Recent content by xxphysics

No my question was if you could relate the momentum equation of an inelastic collision to the change in kinetic energy of that collision

Thank you :) I just thought they were both looking at the change in velocity and in both equations there is a way to account for the differences in mass (before and after collision) so I didn't think was absurd to wonder if there is a connection between the formulas.

In an inelastic collision is the change in kinetic energy equal to the difference of final and initial momentum if one of the objects is initially at rest? For example: m1v = (m1+m2)Vf -----> 0 = (m1+m2)Vf - m1v1 1/2(m1+m2)Vf^2 - 1/2m1v^2 = (m1+m2)Vf - m1v1 Or totally wrong? Thanks!

Yup! Thanks!

Meant -sin(10π*0.6) = 0, my bad

Sorry I thought numbers that didn't have the variable being derived weren't present in the derivation. The reason why you add "+c" to the end of integrations? And since cos(10π*0.6) = 0, whether the rest of the equation was there wouldn't matter since you're multiplying by 0. Anyways thank you :)

Also, thank you!

Ahh forgot the negative. v=35*10π*-sin(10πt) thank you! For acceleration would you just take the derivative of that for a = 35*10π*10π* -cos(10πt) ? Or since in v=35*10π*-sin(10πt) the sine function equaled zero and the rest didn't really matter was is supposed to be v=10π*-sin(10πt) in which...

I haven't taken calc in three years and my physics book doesn't exactly go over chain rule, that's why I came here for help. Would you derive with respect to t? v=35*10π*sin(10πt) ?

Derive then? I'm sorry I'm not really sure what the derivative would be?

I mean v = A(1/2*t^2)(sin10πt)

Intergrate ...? would the equation be v = A(1/2*t^2)(cos10πt) ?

Homework Statement A uniform disk of mass m and radius R lies in a vertical plane and is pivoted about a point a distance ℓcm from its center of mass in (Figure 1) . When given a small rotational displacement about the pivot, the disk undergoes simple harmonic motion. Determine the period of...

Homework Statement The position of a particle undergoing simple harmonic motion is given by x(t)=35cos(10πt), where x is in millimeters and t is in seconds. Determine the x component of velocity of the particle at t = 0.60 s . Homework Equations v = x/t The Attempt at a Solution I correctly...

mArA = mBrB ; mA = 3mB 3mBrA = mBrB 3rA = rB rA + rB = ℓ rA + 3rA = ℓ 4rA = ℓ rA/ℓ =1/4I got the answer of 3 which was correct. Thank you