1000 small balls, each weighing 1 gram, strike one square cm of area per second with a velocity $50…
1000 small balls, each weighing 1 gram, strike one square cm of area per second with a velocity $50 \mathrm{~m} / \mathrm{s}$ in a normal direction and rebound with the same velocity. The value of pressure on the surface will be
$10^7 \mathrm{~N} / \mathrm{m}^2$
$10^6 \mathrm{~N} / \mathrm{m}^2$
$5 \times 10^6 \mathrm{~N} / \mathrm{m}^2$
$2 \times 10^6 \mathrm{~N} / \mathrm{m}^2$
Solution
Given that,
$\begin{aligned}
& \mathrm{N}=10^3, \mathrm{~m}=1 \mathrm{~g}=10^{-3} \mathrm{~kg}, \\
& \mathrm{~A}=1 \mathrm{~cm}^2=10^{-4} \mathrm{~m}^2, \mathrm{v}=50 \mathrm{~m} / \mathrm{s}
\end{aligned}$
Change in momentum in each collision
$=\mathrm{m}[\mathrm{v}-(-\mathrm{v})]=2 \mathrm{mv}$
$\therefore \quad$ Force exerted on the surface
$\begin{aligned}
& =10^3 \times 2 \mathrm{mv} \\
& =10^3 \times 2 \times 10^{-3} \times 50=100 \mathrm{~N}
\end{aligned}$ Now, Pressure $=\frac{F}{A}=\frac{100}{10^{-4}}=10^6 \mathrm{~N} / \mathrm{m}^2$.