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A block of mass $0.50 \mathrm{~kg}$ is moving with a speed of $2.00 \mathrm{~m} / \mathrm{s}$ on a smooth surface. It strikes another mass of $1.00 \mathrm{~kg}$ and then they move together as a single body. The energy loss during the collision is
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The correct answer is:
$0.67 \mathrm{~J}$
$0.67 \mathrm{~J}$
$$
\begin{aligned}
& m_1 u_1+m_2 u_2=\left(m_1+m_2\right) v \\
& v=2 / 3 \mathrm{~m} / \mathrm{s} \\
& \text { Energy loss }=\frac{1}{2}(0.5) \times(2)^2-\frac{1}{2}(1.5) \times\left(\frac{2}{3}\right)^2=0.67 \mathrm{~J}
\end{aligned}
$$
\begin{aligned}
& m_1 u_1+m_2 u_2=\left(m_1+m_2\right) v \\
& v=2 / 3 \mathrm{~m} / \mathrm{s} \\
& \text { Energy loss }=\frac{1}{2}(0.5) \times(2)^2-\frac{1}{2}(1.5) \times\left(\frac{2}{3}\right)^2=0.67 \mathrm{~J}
\end{aligned}
$$
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