A bomb of mass $9\,kg$ explodes into $2$ pieces of mass $3\,kg$ and $6\,kg.$ The velocity of mass $3\,kg$ is $1.6\, m/s$, the K.E. of mass $6\,kg$ is ............ $J$
A bomb of mass $9\,kg$ explodes into $2$ pieces of mass $3\,kg$ and $6\,kg.$ The velocity of mass $3\,kg$ is $1.6\, m/s$, the K.E. of mass $6\,kg$ is ............ $J$
$3.84$
$9.6$
$1.92$
$2.92$
A bomb of mass $9\,kg$ explodes into $2$ pieces of mass $3\,kg$ and $6\,kg.$ The velocity of mass $3\,kg$ is $1.6\, m/s$, the K.E. of mass $6\,kg$ is ............ $J$
As the bomb initially was at rest therefore
Initial momentum of bomb $= 0$
Final momentum of system = ${m_1}{v_1} + {m_2}{v_2}$
As there is no external force
${m_1}{v_1} + {m_2}{v_2} = 0$ ==> $3 \times 1.6 + 6 \times {v_2} = 0$
velocity of 6 kg mass ${v_2} = 0.8\,m/s$ (numerically)
Its kinetic energy$ = \frac{1}{2}{m_2}v_2^2$$ = \frac{1}{2} \times 6 \times {(0.8)^2} = 1.92\,J$
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