Answer:
The raindrop strikes the top surface of the car at u m/sec.
\[\Rightarrow \] Initial velocity of the raindrop striking the top surface of the car = u m/s
Then it comes to rest
\[\Rightarrow \] Final velocity = v = ‘0’ m/s
We know that Force = F = Rate of change of
\[momentum=\frac{Change\,in\,\,momentum}{time}\]
\[=\frac{final\,\,momentum\,\,initial\,\,momentum}{time}\]
\[=\frac{mv-mu}{t}\]
\[\Rightarrow F=\frac{m(v-u)}{t}=\frac{m}{t}(v-u)\] ….. (1)
Rate at which the raindrop strikes on unit area of the top surface of the car \[=\frac{m}{t}=k\,\,kg/\sec \]
Total surface area of the car\[=x\,\,{{m}^{2}}\]
Force acting on the top surface of the car = ?
Substituting above values in (1), we get
\[F=\frac{m}{t}(0-u)=k(-u)\Rightarrow F=(-ku)N\][\[F=4\times {{10}^{6}}\times 4.5\times {{10}^{5}}=\frac{18}{10}=1.8N\]sign indicates that the force is retarding]
\[\Rightarrow \] force exerted on top surface of the car for unit area\[=(-ku)N\]
Therefore, force exerted on total surface area of the car\[=(-kux)N\]
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