A bomb of mass \(9 \mathrm{~kg}\) explodes into two pieces of mass \(3 \mathrm{~kg}\) and \(6 \mathrm{~kg}\)…

A bomb of mass \(9 \mathrm{~kg}\) explodes into two pieces of mass \(3 \mathrm{~kg}\) and \(6 \mathrm{~kg}\). The velocity of mass \(3 \mathrm{~kg}\) is \(16 \mathrm{~m} / \mathrm{s}\). The kinetic energy of mass \(6 \mathrm{~kg}\) (in joule) is
  1. 96
  2. 384
  3. 192
  4. 768

Solution

Given, mass of bomb, \(M=9 \mathrm{~kg}\)
According to conservation of linear momentum, \(\begin{aligned} & M \times 0 & =m_1 v_1+m_2 v_2 \\ \Rightarrow & 9 \times 0 & =3 \times 16+6 \times v_2 \\ \Rightarrow & v_2 & =\frac{-3 \times 16}{6}=-8 \mathrm{~m} / \mathrm{s} \end{aligned}\) \(\therefore\) Kinetic energy of mass, \(m_2=6 \mathrm{~kg}\) (given) \(\begin{aligned} K & =\frac{1}{2} m_2 v_2^2=\frac{1}{2} \times 6 \times(-8)^2 \\ & =\frac{1}{2} \times 6 \times 64=192 \mathrm{~J} \end{aligned}\)

Asked in: AP EAMCET 2020 (21 Sep Shift 2)

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