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
Expert's answer
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,

F=mvmut=mum(u)t=2mut=2u(mt)\qquad\qquad \begin{aligned} \small F &= \small \frac{mv-mu}{t}\\ &= \small \frac{mu-m(-u)}{t}\\ &= \small \frac{2mu}{t}\\ & = \small 2u\big(\frac{m}{t}\big) \end{aligned}

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

=Axρt=Aρ(xt)=Aρu\qquad\qquad \begin{aligned} &= \small \frac{Ax\rho}{t} &= A\rho\big(\frac{x}{t}\big) &=A\rho u \end{aligned}   : ρ=\rho = density of water

  • Therefore,

F=2u×Auρ=2Au2ρ=2×20×104m2×(10ms1)2×103kgm3=40N\qquad\qquad \begin{aligned} \small F &= \small 2u\times Au \rho\\ &= \small 2Au^2 \rho\\ &= \small2\times20\times10^{-4}m^2\times(10ms^{-1})^2 \times10^3kgm^{-3}\\ &= \small \bold{40N} \end{aligned}


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment