Question

a shell at rest at a large height explodes into two parts which move horizontally with speeds v1 and v2. after what time their instantaneous velocities are perpendicular

Answer 1

after time t = √(v1v2)/g their instantaneous velocities are perpendicular.

direction of velocities has not given. so, first of all we have to apply law of conservation of linear momentum to find direction of fragments.

let m1 and m2 are masses of fragments.

initial Momentum = 0

final Momentum = m1v1 + m2v2

as external force = 0

so, initial momentum = final momentum

⇒0 = m1v1 + m2v2

⇒m1v1 = -m2v2

it is clear that, v1 and v2 are in opposite directions.

let after time t, instantaneous velocities of both are perpendicular to each other.

so, v1' = v1i - gt j .....(1)

v2' = -v2 i - gt j .........(2)

v1' and v2' are perpendicular to each other.

so, v1'.v2' = 0

⇒(v1 i - gt j).(-v2 i - gt j) = 0

⇒-v1.v2 + g²t² = 0

⇒t² = v1v2/g²

⇒t = √(v1v2)/g

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Answer 2

$\huge\bold\purple{Answer:-}$

direction of velocities has not given. so, first of all we have to apply law of conservation of linear momentum to find direction of fragments.

let m1 and m2 are masses of fragments.

initial Momentum = 0

final Momentum = m1v1 + m2v2

as external force = 0

so, initial momentum = final momentum

⇒0 = m1v1 + m2v2

⇒m1v1 = -m2v2

it is clear that, v1 and v2 are in opposite directions.

let after time t, instantaneous velocities of both are perpendicular to each other.

so, v1' = v1i - gt j .....(1)

v2' = -v2 i - gt j .........(2)

v1' and v2' are perpendicular to each other.

so, v1'.v2' = 0

⇒(v1 i - gt j).(-v2 i - gt j) = 0

⇒-v1.v2 + g²t² = 0

⇒t² = v1v2/g²

⇒t = √(v1v2)/g