A spring gun is made by compressing a spring in a tube and then latching the spr

Question

A spring gun is made by compressing a spring in a tube and then latching 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 4.47 cm, and it takes a force of 9.12 N to compress the spring to that point. (Note: while the spring is being compressed the ball is not in contact with the spring.)

(a) If the gun is fired vertically, how fast is the pellet moving when it loses contact with the spring? (Include the effect of gravity and assume that the pellet leaves the spring when the spring is back to its relaxed length.)

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

I know answer for (a) is 9.008m/s and answer for (b) 4.185m but I don't know how to get there. I've tried many different formulas and ways but I cannot get to those answers. Any help pls?

Explanation / Answer

a) F = kx

9.12 = k (0.0447)

k = 204.03 N/m

initial spring PE = kx^2 /2 = 204.03 (0.0447^2)/2 = 0.204 J

iniial gravitationalPE = 0 (reference point )

KE = 0

final spring PE = 0

GPE= mgh = 4.97 x 10^-3 kg x 9.8 x 0.0447 = 0.00218 J

using energy conservation,

0.204 + 0 + 0 = 0 + 0.00218 + mv^2/2

v = 9 m/s

b) it will rise until its speed becomes zero.

mgh = mv^2 /2

h = 9^2 / (2 x 9.8) = 4.14 m

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