2. A force of 6.00 N acts on a 3.00 kg object for 10.0 s
a. What is the objects change in momentum
b. What is its change in velocity

Question

2. A force of 6.00 N acts on a 3.00 kg object for 10.0 s 
a. What is the objects change in momentum 
b. What is its change in velocity

theeric · Answer

Do you have an idea for part (a)?
The formula for it should be familiar! We used it before!

anonymous · Answer

is it p=ft?

theeric · Answer

Yep! Specifically, $\Delta p=\bar F\ \Delta t$

anonymous · Answer

isn't it just change of p = ft 
not change of t?

theeric · Answer

Well, if you're plotting $t$, then you start at $t=0$ and go to $t=10.0$, for example.

You can use it as you are, but then just know that then $t$ means the time interval!

theeric · Answer

Either way that you want to look at it is fine.

theeric · Answer

You should go with whatever you are most comfortable with.

anonymous · Answer

okay

anonymous · Answer

so then is it p = 6 x 10

theeric · Answer

Yes!

theeric · Answer

That is the change in momentum.
Also known as $p_2-p_1=\Delta p$ :)

anonymous · Answer

i thought it was just p=mv though

anonymous · Answer

also is the units Newton seconds

theeric · Answer

Those are the correct units!

You can look at it that way, too! In that case, $p$ is the momentum change. So it's the change in $mv$. And mass doesn't change, so $v$ is the velocity change!

Does that help with part (b)?

anonymous · Answer

sort of

anonymous · Answer

would the equation be m times change of V = change of p?

theeric · Answer

Yep! :) and you know $m$ and the change in momentum ($p$), so you can solve for the change in velocity ($v$)! :)

anonymous · Answer

so I get 20 m/s?

theeric · Answer

That's what I got!

anonymous · Answer

thanks

theeric · Answer

You're welcome! You did it; congratulations!