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Momentum final velocity

by mortho
Tags: final, momentum, velocity
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mortho
#1
Jan3-08, 06:48 PM
P: 100
1. The problem statement, all variables and given/known data
A constant force of 4.0 N to the right acts on a 3.0 kg mass for 0.45 s.

a)Find the final velocity of the mass if it is initially at rest.

b)Find the final velocity of the mass if it is initially moving along the x-axis with a velocity of 12 m/s to the left.
2. Relevant equations

F=p/t

3. The attempt at a solution

okay so for the a i got 0.6 using vf=Ft/m and just plugged in. but for b i don't know what to do..need your help thanks!
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rock.freak667
#2
Jan3-08, 06:51 PM
HW Helper
P: 6,202
First part is correct.

Use the formula [itex]F=\frac{mv-mu}{t}[/itex] where m is mass,v is final velocity and u is the initial velocity and t is time.(Newton's 2nd law)

Also remember that velocity is a vector quantity so you should give a direction when you are stating your answers.
mortho
#3
Jan3-08, 07:02 PM
P: 100
ok i used the formula but i have a feeling i calculated it wrong..i got 28 m/s, to the left

rock.freak667
#4
Jan3-08, 07:14 PM
HW Helper
P: 6,202
Momentum final velocity

[tex]v=\frac{Ft+mu}{m}[/tex]

[tex]v=\frac{(4*0.45)+3*-12}{3}[/tex]


is that what you did?

which would give you 11.4 ms[itex]^{-1}[/itex] to the left
mortho
#5
Jan3-08, 07:16 PM
P: 100
so... -11.4 but my teacher said none of the answers are to be negative so would i still have the same answer but just without a negative sign?
rock.freak667
#6
Jan3-08, 07:19 PM
HW Helper
P: 6,202
Well a vector quantity has both magnitude and direction. So I just took the right as +ve and left as -ve. so the magnitude of the velocity is 11.4ms^-1
mortho
#7
Jan3-08, 07:26 PM
P: 100
Thanks! i guess my teacher would just assume left and right without the signs..that was a lot of help!:)


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