text: # Momentum Question
  
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  #1
  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
  0
  10 years ago
  #2
  you can work out the volume, and therefore the mass.. and you have the velocity!
  0
  #3
  (Original post by unamed)
  you can work out the volume, and therefore the mass.. and you have the velocity!
  sorry i feel completely stupid but how do u work out the volume my mind has gone blank :|
  0
  10 years ago
  #4
  0
  10 years ago
  #5
  (Original post by An0nymous)
  sorry i feel completely stupid but how do u work out the volume my mind has gone blank :|
  it's all right.
  It's x-sectional area * speed.
  0
  #6
  (Original post by unamed)
  it's all right.
  It's x-sectional area * speed.
  so what answer do u make it because i keep getting a different answer to the one on the mark scheme ?
  0
  10 years ago
  #7
  (Original post by An0nymous)
  so what answer do u make it because i keep getting a different answer to the one on the mark scheme ?
  2.88 x 10^ -8
  
  although, now I think about it, I might be wrong. >.<
  0
  #8
  (Original post by unamed)
  2.88 x 10^ -8
  
  although, now I think about it, I might be wrong. >.<
  0
  10 years ago
  #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.
  0
  10 years ago
  #10
  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-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 ∝ ρα
  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 ∝ Aλ
  
  So we get:
  p/t = (ρ )α×(V/t)β×(A)λ
  
  And in terms of the units:
  kgms-2 = (kgm-3)α×(m3s-1)β×(m2)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×(m3s-1)2×(m2)λ
  
  kgms-2 = (kgm-3)α×(m6s-2)×(m2)λ
  
  But then we get m6 and we need m1 as on the left. We also need kg1. So α = 1.
  
  kgms-2 = (kgm-3)1×(m6s-2)×(m2)λ
  
  kgms-2 = (kgm-3)×(m6s-2)×(m2)λ
  
  kgms-2 = (kgm3s-2)×(m2)λ
  
  Lets see what we got. The units of mass are OK, the units of time too.
  We only need to get m1 from m3. And we have m there.
  1 = 3+2λ
  -2 = 2λ
  λ = -1
  
  kgms-2 = (kgm3s-2)×(m2)-1
  
  kgms-2 = kgm3s-2×m-2
  
  kgms-2 = kgms-2
  
  So we got:
  p/t = ρ1×(V/t)2×A-1
  or
  p/t = ρ×(V/t)2/A
  
  That 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 m6s-2.
  OK.. Multiply with the density, we're left with kgm3s-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.
  0
  #11
  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-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 ∝ ρα
  and to the volume per second to some power β:
  p/t ∝ V/tβ
  and to the area to some power λ:
  p/t ∝ Aλ
  
  So we get:
  p/t = (ρ )α×(V/t)β×(A)λ
  
  And in terms of the units:
  kgms-2 = (kgm-3)α×(m3s-1)β×(m2)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×(m3s-1)2×(m2)λ
  
  kgms-2 = (kgm-3)α×(m6s-2)×(m2)λ
  
  But then we get m6 and we need m1 as on the left. We also need kg1. So α = 1.
  
  kgms-2 = (kgm-3)1×(m6s-2)×(m2)λ
  
  kgms-2 = (kgm-3)×(m6s-2)×(m2)λ
  
  kgms-2 = (kgm3s-2)×(m2)λ
  
  Lets see what we got. The units of mass are OK, the units of time too.
  We only need to get m1 from m3. And we have m there.
  1 = 3+2λ
  -2 = 2λ
  λ = -1
  
  kgms-2 = (kgm3s-2)×(m2)-1
  
  kgms-2 = kgm3s-2×m-2
  
  kgms-2 = kgms-2
  
  So we got:
  p/t = ρ1×(V/t)2×A-1
  or
  p/t = ρ×(V/t)2/A
  
  That 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 m6.
  OK.. Multiply with the desity, we're left with kgm3s-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.
  lol thanks this was a multiple choice question worth 1 mark!
  1
  4 years ago
  #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)
  1
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