Asked by liz
An internal explosion breaks an object, initially at rest, into two pieces. One piece (m1) is 1.5 times more massive than the other (m2). If 7500 J were released by the explosion, how much KE did each piece acquire?
Answers
Answered by
Damon
|v1| = 2|v2|
v1^2 = 4 v2^2
so
Ke v1 = 4 Ke v2
so
5 Ke2 = 7500 J
so Ke2 = 1500 J
and Ke1 = 6000 J
v1^2 = 4 v2^2
so
Ke v1 = 4 Ke v2
so
5 Ke2 = 7500 J
so Ke2 = 1500 J
and Ke1 = 6000 J
Answered by
Sierra
7500J = KE1 + KE2
Answered by
Sierra
7500J=1/2(M2V^2) +1/2(1.5M2)(V^2)
7500J=1/2 (M2V^2+1.5M2V^2)
15000=M2V^2+1.5M2V^2
15000= X+1.5X
15000=X(2.5)
6000=X
6000 = X = M2V^2
1/2(6000) = KE
3000 = KE
TOTAL PE (7500J) - KElargerobject(3000) = 4500J (KE smaller object)
7500J=1/2 (M2V^2+1.5M2V^2)
15000=M2V^2+1.5M2V^2
15000= X+1.5X
15000=X(2.5)
6000=X
6000 = X = M2V^2
1/2(6000) = KE
3000 = KE
TOTAL PE (7500J) - KElargerobject(3000) = 4500J (KE smaller object)
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