Recent content by hangten1039

Anyone have any help on this problem. I think I have to use conservation of energy but I really am not sure

A mass of 0.5 kg is attached to the end of a massless spring of spring constant 0.40 N/m. It is released from rest from an extended position. After 0.7 s, the speed of the mass is measured to be 1.75 m/s. What is the amplitude of oscillation? What is the total energy (relative to the mass at...

Thanks so much!

I got 6.942 m but that is not correct

well i set up the equation (5500 (x1) + 0) /6621 for the left side when the cannon balls haven't not moved and for the right i set up the equation 5500(x1) +1121(41) / 6621 but then I don't have x1 and also when i figuring that out I don't come up with the right answer

ok i got that, but still I have a problem figuring out what x's to use for each case I would say that the movement of the cannon balls would be 41 m but we don't know that for the car or the cannon

so xcm= total mass of everything + the right side of the car divided by the total mass xcm= 2121kg (x1) + 19kg (41m) / 2121 + 19 and then continue this for each time a cannon ball is fired I am still confused I don't comprehend how this equation works

i mean 1621 instead of 1121

i set up the equations as (5000(x1) + 1121(41)) = xcm but i don't know what to use for the x's and also what to use for the next equation to find the difference between the two

the final is 77% of the initial

so actually I would have to take 23% of the initial kinetic energy

I still have initial velocity as unknown or final velocity, so I'm not sure how to get rid of that

I'm not sure I understand what you are explaining in part 2.. so my equation would be .5m1(v1^2)= .23(.5)(m1+m2)(m1v1^2/(m1+m2))

nevermind i would actually get: .5(m1)((m1+m2)(vf)/m1) = .23(.5)(m1+m2)vf

so would i get for kinetic equation: .5m1(v1^2)= .23(.5)(m1+m2)(m1v1/(m1+m2))