Recent content by joe426

v1f = -v1 + v2 - v2f is really v1f = -v1 + v2 + v2f. .. (m1v1 + m2v2 - m1v1f) / m2 = v2f turns into (m1v1 + m2v2 + m1v1f) / m2 = v2f so v2f = .307m/s

v1f = -v1 + v2 - v2f Plugged that into, (m1v1 + m2v2 - m1v1f) / m2 = v2f v2f = -.368m/s

First use conservation of momentum ( m1v12 + m2v22 - m2v2f2 ) / m1 = v1f2 Then plug this v1f into this helper equation, the third equation listed in first post. v2f = v1 - v2 + [( m1v12 + m2v22 - m2v2f2 ) / m1 ]1/2 And I'm stuck because I can't get both v2f to one side so i can solve...

I don't think any of the velocity equals 0 at any point. Initially they are moving in the same direction. Then since it says its perfectly elastic, one object will bounce back and have a velocity in the opposite direction. I think the 60g tennis ball will bounce back in the negative direction...

Homework Statement Homework Equations 1/2m1v12 + 1/2m2v22 = 1/2m1v1f + 1/2m2v2f m1v1 + m2v2 = m1v1f + m2v2f v1 - v2 = -(v1f - v2f) The Attempt at a Solution So I solved the momentum of conservation for the final velocity of object 1. I then plug that equation into the third...

I am wrong. The inal position is .15m because that's how far the spring is compressed an that's how far the ball moves before its released from the spring V= 8.3m/s for part a Y= 3.6m for part b Thanks for the help!

Using yfinal = 3.63m. I got the velocity to be 7m/s

Ok, I understand this and I've come up with: 1/2mvi2 = mgyf + 1/2mvf2 vf = sqrt [ vi2 - 2(gyf) ] vf = 7.67m/s

oh ok. i understand. the fastest the ball will be going is right after its released from the spring, not at its highest position in the air. So its, 1/2mvi2 + 1/2kxi2 = mgyf vi = sqrt( 2gyf - kxi2 ) But now I'm stuck not knowing what it's final position is.

KEi = 0 because the initial velocity is 0. And would make 1/2mv2 = 0 And PEf = 0 because y is the vertical distance to move the object. we don't need to move the object because its already at its finaly position of 37m. If wrong on this and I should use 37m for y and solve for v, I end up with 0

So, since we can ignore resistance, the energy is conserved and the potential energy at the beginning equals the kinetic energy at the end. PEi = KEf mgy = 1/2mv2 v = sqrt(2gy) v = sqrt(2 * 9.8m/s2 * 37m) v= 26.93m/s

First I found yf, which is the answer to part b. Since energy is being conserved, I came up with PEf = KEi. This seems weird because usually there is no potential energy at the final position but it was the only thing I could come up with. mgyf = 1/2kxi2 y= (1/2kxi2) / g y=3.63m Then...

Homework Statement Homework Equations Since energy is conserved, MEinital = MEfinal The Attempt at a Solution 1/2 mvi2 + mgyi + 1/2kxi2 = 1/2 mvf2 + mgyf + 1/2kxf2 After finding what equals 0, I am left with: 1/2kxi2 = 1/2mvf2 + mgyf v=5.97m/s I found yf by: mgyf = 1/2kxi2...

Homework Statement Homework Equations F=ma W= Fdcosθ The Attempt at a Solution I first found the acceleration: t= d/v t= 37m/28m/s = 1.32s a = v/t a = 28m/s / 1.32s a= 21.21 m/ss Second, the force: Fclub - Fgsinθ = ma Fclub = a +gsinθ Fclub = 21.21m/ss + 9.8m/ss...

v = sqrt (2gy) v = sqrt (2 * 9.8m/ss * 35m) = 26.19 m/s at point 2 But K2 doesn't equal P2 because K2 equaled P1 at a height of 35m, not 28m, right?