A bomb of mass $3.0\, Kg$ explodes in air into two pieces of masses $2.0 \,kg$ and $1.0\, kg$. The smaller mass goes at a speed of $80 m/s$.The total energy imparted to the two fragments is ............. $kJ$
$1.07$
$2.14$
$2.4$
$4.8$
A bomb of mass $3.0\, Kg$ explodes in air into two pieces of masses $2.0 \,kg$ and $1.0\, kg$. The smaller mass goes at a speed of $80 m/s$.The total energy imparted to the two fragments is ............. $kJ$
Both fragment will possess the equal linear momentum ${m_1}{v_1} = {m_2}{v_2}$
==> $1 \times 80 = 2 \times {v_2}$ ==> ${v_2} = 40\;m./s$
Total energy of system$ = \frac{1}{2}{m_1}v_1^2 + \frac{1}{2}{m_2}v_2^2$
= $\frac{1}{2} \times 1 \times {(80)^2} + \frac{1}{2} \times 2 \times {(40)^2}$
$= 4800 J = 4.8 kJ$
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