a bomb of mass 9 kg explodes into two pieces of 3 kg and 6 kg the velocity of 3 kgs is 16 metre per second the kinetic energy of 6 kg peice is
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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
(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
sanidhyasingla:
i dont understand
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I don't understand
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Answer is 192 joule...
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