Average momentum of an avalanche

  • Thread starter Thread starter I_Try_Math
  • Start date Start date
  • Tags Tags
    Momentum
AI Thread Summary
The discussion centers on calculating the average momentum of an avalanche, with a focus on understanding the correct formula and units. The initial calculation provided was incorrect due to a misunderstanding of average momentum, which should be defined as mass times average velocity (mv_avg). Participants noted that the original answer lacked proper units, leading to confusion about its validity. After reevaluating the calculations and incorporating the correct units, the user was able to arrive at the textbook's answer. This highlights the importance of unit consistency in physics calculations.
I_Try_Math
Messages
114
Reaction score
25
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.
 
Physics news on Phys.org
I_Try_Math said:
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?
 
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.
 
PeroK said:
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}##?
 
That would work.
 
  • Like
Likes PeroK, I_Try_Math and MatinSAR
kuruman said:
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.
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
Thread 'Voltmeter readings for this circuit with switches'
TL;DR Summary: I would like to know the voltmeter readings on the two resistors separately in the picture in the following cases , When one of the keys is closed When both of them are opened (Knowing that the battery has negligible internal resistance) My thoughts for the first case , one of them must be 12 volt while the other is 0 The second case we'll I think both voltmeter readings should be 12 volt since they are both parallel to the battery and they involve the key within what the...
Thread 'Trying to understand the logic behind adding vectors with an angle between them'
My initial calculation was to subtract V1 from V2 to show that from the perspective of the second aircraft the first one is -300km/h. So i checked with ChatGPT and it said I cant just subtract them because I have an angle between them. So I dont understand the reasoning of it. Like why should a velocity be dependent on an angle? I was thinking about how it would look like if the planes where parallel to each other, and then how it look like if one is turning away and I dont see it. Since...
Back
Top