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SayedD
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- Homework Statement
- A particle that is moving with velocity ##v## has kinetic energy ##K##. Suddenly the particle gained extra energy externally with magnitude ##3k/16## that causes the particle to split into two pieces. One piece moves at the speed of ##v## relative to the other piece with the direction of motion parallel with the motion before splitting. If the mass of the smaller piece is 1 kg, determine the mass of the other piece.
- Relevant Equations
- Momentum and Energy
My method is silly but here is my attempt
Initially we know that the kinetic energy is k = ##1/2mv^2## and the final kinetic energy at the moment before splitting is equal to ##(k + 3k/16)mv^2_1 = (19k/16)mv^2_1##. Substituting the value k to the second equation gives us ##v_1 = (\sqrt{19}/4)v##
Since the equation only tells us the relative motion between the pieces, I think we should deal the momentum in the reference frame of that piece. And also notice that ##M = m_1 + m_2## with ##m_1 = 1## This gives us the equation. And there is two cases since we do not know which piece is stationary or moving in this reference frame. So we have to check both case.
Case one, ##m_2## is moving with v relative to other piece.
##(m_1 + m_2)(v_1 - v) = m_2 v
\implies v(1 + m_2)(\sqrt{19}/4 - 1) = m_2 v
\implies m_2 = 0.097 ## which is smaller thus wrong
Case two ##m_1## is moving with v relative to other piece.
##(m_1 + m_2)(v_1 - v) = m_1 v
\implies v(1 + m_2)(\sqrt{19}/4 - 1) = m_1 v
\implies m_2 = 10.23 ## which is bigger thus could right
These all argument could be crap but I wanted to see your solutions.
Initially we know that the kinetic energy is k = ##1/2mv^2## and the final kinetic energy at the moment before splitting is equal to ##(k + 3k/16)mv^2_1 = (19k/16)mv^2_1##. Substituting the value k to the second equation gives us ##v_1 = (\sqrt{19}/4)v##
Since the equation only tells us the relative motion between the pieces, I think we should deal the momentum in the reference frame of that piece. And also notice that ##M = m_1 + m_2## with ##m_1 = 1## This gives us the equation. And there is two cases since we do not know which piece is stationary or moving in this reference frame. So we have to check both case.
Case one, ##m_2## is moving with v relative to other piece.
##(m_1 + m_2)(v_1 - v) = m_2 v
\implies v(1 + m_2)(\sqrt{19}/4 - 1) = m_2 v
\implies m_2 = 0.097 ## which is smaller thus wrong
Case two ##m_1## is moving with v relative to other piece.
##(m_1 + m_2)(v_1 - v) = m_1 v
\implies v(1 + m_2)(\sqrt{19}/4 - 1) = m_1 v
\implies m_2 = 10.23 ## which is bigger thus could right
These all argument could be crap but I wanted to see your solutions.
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