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Mechanics

Anish Sinha's Avatar
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12 Apr 2008 23:42:05 IST
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H.C.Verma - CM, Collission, Momentum
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Lots of doubt yaar. not getting anything. pls solve q nos 19,33,40,42,43,44,45,48,49,50,51,52. and cud u pls xplain whats the relation between force & momentum xcept impulse cos in many questions, momentum is given, friction is given & asked to find the distance moved. how to apply newtons laws of motion in those questions??


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Anish Sinha's Avatar

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13 Apr 2008 00:17:54 IST
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neone??? pls
aNdRoMeDa's Avatar

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13 Apr 2008 00:33:00 IST
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33) m1u1+m2u2=m1v1+m2v2
     
    v(t)=u1+t/dt(v1-v2)       
                         
     v2'= m1u1+m2u2-m1v1')/m2
 
    v1'={m1u1+m2u2-m1(u1+t/dt(v1 -u1) }/m2
thus
u get
 v2'=u2-m1t/m2dt (v1-u1)
aNdRoMeDa's Avatar

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13 Apr 2008 00:42:38 IST
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40)
mass=M
-E/c=Mv
 
initial ke=E
final ke=1/2Mv^2
          =1/2ME^2/c^2M^2*2
 
           =-E^2/2*C^2*M
 
       decreased energy is =E+E^2/2MC^2
aNdRoMeDa's Avatar

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13 Apr 2008 00:50:49 IST
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conservation of momemtum
u get
20*500/1000=10*100/1000+10*v(block)
 
.8=v(bloc)
 
v^2=u^2+2as
 
thus substituttin
v=.8
s=20/100
u egt
a as 1.6
 
thsu
kmg=ma..force equation
 
k=coeficient of fric
 
solvin u egt
k=.16
aNdRoMeDa's Avatar

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13 Apr 2008 00:59:39 IST
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e=vel of sepration/vel of apparch
thus v=ue
 
Sx=vxt
t=Sx/vcos@
 
S=ut
t u knw
thsu
u get
S=R2=2usin@cos@e/g
 
thus
R1+R2=u^@sin2@(e+1)/g
 
sorry if ne mistaek
me in deep slumber
Anish Sinha's Avatar

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13 Apr 2008 01:06:39 IST
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hey pls xpalin the last one. i didnt get it. u hv never used e but hv put it in the answer.
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13 Apr 2008 01:36:40 IST
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Q.19)

mass of the man = M & his initial vel =0

mass of the bag = m

Let the bag be thrown towards left with a vel v so that the vel of man towards right might be V.

Then mv = MV =>  V = mv / M ------------------(1)

Let the total time taken to reach the ground be t1 . Now H = building height. So,

H = ( 1/ 2) g t1 2   =>    t1[ ]  ( 2H / g ).

Now h = height from the ground at which he throws the bag. Means when he fall a distance ( H - h ) from top , he throws the bag .

Time he will take to fall this distance ( H - h ) is t2 (say) , where

t2 = [ ]  [ 2 ( H - h ) / g ]

now

    t1 - t2 = [ ]  ( 2H / g )  - [ ]  [ 2 ( H - h ) / g ]

                 = [ ]  ( 2 / g ) *  [ [ ]  H -  [ ] ( H -h ) ]

we call this t which is the time in which the man reaches the pond.

now x = horizontal distance covered.

vel of man towards right =  V.

so x = V * t

V = x / t  and

 v = MV / m =( M / m ) *  ( x / t )

         = ( M / m )* ( x [ ] g ) / [ [ ]  H -  [ ] ( H -h ) ] [ ] 2

so vel of bag to be imparted is

v = Mx[ ] g  /  m [ [ ]  2H -  [ ] 2( H -h ) ]

As there is no external force in the horizontal dirn, the X co-ordinate of COM will remain in the same posn. So,

0 = [ Mx + m ( -x1 ) ] / ( M + m )

or,  x1 = Mx / m  which is the horizontal dist. covered by bag.

So, the bag will reach the bottom at a distance  Mx / m  towards left of the line it falls.

cheers !!!



Anish Sinha's Avatar

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13 Apr 2008 01:44:57 IST
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Gr8. fantastic!!! I worked on this for 2 hrs but didnt get it. thanx a lot. it was bothering me too much.

New kid on the Block

Joined: 28 Oct 2007
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13 Apr 2008 16:43:44 IST
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42) Mass of bullet, m = .02kg
     initial vel. of bullet = V1 = 500m/sec
     Mass of block = M= 10 kg
     initial vel. of block = U2 = 0
     final vel. of bullet = v = 100m/sec
     Let final vel. of block when the bullet emerges out, if block vel is V , then
   using Conservation of energy:
 mV1+ MU2 = Mv + mV
thus, V = 0.8m/sec
 
After moving distance 0.02m it stops.
Change in KE = Work Done
0 - (1/2)*10*(0.8)^2 = M*10*10*0.2
M = 0.16kg
Anish Sinha's Avatar

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14 Apr 2008 01:12:14 IST
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Help me out of this
ashish  gupta's Avatar

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14 Apr 2008 01:14:32 IST
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cant u just do it all 'tomorrow'

many important discussions are taking place here regarding jee008
why dont u go to sleep duuuuuuuuuuuuuuuude....
Anish Sinha's Avatar

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14 Apr 2008 01:21:19 IST
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Am i stopping u to discuss or what??? if that discussion is important to u, this is important for me. if i'm not stopping u to discuss, why r u bothering me?? if u dont wanna solve these, its all right for me but dont tell me what to do
ashish  gupta's Avatar

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14 Apr 2008 01:27:29 IST
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sorry
ashish  gupta's Avatar

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14 Apr 2008 01:28:21 IST
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i didnt meant this

contineu ur study man

i 'll see who interrupts
Ratish's Avatar

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14 Apr 2008 01:29:12 IST
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i think budokai's comp has been hacked..seriously..
ashish  gupta's Avatar

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14 Apr 2008 01:40:06 IST
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no
not until



i'll  email you first if it gets hacked
Anish Sinha's Avatar

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15 Apr 2008 23:08:58 IST
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can nebody solve these???
Anish Sinha's Avatar

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16 Apr 2008 00:16:26 IST
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plsssss plsssss......neone!!! i beg neone. i'm sorry if i hv done anything wrong!!!! itni badi saza mat do yaar. aap logon ko pata nahi mujhe is chapter me kitna doubt hai. abhi maine jitne q diye hai, woh half se bhi kam hai. pls solve these so that i can understand something abt this chapter. plssssssss



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