# College Physics

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1. A spring gun is made by compressing a spring ina tube and then altching the spring at the compressed position. A 4.97-g pellet is placed against the compressed and latched spring. The spring latches at a compression of 4.06 cm, and it takes a force of 9.12 N to compress the spring to that point.

a. If the gun is fired vertically how fast (m/s) is the pellet moving when it loses contact with the spring?

I thought for this you would write out conservation of energy formula:
1/2mv^2 = mgy+1/2kx^2 and solve for V
K = -F2/x=224.631 N/m
y=x=.0406m
m = .00497 kg

The TA said I'm getting it wrong because I have to account for gravitational potential energy at both ends by she's wrong in saying that because it doesn't make a difference in the equation since mgy = mg(y2-y1)

I get v = 8.68 m/s^2 but it's supposed to be 8.59 m/s^2

b. To what maximum height (m) will the pellet rise? (as measure from the original latched position)

For this again I wrote out the formula for energy conservation:
mgYc = mgYa + 1/2kx^2 and solved for Yc
K = -F2/x=224.631 N/m
Ya=x=.0406m
m = .00497 kg

Again I'm off by such a minor amount... I get 3.84 m instead of 3.80.

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