Recent content by mpittma1

Ok so before dividing I have x: M1V1iCOSθ1+M2V2iSINθ2=MtotVfCOS(0) y: M1V1iSINθ1+M2V2iCOSθ2=MtotVfSIN(0) look legit?

so the equation becomes TANΘ1+COTΘ2=0 Θ2=COT-1(-TANθ1) Θ2= -.025°

I can't see that working because the ending vector is directly along the x-axis, there is no sin or cos involved with that vector.

Homework Statement https://scontent-a-sjc.xx.fbcdn.net/hphotos-prn1/t1.0-9/603671_1409914319284787_3441607163537875560_n.jpg Homework Equations conservation of total momentum M1V1+M2V2 = MtotVf The Attempt at a Solution I tried using the COTM as such: x...

thank you! this is what I did: .5Kxc2=.5m1v12 v1=sqrt(kxc2/m1) = 1.02 m/s then ∫Fdt = m1v1= 1.02(2.4) = 2.448Ns Its not the exact same answer as the paper gives though, any thoughts on that?

Homework Statement https://scontent-a-sjc.xx.fbcdn.net/hphotos-frc1/l/t1.0-9/10155336_1407954212814131_2465716609293795371_n.jpg Homework Equations ∫Fdt = mvf-mvi The Attempt at a Solution I have tried integrating kxcdt from 0 - .01 and haven't had any luck. I am not quite...

Its the value of the work done by friction Wf=μk*m*g*d

ugh finally got it! thanks for all the help!

K I am getting close; .5m1v12=wf v1 = sqrt(29.4) = 5.422 and to find v2 v2=(m1v1/m2) = 1.36m/s

disregard that last post its the same mistake as the first time...

Ya Wnet=ΔKE So, set 100N = ΔKE of mass 1 and find the initial velocity produced by the spring?

would impulse momentum help here?

wf = ukmgd = (.75)(2)(9.8)(2) = 29.4 J

Ya i noticed that after i posted. i will try that though

Ya i see that; 100N = .5M1V12 so V1 = 10m/s right?