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!
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...