What are the Impulse and Forces Involved in a Fire Hose Collision?

[maskedup]

Homework Statement

A fire hose sends 20 kg of water per minute against a burning building. The water strikes the building at 40.0 m/s and is deflected 60deg as shown in the figure. (a) What is the rate of change of momentum of the water? (b) What force does the building exert on the water? (c) What force does the water exert on the building? (d) what are the magnitude and direction of the force exerted by the water on the roof?

Homework Equations

momentum = mv
impulse = F * t = mv1-mv0

The Attempt at a Solution

honestly, i dun know how to approach the question at all, hope someone could help me. its under the topic momentum and collisions.

 

[zorro]

The water jet gets deflected by 60 degrees with the vertical or horizontal?

 

[maskedup]

i am not sure about that...as that is all the information given. but I'm assuming its with the horizontal as that's the way it is shown on in the diagram.

 

[zorro]

Mass of water falling on the building per unit time = 20/60 kg/s = 1/3 kg/s
Impulse of force along the plane of the roof will be 0.
Impulse perpendicular to the plane of the roof will be 2mvcos60
Rate of change of momentum = 2mvcos60/t
from the first equation, m/t = 1/3 kg/s

This will be the force exerted by the water jet on the building and vice-versa.
The direction of force is perpendicular to the roof.

 

[maskedup]

hi, thanks for your response.
i have a few further questions.
a)for impulse of water, isn't it 1/3(kg/s) * 40(m/s) ? since it is rate of change in momentum, isn't it momentum/time?

b)why is the impulse on the roof 0? how does it relate to the impulse equation?

c) your impulse equation = 2mvcos60, why did u multiply by 2?

d) is force exerted on a plane, even at an angle, always perpendicular to the plane?

thanks for answering!

 

[zorro]

a) Impulse equals the change in momentum, not rate of change of momentum
b) and c) Impulse of roof on water can be resolved into two perpendicular components- One along the roof and the other perpendicular to it.

You need to draw a figure to understand it.
Impulse along the roof = mvsin60 - mvsin60 = 0
Impulse perpendicular to the roof = mvcos60 - (-mvcos60) [- sign because the direction changes]

d) Impulse of the plane on the colliding body will be always perpendicular to it provided the collision is perfectly elastic.

Force exerted on a plane by the body will be along its direction of velocity.