Q.) A bomb of mass 300kg fragmented into two parts :one having mass 200kg moving with a velocity of 16m/s² and other having mass 100kg.Find out total kinetic energy produced in this bombarding
The speed of second part of a bomb we can fin using the law of conservation of momentum
"0=m_1v_1-m_2v_2."Hence,
"v_2=m_1v_1\/m_2=200\\times 16\/100=32\\:\\rm m\/s."The total kinetic energy
"K=\\frac{m_1v_1^2}{2}+\\frac{m_2v_2^2}{2}=\\\\\n=\\frac{200\\times 16^2}{2}+\\frac{100\\times 32^2}{2}=76800\\:\\rm J."
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