IIT JEE , AIEEE Forum - Physics , Mechanics - explosion of bomb and the distance btw. the particles of bomb

anujmehan

a bomb at a height of 100 m explodes horizontally with a ratio of 3:1 by mass. If the velocity of the lighter particle is 30 m/s what will be the distance between the two bodies when they reach the ground.

i request u to solve and answer this in few hrs. bcauz.tomorrow is my test.

anuj mehan

11 th class surat

anujmehan

pl reply fast

rooney

Momentum is conserved.

m x 30 + 3mv = 0
v = -10
Horizontal dist travelled by m when it hits the ground = time x 30 =(2h/g)^1/2 x 30
Horizontal dist travelled by 3m when it hits the ground = (2h/g)^1/2 x 10

Total distance = 40(2h/g)^1/2 = 80(5)^1/2 m = 178.4m

spideyunlimited

initial momentum = final momentum
0 = mv + MV
-mv = MV
m/M = -V/v
1/3 = - V/30
V = -10m/s

time taken for bodies to reach ground from height of 100m:
S = 1/2 at^2
100 = 1/2 * 10 * t^2
t =sqrt (20) seconds

horizontal distance between these 2 particles when they reach the ground ( that is, after sqrt(20) seconds have passed) :
D = (relative speed of one particle with repect to the other) x (time)
D = (30 - (-10) ) x sqrt(20)
D = 40 x sqrt(20)
D = 40 x 2 sqrt(5)
D = 178.885 m

krishna.gopal

Perfect answers by both of you.Good work

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