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Question

A projectile of mass 4 kg is fired with a velocity 4 m/s at an angle 450 with the horizontal. At the highest position in its flight, it explodes into two fragments of masses 1 kg and 3 kg. The heavier fragment falls vertically with zero initial speed. the horizontal distance between the fragments when they touch the ground is

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Solution

intial velocity u=20m/s
angle of projection=600
velocity at heighest point =ucosθ=u2=10m/s
since the system explodes due to internal forces momentum remains constant
Pi=Pf
Pi=5×10=501
Pf=1×V1+4V22
1=250=V1+4V23
given kinetic energy doubled after explosion
(K.E)f=2(K.E)i(K.E)i=mv22=5×1022=5002=250
(K.E)f=500V12+4V222=5004
from solving 3 and 4 we get V1 and V2 as -10 & 15
time taken by the particle to reach the highest point =T2=usinθg
distance covered by both particles be d
let g=10ms2
d1=V1t=10usinθg=usinθ
d2=V2t=15usinθg=1.5usinθ
total distance between the particles when they reach the ground d=d1+d2=2.5usinθ
d=253=43.25m



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