A bomb of mass 12,kg at rest explodes into tw | vedclass

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(A) Given: Total mass $M = 12\,kg$. Mass ratio $m_1 : m_2 = 1 : 3$.Thus,$m_1 = 3\,kg$ and $m_2 = 9\,kg$.Kinetic energy of the smaller fragment $K_1 =

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$36$

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(A) Given: Total mass $M = 12\,kg$. Mass ratio $m_1 : m_2 = 1 : 3$.Thus,$m_1 = 3\,kg$ and $m_2 = 9\,kg$.Kinetic energy of the smaller fragment $K_1 = 216\,J$.Since the bomb is at rest,the initial momentum is zero. By the law of conservation of momentum,the magnitudes of the momenta of the two fragments must be equal: $p_1 = p_2 = p$.We know that $K = \frac{p^2}{2m}$,so $p = \sqrt{2mK}$.For the smaller fragment: $p = \sqrt{2 	imes 3 	imes 216} = \sqrt{6 	imes 216} = \sqrt{1296} = 36\,kg\cdot m/s$.Since $p_1 = p_2$,the momentum of the heavier fragment is also $36\,kg\cdot m/s$.

$A$ bomb of mass $12\,kg$ at rest explodes into two fragments of masses in the ratio $1 : 3$. The $K.E.$ of the smaller fragment is $216\,J$. The momentum of the heavier fragment is (in $kg\cdot m/s$):

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