Search any question & find its solution
Question:
Answered & Verified by Expert
A bomb of mass $16 \mathrm{~kg}$ explodes into two pieces of masses $4 \mathrm{~kg}$ and $12 \mathrm{~kg}$. The velocity of the $12 \mathrm{~kg}$ mass is $4 \mathrm{~ms}^{-1}$. The kinetic energy of the second piece is
Options:
Solution:
2954 Upvotes
Verified Answer
The correct answer is:
$288 \mathrm{~J}$
The correct option is B $288 \mathrm{~J}$
Let the velocity and mass of $4 \mathrm{~kg}$ piece by $\mathrm{v}_1$ and velocity and mass of $12 \mathrm{~kg}$ piece be $\mathrm{v}_2$ and $\mathrm{m}_2$ Applying conservation of linear momentum
$$
\begin{aligned}
& \mathrm{m}_2 \mathrm{v}_1=\mathrm{m}_1 \mathrm{v}_2 \\
& \Rightarrow \mathrm{v}_1=\frac{12 \times 4}{4}=12 \mathrm{~ms}^{-1}
\end{aligned}
$$
Kinetic energy is
$$
\therefore \mathrm{KE}_1=\frac{1}{2} \mathrm{~m}_1 \mathrm{v}_1^2=\frac{1}{2} \times 4 \times 144=288 \mathrm{~J}
$$
Let the velocity and mass of $4 \mathrm{~kg}$ piece by $\mathrm{v}_1$ and velocity and mass of $12 \mathrm{~kg}$ piece be $\mathrm{v}_2$ and $\mathrm{m}_2$ Applying conservation of linear momentum
$$
\begin{aligned}
& \mathrm{m}_2 \mathrm{v}_1=\mathrm{m}_1 \mathrm{v}_2 \\
& \Rightarrow \mathrm{v}_1=\frac{12 \times 4}{4}=12 \mathrm{~ms}^{-1}
\end{aligned}
$$
Kinetic energy is
$$
\therefore \mathrm{KE}_1=\frac{1}{2} \mathrm{~m}_1 \mathrm{v}_1^2=\frac{1}{2} \times 4 \times 144=288 \mathrm{~J}
$$
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.