Mass = m, momentum is p=mv. In time Δt, momentum changes by Δp, the rate of change of momentum is: Δp/Δt=Δ(mv)/t=mΔv/Δt

clovnerie0q 2022-09-30 Answered
Mass = m , momentum is p = m v. In time Δ t, momentum changes by Δ p, the rate of change of momentum is:
Δ p Δ t = Δ ( m v ) t = m Δ v Δ t
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Answers (1)

ralharn
Answered 2022-10-01 Author has 15 answers
1) Yes indeed, the absence of a Δ in the second expression is just a typo.
2) The last expression is derived assuming that mass is a constant. If it helps, just set the mass equal to 4, or something. If we want to know how the quantity 4v changes, we really only need to know how the quantity v changes. Suppose v changes from v 1 to v 2 . Then
Δ v = v 2 v 1 ,
and what will Δ ( 4 v ) be? Why it will be
Δ ( 4 v ) = 4 v 2 4 v 1 = 4 ( v 2 v 1 ) = 4 Δ v .
So the constant comes out the front, because it doesn't change between the start and end points that you're considering. Hence it can be factored out.
3) It's important that I point out that
Δ p Δ t = m Δ v Δ t
is not Newton's second law. It's just a relationship between the rate of change of momentum and acceleration, and can be derived straight from the definitions of these things. Newton's second law cannot be derived, and is a statement of real physical content --- hence it is called a law. Newton's law can be written as either
F = m Δ v Δ t ,
or
F = Δ p Δ t ,
whichever you prefer. You can show these two are equivalent using the argument in your textbook (although in fact they're not quite equivalent, as I established above --- the first equation only holds when mass is constant; the second is more general, and more universally true).
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