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A body of mass $4 \mathrm{~kg}$ moving with velocity $12 \mathrm{~m} / \mathrm{s}$ collides with another body of mass $6 \mathrm{~kg}$ at rest. If two bodies stick together after collision, then the loss of kinetic energy of system is
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Verified Answer
The correct answer is:
$172.8 \mathrm{~J}$
Loss in K.E. =
$\begin{aligned}
& \frac{m_1 m_2}{2\left(m_1+m_2\right)}\left(u_1-u_2\right)^2 \\
&=\frac{4 \times 6}{2 \times 10} \times(12-0)^2=172.8
\end{aligned}$
$\begin{aligned}
& \frac{m_1 m_2}{2\left(m_1+m_2\right)}\left(u_1-u_2\right)^2 \\
&=\frac{4 \times 6}{2 \times 10} \times(12-0)^2=172.8
\end{aligned}$
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