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A $1 \mathrm{~kg}$ ball moving with a speed of $6 \mathrm{~ms}^{-1}$ collides headon with a $0.5 \mathrm{~kg}$ ball moving in the opposite direction with a speed of $9 \mathrm{~ms}^{-1}$. If the coefficient of restitution is $\frac{1}{3}$, then the energy lost in the collision is
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The correct answer is:
$33.3\ J$
Using $\Delta \mathrm{KE}=\frac{m_1 m_2}{2\left(m_1+m_2\right)}\left(1-e^2\right)\left(u_1+u_2\right)^2$
Given: $m_1=1 \mathrm{~kg}$
$\begin{aligned}
& m_2=\frac{1}{2} \mathrm{~kg} ; \\
& u_1=6 \mathrm{~ms}^{-1} ; \quad u_2=9 \mathrm{~ms}^{-1} \text { and } e=\frac{1}{3} \\
& \therefore \Delta \mathrm{KE}=\frac{1 \times \frac{1}{2}}{2\left(1+\frac{1}{2}\right)}\left[1-\left(\frac{1}{3}\right)^2\right](6+9)^2 \\
& =\frac{1}{2} \times \frac{2}{6} \times \frac{8}{9} \times(15)^2 \\
& =\frac{1}{6} \times \frac{8}{9} \times 225 \\
& =33.33 \mathrm{~J}
\end{aligned}$
Given: $m_1=1 \mathrm{~kg}$
$\begin{aligned}
& m_2=\frac{1}{2} \mathrm{~kg} ; \\
& u_1=6 \mathrm{~ms}^{-1} ; \quad u_2=9 \mathrm{~ms}^{-1} \text { and } e=\frac{1}{3} \\
& \therefore \Delta \mathrm{KE}=\frac{1 \times \frac{1}{2}}{2\left(1+\frac{1}{2}\right)}\left[1-\left(\frac{1}{3}\right)^2\right](6+9)^2 \\
& =\frac{1}{2} \times \frac{2}{6} \times \frac{8}{9} \times(15)^2 \\
& =\frac{1}{6} \times \frac{8}{9} \times 225 \\
& =33.33 \mathrm{~J}
\end{aligned}$
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