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I know its physics but can someone help me out on this? Please...What is the resultant momentum of a system of two particles with these masses and velocities: 3.75kg m/s moving North and 4.2 kg moving Northwest at 2.3 m/s?

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is the 3.75kg also traveling at 3.75m/s?
wait I forgot...3.75 is moving at 5.6 m/s
I don't totally understand the concept of the question, but it might just be that you need to add the two momentum vectors together. Step 1: draw a picture|dw:1318046019675:dw|

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Now calculate the momemtum vectors.|dw:1318046160525:dw| Now add the two vectors.
p=mv P1=3.75*5.6=21N P2=4.2*2.3=9.66NW since its NW you know its 45degrees, so find normal vectors at that angle 9.66cos45 = 5.07 (i dont have calc) so your momentum will be 21N+5.07N = 26.07N
yes i need to add the momentum together but i can't get how. I thought I have to use law of sines and cosines but I am not confident dealing with it.
or you can also calculate NW vector based on your N vector and have a resulting momentum in direction of NW rather than N, however both answers should be correct.
To add vectors together they need to be in rectangular form, not polar form. If you have a vector in polar form as (r, theta), then its rectangular coordinates are (r cos(theta), r sin(theta)).
still confused?
You said you got: In respect to the other post I used Law of Sines and Cosines and got an answer of 17.26kgm/s at 35.45 degree Nof W
|dw:1318053871126:dw| When you add vectors together, the result will be the diagonal of the parallelogram they create. So, the result has to be between the two vectors. Where is your answer vector pointing?
Let P=9.66 Q=21 x=angle between the vectors which is 45 then magnitude of resultant is \[R=\sqrt{P^2+Q^2+2PQcos(x)}\]
and angle with respect to N-direction can be obtained from: \[R\times\cos\theta=Q+P\times\cos45\]

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