IIT JEE , AIEEE Forum - Physics , Mechanics

maverick83

let A bomb of mass m is thrown vertically in the air ... it explodes into two equal parts when its velocity is 21m/sec. If one part moves upto the height of 40m from the pt. of explosion then what is the position of second part fom pt. of explosion.

netkid07

momentum before explosion and momentum after explosion must be same bcoz no external force is acting on the system...

Intial velocity of first part can be found by:

v^2 - u^2 =2as

where v=0

a=10

s=40

therefore,

u=20root2

applying law of conservation of linear momentum

let x be the velocity of second part

M21=m20root2 + mx

as

M=2m

therefore the equation becomes

42=20root2 + x

x=42 - 20root2=11.72

now,u know the velocity of second part

so position can be found by applying same formula

u get s= -9.76

minus sign here means that other part is going downwards

rate my efforts if you find them useful

maverick83

numericaly your ans is correct but i think that it should be with positive sign...

netkid07

sorry that was a mistake

you are right

the answer is with positive sign

elessar_iitkgp

Good work maverick

joyfrancis

After the explosion , the part of the mass which moves up to a height of 40m needn't go directly upwards. It can also take a parabolic path and reach a maximum height of 40m from the point of explosion. The question is incomplete in itself, it should be mentioned that the part of the bomb goes DIRECTLY upwards.P.S this can also be solved in much less time using conservation of momentum. If the time of explosion is infinetely small then dp becomes almost zero and hence momentum is conserved just befor and after the collision.

maha_hell3

momentum before collision=momentum aftr collisison.

devesh_l2k007

mv= m/2v1 + m/2v2
v1=root(2g40)
on solving get v2
put v2sq=2gh
to get h

maverick83

thanx everbody you all are right i will RATE YOU ALL

elessar_iitkgp

I would like to add a few things
System: The particle
The only external force acting on the system before and after the explosion are the gravitational forces. Now, assuming the time taken for the explosion to be small, the impulse generated by these gravitational forces is negligible. Hence linear momentum can be assumed to be constant in the vertical and horizontal directions.
In the horizontal direction,
0 = (m/2)v_1x + (m/2)v_2x
v_1x = -v_2x-----------(1)
As there is no acceleration along the X axis, the pieces move equal distances from the point of explosion, in equal time intervals along the X axis.

In the vertical direction,
mv₀ = (m/2)v_1y + (m/2)v_2y
2v₀ = v_1y + v_2y -----------(2)

Also,
v_1y²/2g = h₁
v_1y =

2gh₁ ---------(3)

From (2) and (3)
v_2y = 2v₀ -

2gh₁

Height to which the second piece rises
h₂ = v_2y²/2g
h₂ = (2v₀ -

2gh₁)²/2g

This is the height to which the second piece rises, and horizontally, it is at the same distance from the projection point as the first piece. This completely determines the position of the first piece.

Substitute
v₀ = 21 m/s
h₁ = 40 m
to get the numerical answer.

maverick83

Good explanation elessar...........

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