# Momentum Question

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Water of density 1000kgm^-3 flows out of a garden hose of cross sectional area 7.2 x 10^-4 m^2 at a rate of 2 x 10^-4 m^3 per second. How much momentum is carried by the water leaving the hose per second?

Can someone explain how you would work this out plz thanks

Can someone explain how you would work this out plz thanks

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(Original post by

you can work out the volume, and therefore the mass.. and you have the velocity!

**unamed**)you can work out the volume, and therefore the mass.. and you have the velocity!

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#5

(Original post by

sorry i feel completely stupid but how do u work out the volume my mind has gone blank :|

**An0nymous**)sorry i feel completely stupid but how do u work out the volume my mind has gone blank :|

It's x-sectional area * speed.

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(Original post by

it's all right.

It's x-sectional area * speed.

**unamed**)it's all right.

It's x-sectional area * speed.

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#7

(Original post by

so what answer do u make it because i keep getting a different answer to the one on the mark scheme ?

**An0nymous**)so what answer do u make it because i keep getting a different answer to the one on the mark scheme ?

although, now I think about it, I might be wrong. >.<

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(Original post by

2.88 x 10^ -8

although, now I think about it, I might be wrong. >.<

**unamed**)2.88 x 10^ -8

although, now I think about it, I might be wrong. >.<

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#9

Careful, you don't have the actual speed, but just how much volume comes out each second. By using dimensional analysis (i.e. look at the units) its quite simple to calculate the speed. So first, calculate the mass of water present in that amount of volume, and then find the speed of the water flowing out.

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#10

The dimensional analysis the way I know it is a bit long..

A shorter way, but the same idea.

We need momentum per second, that's kgms

We have seconds in Volume/Time given only, so I'll square that to get s

We get m

OK.. Multiply with the density, we're left with kgm

Divide by the area and we get kgms

I dont really know what (V/T)

Spoiler:

Show

We need to know Momentum Per Second (Momentum/Time => p/t => mv/t => mst

So suppose this momentum per second is proportional to the density of water to some power α:

p/t ∝ ρ

and to the volume per second to some power β:

p/t ∝ (V/t)

and to the area to some power λ:

p/t ∝ A

So we get:

p/t = (ρ )

And in terms of the units:

kgms

OK then.. we need s

So β = 2 as there is no other second on the right side:

kgms

kgms

But then we get m

kgms

kgms

kgms

Lets see what we got. The units of mass are OK, the units of time too.

We only need to get m

1 = 3+2λ

-2 = 2λ

λ = -1

kgms

kgms

kgms

So we got:

p/t = ρ

or

p/t = ρ×(V/t)

That gives 5.6×10

^{-2}(kgms^{-2}, thats Newtons by the way)).So suppose this momentum per second is proportional to the density of water to some power α:

p/t ∝ ρ

^{α}and to the volume per second to some power β:

p/t ∝ (V/t)

^{β}and to the area to some power λ:

p/t ∝ A

^{λ}So we get:

p/t = (ρ )

^{α}×(V/t)^{β}×(A)^{λ}And in terms of the units:

kgms

^{-2}= (kgm^{-3})^{α}×(m^{3}s^{-1})^{β}×(m^{2})^{λ}OK then.. we need s

^{-2}on the right side, as it is on the left.So β = 2 as there is no other second on the right side:

kgms

^{-2}= (kgm^{-3})^{α}×(m^{3}s^{-1})^{2}×(m^{2})^{λ}kgms

^{-2}= (kgm^{-3})^{α}×(m^{6}s^{-2})×(m^{2})^{λ}But then we get m

^{6}and we need m^{1}as on the left. We also need kg^{1}. So α = 1.kgms

^{-2}= (kgm^{-3})^{1}×(m^{6}s^{-2})×(m^{2})^{λ}kgms

^{-2}= (kgm^{-3})×(m^{6}s^{-2})×(m^{2})^{λ}kgms

^{-2}= (kgm^{3}s^{-2})×(m^{2})^{λ}Lets see what we got. The units of mass are OK, the units of time too.

We only need to get m

^{1}from m^{3}. And we have m^{2λ}there.1 = 3+2λ

-2 = 2λ

λ = -1

kgms

^{-2}= (kgm^{3}s^{-2})×(m^{2})^{-1}kgms

^{-2}= kgm^{3}s^{-2}×m^{-2}kgms

^{-2}= kgms^{-2}So we got:

p/t = ρ

^{1}×(V/t)^{2}×A^{-1}or

p/t = ρ×(V/t)

^{2}/AThat gives 5.6×10

^{-2}.A shorter way, but the same idea.

