$\frac{\mathrm{\Delta}p}{\mathrm{\Delta}t}=\frac{\mathrm{\Delta}(mv)}{t}=m\frac{\mathrm{\Delta}v}{\mathrm{\Delta}t}$

clovnerie0q
2022-09-30
Answered

Mass = m , momentum is $p=mv$. In time $\mathrm{\Delta}t$, momentum changes by $\mathrm{\Delta}p$, the rate of change of momentum is:

$\frac{\mathrm{\Delta}p}{\mathrm{\Delta}t}=\frac{\mathrm{\Delta}(mv)}{t}=m\frac{\mathrm{\Delta}v}{\mathrm{\Delta}t}$

$\frac{\mathrm{\Delta}p}{\mathrm{\Delta}t}=\frac{\mathrm{\Delta}(mv)}{t}=m\frac{\mathrm{\Delta}v}{\mathrm{\Delta}t}$

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ralharn

Answered 2022-10-01
Author has **15** answers

1) Yes indeed, the absence of a $\mathrm{\Delta}$ 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

$\mathrm{\Delta}v={v}_{2}-{v}_{1}\phantom{\rule{thinmathspace}{0ex}},$

and what will $\mathrm{\Delta}(4v)$ be? Why it will be

$\mathrm{\Delta}(4v)=4{v}_{2}-4{v}_{1}=4({v}_{2}-{v}_{1})=4\mathrm{\Delta}v\phantom{\rule{thinmathspace}{0ex}}.$

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

$\frac{\mathrm{\Delta}p}{\mathrm{\Delta}t}=m\frac{\mathrm{\Delta}v}{\mathrm{\Delta}t}$

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

$F=m\frac{\mathrm{\Delta}v}{\mathrm{\Delta}t}\phantom{\rule{thinmathspace}{0ex}},$

or

$F=\frac{\mathrm{\Delta}p}{\mathrm{\Delta}t}\phantom{\rule{thinmathspace}{0ex}},$

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).

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

$\mathrm{\Delta}v={v}_{2}-{v}_{1}\phantom{\rule{thinmathspace}{0ex}},$

and what will $\mathrm{\Delta}(4v)$ be? Why it will be

$\mathrm{\Delta}(4v)=4{v}_{2}-4{v}_{1}=4({v}_{2}-{v}_{1})=4\mathrm{\Delta}v\phantom{\rule{thinmathspace}{0ex}}.$

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

$\frac{\mathrm{\Delta}p}{\mathrm{\Delta}t}=m\frac{\mathrm{\Delta}v}{\mathrm{\Delta}t}$

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

$F=m\frac{\mathrm{\Delta}v}{\mathrm{\Delta}t}\phantom{\rule{thinmathspace}{0ex}},$

or

$F=\frac{\mathrm{\Delta}p}{\mathrm{\Delta}t}\phantom{\rule{thinmathspace}{0ex}},$

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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