Question

a bomb of mass 9kg exploded into two pieces of 3 kg and 6 kg . The velocity of 3kg mass is 16m/seconds. what is the K.E of 6kg mass​

Answer 1

Given, m1=3kg , m2 = 6kg , v1 = 16m/s , K.E 2 = ? According to law of conservation of momentum :
(m1 × v1) + (m2 × v2) = (m1 + m2) ×v
(3×16) + (6×(-v2)) = (3+6) × 0( -v2 because both are moving in opposite directions
48 - (6 × v2) = 0
48 = 6 ×v2
therefore, v2 = 8m/s
K.E2 = 1/2 × m2 × v2 square
= 1/2 ×6 x 8 square
= 3 × 64 = 192

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Answer 2

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Here is your answer

Total mass of the bomb=9 kg

Initial velocity of the bomb = 0 m/s

Mass of first bomb piece = 3kg (m₁)

Mass of second bomb peice = 6kg (m₂)

Let velocity of m₁ = v₁

Let velocity of m₂ = v₂

(3 x 16) + (6 x v₂) = 9 x 0

48 + 6v₂ = 0

6v₂ = -48

v₂ = -48/6 = -8 m/s

Now, KE = 1/2 mv² = 1/2 x 6 x 8² = 192 joules

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