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momentum

here it says that the dart and the asteroid will move of together after collision
so i was using
m1v1 + m2v2 = v (m1+m2)
but this wasnt the case here
DF69228E-DF55-4FA8-B259-696613E6227B.jpg229.8KB

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@Roha125 helppppppppp
Reply 2
Avatar for mqb2766
mqb2766
21
Original post by Ailurophile03
here it says that the dart and the asteroid will move of together after collision
so i was using
m1v1 + m2v2 = v (m1+m2)
but this wasnt the case here

It works for me, what numbers did you substitute?
Reply 3
Avatar for Roha125
Roha125
15
Well I tried to do your way and couldn't do it,fell for the same trap

Then I realised why they made this question 'show that',you see mass of DART is really low in comparison to asteroid's mass, so taking mass of dart 0 after collision is okay,this helps to remove one the unknown variables
(edited 4 years ago)
Reply 4
Avatar for mqb2766
mqb2766
21
Original post by Roha125
Well I tried to do your way and couldn't do it,fell for the same trap

Then I realised why they made this question 'show that',you see mass of DART is really low in comparison to asteroid's mass, so taking mass of dart 0 after collision is okay,this helps to remove one the unknown variables

Please delete. Hints not solutions.
Reply 5
Avatar for Roha125
Roha125
15
Original post by mqb2766
Please delete. Hints not solutions.

Yeah yeah did it
Sometimes explaining complicated maths gets hard and hints becomes more confusing so i go with doing maths,if a student is good they will not cheat but understand the math even more but I know I can't know which student is which,so sorry I will comply with yours 'rules'

The first time I read those 'math solution rules' I said a movie quote from Us, out loud "What kind of White **** is this!"
(edited 4 years ago)
Reply 6
Avatar for mqb2766
mqb2766
21
Original post by Roha125
Yeah yeah did it
Sometimes explaining complicated maths gets hard and hints becomes more confusing so i go with doing maths,if a student is good they will not cheat but understand the math even more but I know I can't know which student is which,so sorry I will comply with yours 'rules'

The first time I read those 'math solution rules' I said a movie quote out loud "What kind of White **** is this!"


Thanks. Generally hints are better for understanding in the long run. The forum isn't a homework service and its not to do with colour.
(edited 4 years ago)
Original post by Roha125
Well I tried to do your way and couldn't do it,fell for the same trap

Then I realised why they made this question 'show that',you see mass of DART is really low in comparison to asteroid's mass, so taking mass of dart 0 after collision is okay,this helps to remove one the unknown variables


do i have to use conservation of momentum formula?😕
i mean in the exam i dont thin i will assume mass of dart is 0
Reply 8
Avatar for mqb2766
mqb2766
21
Original post by Ailurophile03
do i have to use conservation of momentum formula?😕
i mean in the exam i dont thin i will assume mass of dart is 0

The momentum formula is fine. What did you get when you try and put the numbers in?
Original post by mqb2766
The momentum formula is fine. What did you get when you try and put the numbers in?

image.jpg
6250 is the relative velocity of the dart and 0.0004 is the increase in velocity.
Have a think about what these terms mean and how you represent them.
You don't need to know the actual velocity, pre impact, of the asteroid.
Original post by Ailurophile03
image.jpg


thats why you have to assume that speed of dart to be zero to remove that constant

or else you can think that all momentum of dart has been lost and gained by asteroid(in this case momentum before after isn't required)

if you check answer by putting all known values you will see the 300 doesn't make a difference
Original post by Roha125
Well I tried to do your way and couldn't do it,fell for the same trap

Then I realised why they made this question 'show that',you see mass of DART is really low in comparison to asteroid's mass, so taking mass of dart 0 after collision is okay,this helps to remove one the unknown variables

Did you note the word 'relative' velocity?
Original post by Muttley79
Did you note the word 'relative' velocity?

Explain it then
I might learn something new or just an 'old thing' and know what it is called
Original post by Roha125
Explain it then
I might learn something new or just an 'old thing' and know what it is called

Why not Google it, if you'd like an explanation? Top link is a wiki page.
Original post by Roha125
Explain it then
I might learn something new or just an 'old thing' and know what it is called

If I'm travelling at 30km/h and you are travelling at 20 km/h in the same direction our relative velocity is 10 km/h [30 - 20] - if we were going in opposite direction then our relative velocity would be 50km/h
Original post by Muttley79
If I'm travelling at 30km/h and you are travelling at 20 km/h in the same direction our relative velocity is 10 km/h [30 - 20] - if we were going in opposite direction then our relative velocity would be 50km/h


oh this thing
i knew that as speed of approach = speed of seperation
Original post by mqb2766
Why not Google it, if you'd like an explanation? Top link is a wiki page.


did that at first and everything went over my head

so came to tsr cuz great people exist here
Original post by Roha125
did that at first and everything went over my head

so came to tsr cuz great people exist here

Do you understand now?
hmm
but is @Ailurophile03 's confusion solved

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