at a certain point body at rest explodes into two equal fragments are within fragment in receiving a horizontal velocity 10 metre per second the horizontal distance between the two fragments of the displacement vectors are perpendicular to each other is
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Answer:
The time interval after which both the velocity vector will be perpendicular to each other will be 1 sec
Explanation:
From the law of conservation of momentum
m x 0 = mv/2 + mu/2
=> v = -u
where u and v are velocity of both the fragments.
Given u = 10 m/s
=> |v| = |u| = 10 m/s
If Ф is the angle between both the velocity component, then
t = (√uv)/g x cotФ/2
given,
u = v = 10 m/s
g = 10 m/s²
Ф = 90 => cotФ/2 = cot 45 = 1
hence
t = (√10 x 10)/10 x 1
=> t = 10/10 = 1 sec
Hence the time interval after which both the velocity vector will be perpendicular to each other will be 1 sec
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