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A bullet of mass 4.2 xx 10^(-2) kg, moving at a speed of 300 ms^(- 1), gets stuck into a block with a mass 9 times that of the bullet. If the block is free to move without any kind of friction, the heat generated in the process will be |
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Answer» 45 cal By the law of conservation of momentum, `MV+Mxx0=(M+m)V` or `V=(mv)/(m+M)` …(i) LOSS of KINETIC energy= Heat generated in the process `thereforeDeltaK=1/2mv^(2)-1/2(M+m)V^(2)` [From eqn. (i) ] `=1/2mv^(2)-1/3(M+m)(m^(2)v^(2))/((M+m)^(2))` `=1/2mv^(2)-1/2(m^(2)v^(2))/((M+m))=1/2mv^(2)(1-m/(M+m))` `=1/2mv^(2)((M+m-m)/(M+m))` `=1/2mv^(2)(M/(m+m))=(mMv^(2))/(2(M+m))` ...(ii) `becausem=4.2xx10^(-2)kg,v=300ms^(-1)` `M=9m=9xx4.2xx10^(-2)kg` Substituting the values in eqn. (ii), we get `DeltaK=1701J=1701/(4.2)` cal=405 cal |
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