We need momentum per second, that's kgms

^{-2}.We have seconds in Volume/Time given only, so I'll square that to get s

^{-2}.We get m

^{6}s^{-2}.OK.. Multiply with the density, we're left with kgm

^{3}s^{-2}.Divide by the area and we get kgms

^{-2}.I dont really know what (V/T)

^{2}means, but that gives the correct answer.
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(Original post by

The dimensional analysis the way I know it is a bit long..

A shorter way, but the same idea.

We need momentum per second, that's kgms

We have seconds in Volume/Time given only, so I'll square that to get s

We get m

OK.. Multiply with the desity, we're left with kgm

Divide by the area and we get kgms

I dont really know what (V/T)

**yadas**)The dimensional analysis the way I know it is a bit long..

Spoiler:

Show

We need to know Momentum Per Second (Momentum/Time => p/t => mv/t => mst

So suppose this momentum per second is proportional to the density of water to some power α:

p/t ∝ ρ

and to the volume per second to some power β:

p/t ∝ V/t

and to the area to some power λ:

p/t ∝ A

So we get:

p/t = (ρ )

And in terms of the units:

kgms

OK then.. we need s

So β = 2 as there is no other second on the right side:

kgms

kgms

But then we get m

kgms

kgms

kgms

Lets see what we got. The units of mass are OK, the units of time too.

We only need to get m

1 = 3+2λ

-2 = 2λ

λ = -1

kgms

kgms

kgms

So we got:

p/t = ρ

or

p/t = ρ×(V/t)

That gives 5.6×10

^{-2}(kgms^{-2}, thats Newtons by the way)).So suppose this momentum per second is proportional to the density of water to some power α:

p/t ∝ ρ

^{α}and to the volume per second to some power β:

p/t ∝ V/t

^{β}and to the area to some power λ:

p/t ∝ A

^{λ}So we get:

p/t = (ρ )

^{α}×(V/t)^{β}×(A)^{λ}And in terms of the units:

kgms

^{-2}= (kgm^{-3})^{α}×(m^{3}s^{-1})^{β}×(m^{2})^{λ}OK then.. we need s

^{-2}on the right side, as it is on the left.So β = 2 as there is no other second on the right side:

kgms

^{-2}= (kgm^{-3})^{α}×(m^{3}s^{-1})^{2}×(m^{2})^{λ}kgms

^{-2}= (kgm^{-3})^{α}×(m^{6}s^{-2})×(m^{2})^{λ}But then we get m

^{6}and we need m^{1}as on the left. We also need kg^{1}. So α = 1.kgms

^{-2}= (kgm^{-3})^{1}×(m^{6}s^{-2})×(m^{2})^{λ}kgms

^{-2}= (kgm^{-3})×(m^{6}s^{-2})×(m^{2})^{λ}kgms

^{-2}= (kgm^{3}s^{-2})×(m^{2})^{λ}Lets see what we got. The units of mass are OK, the units of time too.

We only need to get m

^{1}from m^{3}. And we have m^{2λ}there.1 = 3+2λ

-2 = 2λ

λ = -1

kgms

^{-2}= (kgm^{3}s^{-2})×(m^{2})^{-1}kgms

^{-2}= kgm^{3}s^{-2}×m^{-2}kgms

^{-2}= kgms^{-2}So we got:

p/t = ρ

^{1}×(V/t)^{2}×A^{-1}or

p/t = ρ×(V/t)

^{2}/AThat gives 5.6×10

^{-2}.A shorter way, but the same idea.

We need momentum per second, that's kgms

^{-2}.We have seconds in Volume/Time given only, so I'll square that to get s

^{-2}.We get m

^{6}.OK.. Multiply with the desity, we're left with kgm

^{3}s^{-2}.Divide by the area and we get kgms

^{-2}.I dont really know what (V/T)

^{2}means, but that gives the correct answer.
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#12

p = mv

m = density x volume.

= 1000 x 2.0 x 10^-4

v = d/t. d being the 'length component' of the water coming out the house

volume of cylinder = area x length(d)

(d) = volume / area

= 2.0 x 10^-4 / 7.2 x 10^-4

= 2 /7.2

p = mv, p = 1000 x 2.0 x 10^-4 x 2 /7.2

= 5.5(recurring) x 10^-2

= 5.6 x 10^-2 (approximately)

m = density x volume.

= 1000 x 2.0 x 10^-4

v = d/t. d being the 'length component' of the water coming out the house

volume of cylinder = area x length(d)

(d) = volume / area

= 2.0 x 10^-4 / 7.2 x 10^-4

= 2 /7.2

p = mv, p = 1000 x 2.0 x 10^-4 x 2 /7.2

= 5.5(recurring) x 10^-2

= 5.6 x 10^-2 (approximately)

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