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.
To solve part (a) of the problem, you correctly used the conservation of energy equation: 1/2mv^2 = mgy + 1/2kx^2. However, you mentioned that the TA claimed you need to account for gravitational potential energy at both ends. This is not necessary in this case because the potential energy is the same at both ends of the motion.
Plugging in the given values:
m = 0.00497 kg
y = x = 0.0406 m
k = -F2/x = 224.631 N/m
Now, let's solve for v. Rearrange the equation and substitute the given values:
1/2mv^2 = mgy + 1/2kx^2
1/2 * 0.00497 kg * v^2 = 0.00497 kg * 9.8 m/s^2 * 0.0406 m + 1/2 * 224.631 N/m * (0.0406 m)^2
0.00248 * v^2 = 0.020036 + 0.029329
0.00248 * v^2 = 0.049365
v^2 = 19.876612903225806
v = √19.876612903225806
v ≈ 4.46 m/s
Therefore, the correct velocity of the pellet when it loses contact with the spring is approximately 4.46 m/s, which is different from your original calculation.
For part (b), to find the maximum height the pellet will rise, you again correctly applied the conservation of energy equation: mgYc = mgYa + 1/2kx^2.
Using the same given values:
m = 0.00497 kg
Ya = x = 0.0406 m
k = -F2/x = 224.631 N/m
Rearrange the equation and substitute the given values:
mgYc = mgYa + 1/2kx^2
0.00497 kg * 9.8 m/s^2 * Yc = 0.00497 kg * 9.8 m/s^2 * 0.0406 m + 1/2 * 224.631 N/m * (0.0406 m)^2
0.048703 * Yc = 0.019421 + 0.029329
0.048703 * Yc = 0.04875
Yc ≈ 0.997 m
Therefore, the correct maximum height the pellet will rise from the original latched position is approximately 0.997 m, which is slightly different from your initial calculation.
It seems like your calculations were very close to the correct answers, but there might have been rounding errors or small mistakes in the calculations. Double-checking the calculations and being careful with significant figures should help you improve the accuracy of your final answers.