How Do You Calculate Jet Speed, Momentum, and Forces in Physics Problems?

  • Thread starter zephyrus777
  • Start date
You can't use momentum here because the hose is attached to the ground, so there is an upward force on the hose. That upward force is equal to the downward force of the water jet (Newton's third law). So you have to use conservation of energy. You know the maximum height of the water, so you can calculate the potential energy at that point. From that you can get the kinetic energy at that point. Then use conservation of energy to find the speed of the water at the nozzle. In summary, the minimum speed of water from the hose is calculated by using conservation of energy and knowing the height it reaches.
  • #1
zephyrus777
2
0

Homework Statement



A jet of water from a fire hose is capable of reaching a height of 20m. What is the minimum speed of water from the hose? Given the area of cross section of the hose outlet is 4 x 10^(-4) m squared . calculate

a) The mass of water leaving the hose each second.
b)The momentum leaving the hose each second
c) the force on the hose due to the water jet. the density of water is 1000 kg/m cube

Homework Equations



f= dmv/dt
f= v x (dm/dt) (d is delta/small change)

P=mv

The Attempt at a Solution



I got stucked on first part.. no idea at all how to get the speed to 20 m/s (answer from my book says so :( )

----------------------------------------------------------------------------------

2) A car of mass 1200 kg traveling at high speed collides with a stationary car of mass 800kg. The two cars lock together on impact and slide a distance of 120m before stopping. Tests on the road surface show that the frictional force in the slide was 0.2 x the weight of the vehicles. Calculate
a) The decceleration of the two vehicles after impact.
b) the speed of the two vehicles immediately after impact.
c) the speed of the 1200kg car just before impact.

Homework Equations



m1u1 + m2u2 = (m1 + m2) v
the four linear dynamics equation..


The Attempt at a Solution



1200(u1) + 800(0) = (1200+800) v

then i make v as the subject and subtitute into the linear equations.. i got stucked as i got the equation in multiple terms and can't be simutanous equated.

thanks in advande for answering my questions, hopefully :D
 
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  • #2
zephyrus777 said:

Homework Statement



A jet of water from a fire hose is capable of reaching a height of 20m. What is the minimum speed of water from the hose?

...

I got stucked on first part.. no idea at all how to get the speed to 20 m/s (answer from my book says so :( )

Use conservation of energy. Just to fix the notation:

Let [itex]m[/itex] be the mass of water leaving the hose.
Let [itex]v[/itex] be the velocity of the water leaving the hose.
Let [itex]h[/itex] be the maximum height that the water reaches.

Taking the location of the hose to be [itex]y=0[/itex], can you write down the law of conservation of energy for this problem?
 
  • #3
1/2 m v^2 = mgh , when h= 20, find v.. oh thanks dude :D


how bout (b) :( sigh..
 
Last edited:
  • #4
That should be easy. You know v and you know the cross sectional area of the nozzle. That's all you need.
 

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