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A raindrop of mass $1.00 \mathrm{~g}$ falling from a height of $1 \mathrm{~km}$ hits the ground with a speed of $50 \mathrm{~ms}^{-1}$. Calculate
(a) the loss of $\mathrm{PE}$ of the drop.
(b) the gain in $\mathrm{KE}$ of the drop.
(c) Is the gain in KE equal to loss of PE? If not why? Take, $g=10 \mathrm{~ms}^{-2}$.
(a) the loss of $\mathrm{PE}$ of the drop.
(b) the gain in $\mathrm{KE}$ of the drop.
(c) Is the gain in KE equal to loss of PE? If not why? Take, $g=10 \mathrm{~ms}^{-2}$.
Solution:
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Verified Answer
As given that, mass of the rain drop $(m)$
$$
=1.00 \mathrm{~g}=1 \times 10^{-3} \mathrm{~kg}
$$
Height of falling $(h)=1 \mathrm{~km}=10^3 \mathrm{~m}$
$$
g=10 \mathrm{~m} / \mathrm{s}^2
$$
Speed of the rain drop $(v)=50 \mathrm{~m} / \mathrm{s}, u=0$.
(a) Loss of $\mathrm{PE}$ (at highest point) of the drop $=m g h=$ $1 \times 10^{-3} \times 10 \times 10^3$
So loss of $P E=10 \mathrm{~J}$
(b) Gain in KE of the drop $=\frac{1}{2} m v^2$
$$
\begin{aligned}
&=\frac{1}{2} \times 1 \times 10^{-3} \times(50)^2 \\
&=\frac{1}{2} \times 10^{-3} \times 2500 \\
&=1.250 \mathrm{~J}
\end{aligned}
$$
$$
\text { Gain in } \mathrm{KE}=1.250 \mathrm{~J}
$$
(c) Gain in KE is not equal to the loss in PE. It is due to resistance or dragging force of air.
$$
=1.00 \mathrm{~g}=1 \times 10^{-3} \mathrm{~kg}
$$
Height of falling $(h)=1 \mathrm{~km}=10^3 \mathrm{~m}$
$$
g=10 \mathrm{~m} / \mathrm{s}^2
$$
Speed of the rain drop $(v)=50 \mathrm{~m} / \mathrm{s}, u=0$.
(a) Loss of $\mathrm{PE}$ (at highest point) of the drop $=m g h=$ $1 \times 10^{-3} \times 10 \times 10^3$
So loss of $P E=10 \mathrm{~J}$
(b) Gain in KE of the drop $=\frac{1}{2} m v^2$
$$
\begin{aligned}
&=\frac{1}{2} \times 1 \times 10^{-3} \times(50)^2 \\
&=\frac{1}{2} \times 10^{-3} \times 2500 \\
&=1.250 \mathrm{~J}
\end{aligned}
$$
$$
\text { Gain in } \mathrm{KE}=1.250 \mathrm{~J}
$$
(c) Gain in KE is not equal to the loss in PE. It is due to resistance or dragging force of air.
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