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Mechanics
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
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Joined: 10 Feb 2008
Posts: 57
13 Apr 2008 00:17:54 IST
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neone??? pls
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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
=-E^2/2*C^2*M
decreased energy is =E+E^2/2MC^2
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

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 !!!
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 !!!
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
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
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





..seriously..







