Asked this in the chat thread, but it's probably going to be drained out by people talking about asian chicks and "nights on the town."
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At the top of a 3.5
meter tall hill, a cart of mass 3.5 kg has an initial velocity of 2 m/s.
It collides with another cart of mass 5.0 kg at the bottom of the hill
that is stationary, the collision is cushioned by a spring that is
compressed 1.0 cm when the first cart is still traveling 2.5 m/s.
Calculate a) the spring constant of the spring, and b) the maximum
possible compression of the spring. Assume no energy is lost to heat,
sound, etc.
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So what I did was:
(Note the random 1s and 2s are subscripts)
Etot (top of hill) = Ek1 + Eg1 + Ee1
No elastic potential energy obviously so...
Etot1 = 1/2MV1^2 + MGH
= (1/2)(3.5 kg)(2 m/s)^2 + (3.5 kg)(9.8 m/s^2)(3.5 m)
= sum value
Then I found the speed once it reaches the bottom of the hill (which will be V1 for the collision)
Etot2 = Ek2 + Eg2 + Ee2
No elastic or gravitation potential energy.
Etot2 = 1/2MV2^2
Etot2 = Etot1 (Conservation of Energy)
sum value = 1/2MV2^2
V2 = SQUAREROOT((2 * sumvalue)/(m))
= another value.
Got
kind of stumped from there. Conservation of momentum has got to be in
there somewhere, just not sure how to tie it in with either elastics
potential energy or F= kx
_______________________________________________________________________________________________________________________________________
At the top of a 3.5
meter tall hill, a cart of mass 3.5 kg has an initial velocity of 2 m/s.
It collides with another cart of mass 5.0 kg at the bottom of the hill
that is stationary, the collision is cushioned by a spring that is
compressed 1.0 cm when the first cart is still traveling 2.5 m/s.
Calculate a) the spring constant of the spring, and b) the maximum
possible compression of the spring. Assume no energy is lost to heat,
sound, etc.
________________________________________________________________________________________________________________________________________
So what I did was:
(Note the random 1s and 2s are subscripts)
Etot (top of hill) = Ek1 + Eg1 + Ee1
No elastic potential energy obviously so...
Etot1 = 1/2MV1^2 + MGH
= (1/2)(3.5 kg)(2 m/s)^2 + (3.5 kg)(9.8 m/s^2)(3.5 m)
= sum value
Then I found the speed once it reaches the bottom of the hill (which will be V1 for the collision)
Etot2 = Ek2 + Eg2 + Ee2
No elastic or gravitation potential energy.
Etot2 = 1/2MV2^2
Etot2 = Etot1 (Conservation of Energy)
sum value = 1/2MV2^2
V2 = SQUAREROOT((2 * sumvalue)/(m))
= another value.
Got
kind of stumped from there. Conservation of momentum has got to be in
there somewhere, just not sure how to tie it in with either elastics
potential energy or F= kx
