question_answer

A raindrop is striking the top surface of a car at \[um/\sec \]and comes to rest after striking. The mass of rain falling per second is k kg/sec on unit area of the car. Find the force exerted by the rain on top surface of the car, if the car has surface area of \[x\,{{\text{m}}^{\text{2}}}\].

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\[=(-kux)N\]