The distance covered by a body of mass $5 \mathrm{~g}$ having linear momentum $0.3 \mathrm{~kg} \mathrm{~m} / \mathrm{s}$ in $5 \mathrm{~s}$ is:

Question

Aicharya · Accepted Answer

Given,
$\begin{aligned}
m=5 \mathrm{~g} & =0.005 \mathrm{~kg}, \\
\mathrm{P}=m v & =0.3 \mathrm{~kg} \mathrm{~m} / \mathrm{s}, \text { and } \\
t & =5 \mathrm{~s},
\end{aligned}$
(where, symbols have their usual meanings.)
$\mathrm{P}=m v=0.3$
Hence, $\mathrm{V}=\frac{0.3}{m}=\frac{0.3}{0.005}=60 \mathrm{~m} / \mathrm{s}$
Hence, the distance covered $=$ speed $\times$ time
$=60 \times 5=300 \mathrm{~m}$

The distance covered by a body of mass $5 \mathrm{~g}$ having linear momentum $0.3 \mathrm{~kg} \mathrm{~m}…

Solution