Average momentum of an avalanche

[I_Try_Math]

Homework Statement

What is the average momentum of an avalanche that moves a 40-cm-thick layer of snow over an area of 100 m by 500 m over a distance of 1 km down a hill in 5.5 s? Assume a density of 350 kg/ m^3 for the snow.

Relevant Equations

$\rho$ = mv

$m_s$ = mass of snow
$V_s$ = volume of snow
$\vec{v}$ = velocity of snow
D = density of snow

$\rho_{avg} = \frac{\rho}{V_s}$
$=\frac{m_s \vec{v}}{V_s}$
$=\frac{V_s D \vec{v}}{V_s}$
$=D \vec{v}$
$=350 \frac{1,000}{5.5}$
$=63,636.4$
The textbook's answer is $1.3 \times 10^9$. I guess I must not be understanding what's meant by "average momentum" since my answer is wrong. Any help is appreciated.

 

[PeroK]

Homework Statement: What is the average momentum of an avalanche that moves a 40-cm-thick layer of snow over an area of 100 m by 500 m over a distance of 1 km down a hill in 5.5 s? Assume a density of 350 kg/ m^3 for the snow.
Relevant Equations: $\rho$ = mv

$m_s$ = mass of snow
$V_s$ = volume of snow
$\vec{v}$ = velocity of snow
D = density of snow

$\rho_{avg} = \frac{\rho}{V_s}$
$=\frac{m_s \vec{v}}{V_s}$
$=\frac{V_s D \vec{v}}{V_s}$
$=D \vec{v}$
$=350 \frac{1,000}{5.5}$
$=63,636.4$
The textbook's answer is $1.3 \times 10^9$. I guess I must not be understanding what's meant by "average momentum" since my answer is wrong. Any help is appreciated.

That calculation doesn't make a lot of sense. First things first: How would you define average momentum for an avalanche?

 

[kuruman]

How do you know that your answer is wrong? You quote numbers without units so both you and the texbook can be correct in principle. Had you used units, you would have seen that momentum over volume, your starting equation, does not have units of momentum because it is divided by volume.

 

[I_Try_Math]

That calculation doesn't make a lot of sense. First things first: How would you define average momentum for an avalanche?

I suppose you could define it as $mv_{avg}$?

 

[kuruman]

That would work.

 

[I_Try_Math]

How do you know that your answer is wrong? You quote numbers without units so both you and the texbook can be correct in principle. Had you used units, you would have seen that momentum over volume, your starting equation, does not have units of momentum because it is divided by volume.

Right I will keep that in mind. Found out my original answer has units $\frac{kg}{m^2s}$. Obviously incorrect. I worked through it and got the correct answer.