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# Answer to Question #127893 in Mechanics | Relativity for Atharv Shukla

Question #127893
A horizontal jet of water coming out of a pipe of area of cross section 20cm hits a vertical wallwith a velocity of 10ms and rebounds with the same speed . The force exerted by the water on the wall is
1
2020-07-30T10:38:30-0400

Explanations & Calculations

• To do this sum consider about some volume of water having a mass “m(kg)” hitting the wall at the given velocity “v” and rebounds after a period “t (s)”.
• Due to the change of momentum within this period, a force (F) is exerted on that volume of water (same is exerted on the wall at the point of hit).
• Applying the equation related to momentum change,

"\\qquad\\qquad\n\\begin{aligned}\n\\small F &= \\small \\frac{mv-mu}{t}\\\\\n&= \\small \\frac{mu-m(-u)}{t}\\\\\n&= \\small \\frac{2mu}{t}\\\\\n& = \\small 2u\\big(\\frac{m}{t}\\big)\n\\end{aligned}"

• This "\\large \\frac{m}{t}" is called the rate of mass flow which equals to,

"\\qquad\\qquad\n\\begin{aligned}\n&= \\small \\frac{Ax\\rho}{t} \n&= A\\rho\\big(\\frac{x}{t}\\big)\n&=A\\rho u\n\\end{aligned}"   : "\\rho =" density of water

• Therefore,

"\\qquad\\qquad\n\\begin{aligned}\n\\small F &= \\small 2u\\times Au \\rho\\\\\n&= \\small 2Au^2 \\rho\\\\\n&= \\small2\\times20\\times10^{-4}m^2\\times(10ms^{-1})^2\n\\times10^3kgm^{-3}\\\\\n&= \\small \\bold{40N}\n\\end{aligned}"

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