Part A) A rollercoaster (m=300 kg) starts a the top of a track that is 25.0 mete

Question

Part A)

A rollercoaster (m=300 kg) starts a the top of a track that is 25.0 meters above the ground. It moves down to the bottom (ground level) of the frictionless track and then gains another 10.0 meters of height and runs into a spring. If this spring brings the roller coaster to a stop over a horizontal distance of 5.00 meters, what is the spring constant?

Part B)

What if the rollercoaster (m=300 kg) starts at the top of the track that is 25.0 meters above the ground and then runs into an identical rollercoaster cart (m=300 kg)? What is the final velocity of these two if they stick together at ground level?

Explanation / Answer

(A) Applying energy conservation,

PEi + KEi = PEf + KEf

(300 x 9.8 x 25 + 0 ) + 0 = ((300 x9.8 x 10) + (k (5^2 / 2))) + 0

73500 = 29400 + 12.5k

k = 3528 N/m .......Ans

(B) Applying energy conservation to find its speed at bottom,

(300 x 9.8 x 25) = 300 v^2 / 2

v = 22.1 m/s

NOw applying momenetum conservation,

(300 x 22.1) = (300 + 300) vf

vf = 11 m/s

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