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An object of mass 5kg is projected with a velocity of 20m/s at an angle of 60o to the horizontal.At the highest point of its path the projectile explodes and breaks up into two fragments of masses 1kg and 4kg. The fragments seperate horizontally after the exposion.The explosion releases internal energy such that the kinetic energy of the system at the highest point is doubled. calculate the seperation between the two fragments when they reach the ground(g=10m/s2).???
initial velocity = 20m/s
@ of projection = 60o
velocity at highest point = ucos@ = u/2 = 10m/s
v= 10
since the system explodes due to internal forces do momentam will remain conserved ...
Pi = Pf ..........1
Pi = Mv = 5(10) = 50 ............2
finally after explosion both particle are moving in horizontally , let v1 , v2 are the velocity
of particles of mass 1kg & 4kg respectively ...
final momentam (Pf) = v1 + 4v2 ...............3
momentam is conserved so from eq 1
v1 + 4v2 = 50 ...........4
it is given that after explosion KE of particle is doubled
(KE)f = 2(KE)i ( ke before explosion)
(KE)i = Mv2/2 = 5(10)2/2 = 250J so
(KE)f = 500 j =(v12 + 4v22)/2 ...............5
solving 4 & 5 we get v
v1 ,v2 = (-10,15)
now time taken by particle to reach the highest point = T/2 = usin@/g = time taken by particle to reach to ground
distance covered by both particles = d = (speed)(time)
taking g = 10m/s2
d1 = v1t = 10(usin@)/g = usin@ (distance covered by particle 1)
d2 = v2t = 15(usin@)/g = 1.5usin@ (distance covered vy particle 2)
total distance bw particle when they reach the ground = d1+d2 = 2.5usin@
d = 25root3 meters
=43.25m
thnku
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