhodip ·2010-10-05 01:46:05

1st one is very easy...

so (v1 i + gt j) (v2 i + gt j) = 0 ( as 2 vectors r perpendicular)

so t =âˆšv1 v2 /g
distance = âˆšv1 v2 /g (v1 + v2)

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23

qwerty ·2010-10-05 00:13:59

are ha , dont confuse x with horizontal coordinate, i have taken it as vertical coordinate only

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23

qwerty ·2010-10-05 00:09:57

i didnt get wich is the surface?

any way

while going up , net force = - (mg+cu²) ( j )

and while going down net force = (cu² - mg ) ( j )

now while going up ,

accⁿ a = mg+ cu²m = - dudt= - ududx

so

mudumg+cu² = -dx

i.e mclog(mg+cu²)= -x

where u changes from u_o to u , and x changes from 0 to x

(x measured from lowest point )

so, - cx = mlog(mg+cu²mg+cu_o²)

i m assuming that the surface is the one frm which the particle is thrown

while going down

accⁿ a = mg-cu²m = +dudt= + ududx

mudumg-cu² = dx

i.e
- mclog(mg-cu²) = x

where u changes from 0 to some u

and x changes from 0 to some x (measured from highest point )

so - mclog(mg-cu²mg) = x
so we got 2 eqns

- x = mclog(mg+cu²mg+cu_o²) ........(1)

- mclog(mg-cu²mg) = x ...........(2)

in eqn 1 , put u = 0 , and x =x_max

in eqn 2 , u is the u required , and put x = x_max

and equate x_max in the 2 eqns

so mgmg+cu_o² = mg-cu²mg

m²g² = m²g² +cmg(u_o²-u²) - c²u_o²u²

so mg(u_o²-u²)=cu_o²u², i.e mgu_o²=u²(cu_o²+mg)

u² = mgu_o²cu_o²+mg

wats the ans given ?

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AKHIL ·2010-10-04 00:54:11

ya....

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scintillating dev ·2010-10-03 20:48:32

is akhil right?

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scintillating dev ·2010-10-03 09:45:25

u sure???

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AKHIL ·2010-10-03 09:43:30

while going up in direcn of W
and while coming down opp to it...

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scintillating dev ·2010-10-03 08:16:33

THANX PROPHET SIR. PLEASE TRY THE NEXT ONE TOO

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scintillating dev ·2010-10-02 23:36:29

Some1 try these out, yaar, pleaaaase.

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scintillating dev ·2010-10-03 02:13:39

Some1 try this out...

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scintillating dev ·2010-10-03 01:26:15

air drag in the direction of W or opposite??

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AKHIL ·2010-10-03 01:24:14

for 2nd draw the FBD
write the corresponding eqns
only 2 forces r acting
W and air drag
find the acc
and hence the vel..

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scintillating dev ·2010-10-03 01:19:56

anyway, can u solve the second one??

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scintillating dev ·2010-10-03 01:18:35

i found it in the FIITJEE package( of one of my friends).

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Euclid ·2010-10-03 01:07:56

whats the source of ur quest?

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scintillating dev ·2010-10-03 01:07:00

Yeah, i see. it's dimensionally incorrect. But this is what the ans is given

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Euclid ·2010-10-03 00:56:09

for 1st the ans that u hav posted is dimensionally not correct.... plz check out

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