IIT JEE , AIEEE Forum - Physics , Mechanics - H.C.Verma

anish292

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??

anish292

neone??? pls

deedee

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)

deedee

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

deedee

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

deedee

e=vel of sepration/vel of apparch

thus v=ue

Sx=vxt

t=Sx/vcos@

=2usin@e/g................v=ue

Sx=R1=u^2sin2@/g

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

anish292

hey pls xpalin the last one. i didnt get it. u hv never used e but hv put it in the answer.

ramyani

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 !!!

anish292

Gr8. fantastic!!! I worked on this for 2 hrs but didnt get it. thanx a lot. it was bothering me too much.

shrey_iit

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

anish292

Help me out of this

budokai_tenkaichi

cant u just do it all 'tomorrow'

many important discussions are taking place here regarding jee008
why dont u go to sleep duuuuuuuuuuuuuuuude....

anish292

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

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