MHT CET · Physics · Center of Mass Momentum and Collision
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
- A \(10^7 \mathrm{~N} / \mathrm{m}^2\)
- B \(10^6 \mathrm{~N} / \mathrm{m}^2\)
- C \(5 \times 10^6 \mathrm{~N} / \mathrm{m}^2\)
- D \(2 \times 10^6 \mathrm{~N} / \mathrm{m}^2\)
Answer & Solution
Correct Answer
(B) \(10^6 \mathrm{~N} / \mathrm{m}^2\)
Step-by-step Solution
Detailed explanation
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 \(=\frac{F}{A}=\frac{100}{10^{-4}}=10^6 \mathrm{~N} / \mathrm{m}^2\).
\(\begin{aligned}
& \mathrm{N}=10^3, \mathrm{~m}=1 \mathrm{~g}=10^{-3} \mathrm{~kg}, \\
& \mathrm{~A}=1 \mathrm{~cm}^2=10^{-4} \mathrm{~m}^2, \mathrm{v}=50 \mathrm{~m} / \mathrm{s}
\end{aligned}\)
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 \(=\frac{F}{A}=\frac{100}{10^{-4}}=10^6 \mathrm{~N} / \mathrm{m}^2\).
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