A metal ball of mass $2 \mathrm{~kg}$ moving with a speed of $10 \mathrm{~ms}^{-1}$ had a head-on collision…
A metal ball of mass $2 \mathrm{~kg}$ moving with a speed of $10 \mathrm{~ms}^{-1}$ had a head-on collision
with a stationary ball of mass $3 \mathrm{~kg}$. If after collision, both the balls move together,
then the loss in kinetic energy due to collision is
$60 \mathrm{~J} .$
$100 \mathrm{~J} .$
$140 \mathrm{~J} .$
$40 \mathrm{~J} .$
Solution
Apply conservation of momentum,
$\begin{array}{l}
\mathrm{m}_{1} \mathrm{v}_{1}=\left(\mathrm{m}_{1}+\mathrm{m}_{2}\right) \mathrm{v} \\
\mathrm{v}=\frac{\mathrm{m}_{1} \mathrm{v}_{1}}{\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)}
\end{array}$
Here $\mathrm{v}_{1}=36 \mathrm{~km} / \mathrm{hr}=10 \mathrm{~m} / \mathrm{s}$,
$\begin{array}{l}
\mathrm{m}_{1}=2 \mathrm{~kg}, \mathrm{~m}_{2}=3 \mathrm{~kg} \\
\mathrm{v}=\frac{10 \times 2}{5}=4 \mathrm{~m} / \mathrm{s}
\end{array}$
K.E. (initial) $=\frac{1}{2} \times 2 \times(10)^{2}=100 \mathrm{~J}$
K.E. (Final) $=\frac{1}{2} \times(3+2) \times(4)^{2}=40 \mathrm{~J}$
Loss in K.E. $=100-40=60 \mathrm{~J}$
Alternatively use the formula
$-\Delta \mathrm{E}_{\mathrm{k}}=\frac{1}{2} \frac{\mathrm{m}_{1} \mathrm{~m}_{2}}{\left(\mathrm{~m}_{1}+\mathrm{m}_{2}\right)}\left(\mathrm{u}_{1}-\mathrm{u}_{2}\right)^{2}$