The Student Room Group

Unit 4 Physics PP question!

Hello I have a few questions which I can't start of, can anyone help me out..

(Q1) A hose pipe is used to water a garden. The supply delivers water at a rate of 0.31 kg/s to the nozzle which has cross-sectional area of

7.3 x 10^-5 m².

(i) Show that water leaves the nozzle at a speed of about 4 m/s.

(density of water = 1000 kg/m³)


(ii) Before it leaves the hose, the water has a speed of 0.68 m/s. Calculate the force on the hose.

The water from the hose is sprayed onto a brick wall the base of which is firmly embedded in the ground. Explain why there is no overall effect on the rotation of the Earth.

I tried answering this by saying that momentum from water to brick wall is absorbed by the wall and doesnt provide enough force to make it move...? It common sense I mean how would the Earth rotate with such a small force..Maybe I'm not understanding what they are trying to say by rotation of the Earth.


Thanks a lot for your time
For the first one. Consider one second. In that time, a cylinder of water of length v will have left the hose. You know the mass of it because you know the mass per second. Then use ρ=mV\rho = \frac{m}{V}


For the second one you need to use F=Δ(mv)ΔtF = \frac{\Delta (mv)}{\Delta t}, and Δm\Delta m is given in the question.


I am not at all sure about the third question, but it might be something to do with the fact that the wall is also rotating?
is this from an AQA A paper?
physicsgirlie

I am not at all sure about the third question, but it might be something to do with the fact that the wall is also rotating?

both are attached to the earth, so given that momentum is conserved, the water droplets rotate the earth one way but the reaction felt by the hose balances it by rotating it the other way.
Reply 4
So both pair of forces cancel out eachother and stays in equilibrium then?

Im not getting question 1..

p = m / V I know density and mass so i work out volume then after that what do I do

V= Al

V/A = l thus this is the distance the water has travelled in one second?
Yes that is right, so the velocity is l metres per second.
Reply 6
Thanks a lot
Reply 7
Original post by physicsgirlie
For the first one. Consider one second. In that time, a cylinder of water of length v will have left the hose. You know the mass of it because you know the mass per second. Then use ρ=mV\rho = \frac{m}{V}


For the second one you need to use F=Δ(mv)ΔtF = \frac{\Delta (mv)}{\Delta t}, and Δm\Delta m is given in the question.


I am not at all sure about the third question, but it might be something to do with the fact that the wall is also rotating?

Hi, if you're still able to respond to this, how would you find the change in time? Would you do 4/0.68 or just use 1?