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# M1 Help. watch

1. Why we can't apply Impulse-Momentum principle to a fixed wall ?
2. (Original post by FZka)
Why we can't apply Impulse-Momentum principle to a fixed wall ?
Momentum = mass times velocity.

Question: What is the mass of a FIXED wall?
3. (Original post by FZka)
Why we can't apply Impulse-Momentum principle to a fixed wall ?
are you doing m1 tomorrow morning? i just got that question. If you did apply momentum principle, the speeds of the particle are exactly the same before and after because the wall's velocity is always 0.
4. Take m1 as a ball and m2 as the wall.

If you use m1v1 +m1u1 = m2v2 + m2u2 you'd get that the m1v1= -m1u1 as the wall can not have a velocity.

Therefore you can't use the impulse momentum as if you use

Impulse= m1v1-m1u1 you'd get 0.
5. (Original post by SirMasterKey)
Take m1 as a ball and m2 as the wall.

If you use m1v1 +m1u1 = m2v2 + m2u2 you'd get that the m1v1= -m1u1 as the wall can not have a velocity.

Therefore you can't use the impulse momentum as if you use

Impulse= m1v1-m1u1 you'd get 0
From what I understood i think you meant I = m2v2-m2u2 = 0 ?

Plus. why would I use m1v1 +m1u1 = m2v2 + m2u2 ?
6. Momentum= m1(v1-u1)

since it's fixed, v1-u1 = 0

m1*0 = 0

Therefore Momentum = 0

Does that answer that bit?

Then impulse is rate of change of momentum (if I remember correctly). So it's 0 divided by the duration of the contact which in turn equals 0

Does that make sense?
7. (Original post by FZka)
From what I understood i think you meant I = m2v2-m2u2 = 0 ?

Plus. why would I use m1v1 +m1u1 = m2v2 + m2u2 ?
Yes as long as m2 is the mass of the wall.

Well I was simply assuming something was thrown at the wall otherwise why would there be need to find out the momentum of a wall.

It is the conversation of momentum and is using the same principle as the impulse= m1v1-m1u1.
8. (Original post by SirMasterKey)
Yes as long as m2 is the mass of the wall.

Well I was simply assuming something was thrown at the wall otherwise why would there be need to find out the momentum of a wall.

It is the conversation of momentum and is using the same principle as the impulse= m1v1-m1u1.
It's clear now. Thanks!
9. (Original post by Snoodle)
Momentum= m1(v1-u1)

since it's fixed, v1-u1 = 0

m1*0 = 0

Therefore Momentum = 0

Does that answer that bit?

Then impulse is rate of change of momentum (if I remember correctly). So it's 0 divided by the duration of the contact which in turn equals 0

Does that make sense?
Impulse is just the change in momentum.
10. (Original post by FZka)
Impulse is just the change in momentum.
so your OK now?

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