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Laws of Motion

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Question : 28 of 40
Marks: +1, -0
A stream of water flowing horizontally with a speed of 15 m s−115\,\mathrm{m}\,\mathrm{s}^{-1} gushes out of a tube of cross-sectional area 10−2 m210^{-2}\,\mathrm{m}^2, and hits a vertical wall nearby. What is the force exerted on the wall by the impact of water, assuming it does not rebound?
Solution:  
Here, v=15 m s−1v = 15\,\mathrm{m}\,\mathrm{s}^{-1}
Area of cross section, A=10−2 m2,F=?A = 10^{-2}\,\mathrm{m}^2, F = ?
Volume of water pushing out per second =A×v=10−2×15 m3 s−1= A \times v = 10^{-2} \times 15\,\mathrm{m}^3\,\mathrm{s}^{-1}
As density of water is 103 kg m−310^3\,\mathrm{kg}\,\mathrm{m}^{-3} therefore, mass of water striking the wall per second is
m=(15×10−2)×103=150 kg s−1m = (15 \times 10^{-2}) \times 10^3 = 150\,\mathrm{kg}\,\mathrm{s}^{-1}
As F=change in linear momentum  time F = \frac{\text{change in linear momentum }}{\text{ time }}
∴F=m×vt\therefore F = \frac{m \times v}{t}
=150×151=2250 N= \frac{150 \times 15}{1} = 2250\,\mathrm{N}
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