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A bomb of mass $3.0 \mathrm{~kg}$ explodes in air into two pieces of masses $2.0 \mathrm{~kg}$ and $1.0 \mathrm{~kg}$. The smaller mass goes at a speed of $80 \mathrm{~m} / \mathrm{s}$. The total energy imparted to the two fragments is
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Verified Answer
The correct answer is:
$4.8 \mathrm{~kJ}$.
Velocity of second piece can be find out using conservation of momentum.
$$
\begin{aligned}
& \mathrm{O}=m_1 v_1+m_2 v_2=2 \times v_1+1 \times 80 \\
\Rightarrow \quad & v_1=-40 \mathrm{~m} / \mathrm{s} .
\end{aligned}
$$
Negative sign showing that particle is moving in opposite direction of other particle direction.
Energy imparted to the fragments are converted into their kinetic energy.
$$
\begin{aligned}
\therefore \quad \text { Total energy } & =\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2 \\
& =\frac{1}{2} \times 2 \times 1600+\frac{1}{2} \times 1 \times 6400 \\
& =4800 \mathrm{~J}=4.8 \mathrm{~kJ} .
\end{aligned}
$$
$$
\begin{aligned}
& \mathrm{O}=m_1 v_1+m_2 v_2=2 \times v_1+1 \times 80 \\
\Rightarrow \quad & v_1=-40 \mathrm{~m} / \mathrm{s} .
\end{aligned}
$$
Negative sign showing that particle is moving in opposite direction of other particle direction.
Energy imparted to the fragments are converted into their kinetic energy.
$$
\begin{aligned}
\therefore \quad \text { Total energy } & =\frac{1}{2} m_1 v_1^2+\frac{1}{2} m_2 v_2^2 \\
& =\frac{1}{2} \times 2 \times 1600+\frac{1}{2} \times 1 \times 6400 \\
& =4800 \mathrm{~J}=4.8 \mathrm{~kJ} .
\end{aligned}
$$
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