Question 315: Centre of Mass and Rotational Motion

Question

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

Accepted Answer

$$\begin{aligned}
&	ext { Given that, }\
&\begin{aligned}
& \mathrm{N}=10^3, \mathrm{~m}=1 \mathrm{~g}=10^{-3} \mathrm{~kg}, \
& \mathrm{~A}=1 \mathrm{~cm}^2=10^{-4} \mathrm{~m}^2, \mathrm{v}=50 \mathrm{~m} / \mathrm{s}
\end{aligned}
\end{aligned}$$
Change in momentum in each collision $=\mathrm{m}[\mathrm{v}-(-\mathrm{v})]=2 \mathrm{mv}$
$	herefore \quad$ Force exerted on the surface
$$\begin{aligned}
& =10^3 	imes 2 \mathrm{mv} \
& =10^3 	imes 2 	imes 10^{-3} 	imes 50=100 \mathrm{~N} \
& 	ext { Now, Pressure }=\frac{\mathrm{F}}{\mathrm{~A}}=\frac{100}{10^{-4}}=10^6 \mathrm{~N} / \mathrm{m}^2
\end{aligned}$$

Answer

$10^7 \mathrm{~N} / \mathrm{m}^2$

Answer

$10^6 \mathrm{~N} / \mathrm{m}^2$

Answer

$5 	imes 10^6 \mathrm{~N} / \mathrm{m}^2$

Answer

$2 	imes 10^6 \mathrm{~N} / \mathrm{m}^2$

Centre of Mass and Rotational Motion

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

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