A hose pipe of cross-sectional area 0.08m2 has water flowing out of it with a speed of 4ms^-1. The water travels horizontally and impacts a vertical wall soon after leaving the pipe.

Calculate the force exerted by the water on the wall.

I know how to calculate the momentum before i.e. 0.08 * 1000 = 80Kg of water
Momentum = 80 * 4 = 320Kgms^-1.

My question is to calculate the force you need to know the momentum afterwards. Assuming that water bounces back elastically we still don't know the time of collision. My tutor is saying that time of collision is 1second but I don't agree. If 320Kg of water are coming out travelling 4 m per second it doesn't mean that it takes 1 second to rebound. The wall may be course and rough and time of collision would be larger.

Could you please answer me and tell me if I am right?

I know how to calculate the momentum before i.e. 0.08 * 1000 = 80Kg of water
Momentum = 80 * 4 = 320Kgms^-1.


This is not momentum - check your units: 0.08 m^2 * 1000 Kg/m^3 = 80 Kg/m (not Kg). 80 Kg/m * 4 m/s = 320 Kg/s. This is the mass of water that must be turned around every second. The force required to stop a 320 Kg mass from 4 m/s to 0 m/s is


F = m \frac {\Delta v}{\Delta t} = 320 Kg \frac {4 m/s} {1 s} = 1280 N


Note that this assumes the water hits the wall and stops with no rebound. If you assume instead that the water splashes back at 4 m/s then the force exerted is double this amount.



My tutor is saying that time of collision is 1second but I don't agree. If 320Kg of water are coming out travelling 4 m per second it doesn't mean that it takes 1 second to rebound.

It's not that that the collision takes one second for each drop of water, but rather that every second a mass of 320 Kg is being slowed from 4 m/s to 0 m/s (or to -4m/s).

I Don't agree. Sorry.

Every second a mass of 320Kg is coming out of the hose pipe but it may take an x amount of seconds to slow down.
I compare this to a ball - it may be 1Kg heavy and travelling at 5m/s but it then may take an x amount of seconds to be in contact with a wall and rebound.

This time instead of a ball, there is water.

Actually the 320Kg must take exactly one second to come to a stop. Think of conservation of mass flow - if it took longer than that the wall water would start to pile up on the wall. And it can't be shorter than that because the water comes out of the pipe at that rate. To use the ball analogy - if you throw 10 balls per second at the wall you expect 10 balls per second to rebound back, no more nor less. Here we throw 320 Kg of water at the wall per second and so 320Kg of water per second bounces back.

In this way you are assuming that the collision is elastic. But it may not be elastic and there would still be conservation of momentum.
For example if I throw 1 ball at a wall at 4ms^-1 and its mass is 1Kg for example then momentum before is 1Kg * 4ms^-1;
it may not return with that speed but momentum is still conserved because the wall may move a bit upon collision.

Secondly, if it is as you are saying if it takes 1 second for 320Kg of water to reach the wall and 1 second for 320Kg for the water to bounce back then the time during which the wall is in contact with the ball is 0 which is not what you said because you said that it takes 1 second. By time of collision I mean how long the two objects remain in contact and not the time the object takes to return.

according to your explanation, any substance would return at the same time. If it is oil for example I think it would take a longer time to bounce back - it would stick longer with wall before moving out.

So according to this reasoning, (sorry for this phrase, I don't mean to appear rude) time of collision is always 1 second no matter what?

In this way you are assuming that the collision is elastic. But it may not be elastic and there would still be conservation of momentum.
For example if I throw 1 ball at a wall at 4ms^-1 and its mass is 1Kg for example then momentum before is 1Kg * 4ms^-1;
it may not return with that speed but momentum is still conserved because the wall may move a bit upon collision.

If the wall moves then this is a completely different problem. As for whether it's an elastic collision or not - I am not assuming that it's elastic. Like I said earlier it may not be - if the water hits the wall and drops down then momentuin is NOT conserved, as the momentum of the water after it hits the wall is 0. For an inelastic collision use \Delta v = 4 m/s. For elastic collision use \Delta v = 8 m/s.


So according to this reasoning, (sorry for this phrase, I don't mean to appear rude) time of collision is always 1 second no matter what?

I'm afraid you still misunderstand. Let me try a different tact. Let's model the situation as consisting of many impacts of water (or balls) drops per second, each of which results in an impact force that lasts a tiny fraction of a second. What we want to find out is the average force, over time, that these impacts cause. Each impact causes a spike of force, but in the time between impacts there is 0 force, so we need to average this out.

Here's an exampel of how to calculate this. Suppose I throw ten 1 Kg balls per second at the wall, that the time of impact for each ball is 0.01 s, and that the collisions are inelastic, so that \Delta v = 4 m/s. Then using F = m \Delta v / ( \Delta t) you find that the force of each impact is F = 1 Kg x 4 m/s /(0.01s) = 400N per impact. Over a period of 1 second there are ten such impacts, each taking 0.01 s, so the average force the wall feels over a period of T seconds is (400N x 10T impacts x 0.01 sec/impact )/ T sec= 40 N. Thus over a period of T seconds the avarage force experienced is 40 N. The attached graph illustrates how this works. If you try this using a different value for \Delta t you get the same answer. Note how this average force is equal to the mass of balls per second times their change in velocity:


F_{avg} = 10 \frac {Kg} {s} \times 4 \frac m s = 40 N


This is precisely the same answer I gave earlier, only arrived at in a slightly different way.

We can derive a general equation for this situation as follows. Let m = mass of each ball, T = time over which we want to measure the average force, and N = number of balls during that time T:


F_{avg} = \frac {m \Delta v}{\Delta t} \times \frac {N \Delta t}{T} = m \Delta v \frac N T


Note that the quantity Nm/T is equal to the rate of mass flow of balls, in Kg per second. If we let \phi = Nm/T then this becomes:


F_{avg} = \phi \Delta v


Conclusion: when loking at the average force the wall experiences it only depends on the rate of mass flow of balls and change in velocity of each ball.

For the case of water hitting the wall you could model it as one big hunk of 320Kg water hitting once per second, or as a million drops of water each of mass 0.32 g hitting every 1 millionth of a second - the answer comes out the same.

Hope this helps.

I can now understand how the time of collision cancels out. But I think you wanted to say that time of collision is 0.1 seconds not 0.01.

SImply great thank you! The average part was the main part.

So in this question the time over which we are taking the average can be any time. Only that in the question he told you that 320Kg were hitting the wall every second so T must be 1 second; just like 100 balls hit the wall every 1 second

SImply great thank you! The average part was the main part.

Glad to help!


So in this question the time over which we are taking the average can be any time. Only that in the question he told you that 320Kg were hitting the wall every second so T must be 1 second; just like 100 balls hit the wall every 1 second

You can use 320 Kg of water every second, or 640 Kg every two seconds, or 32 Kg every tenth of a second - the rate is the same and you get the same answer.