How High Does the Dart Travel When the Spring is Compressed Half as Far?

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When a dart is shot from a spring-loaded gun, the maximum height achieved is directly related to the compression of the spring, following Hooke's law. The elastic potential energy stored in the spring is calculated using the formula 0.5(k)x^2, while the gravitational potential energy at the dart's peak height is given by GPE = mgh. If the spring is compressed half as far, the potential energy becomes one-fourth of the original, leading to a proportional decrease in height. Therefore, when the spring is compressed half as much, the dart reaches a height of 6 meters. This illustrates the relationship between spring compression and the resulting height of the dart.
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Homework Statement


A spring-loaded dart gun is used to shoot a dart straight up into the air, and the dart reaches a maximum height of 24 meters. The same dart is shot up a second time from the same gun, but this time the spring is compressed only half as far (compared to the first shot). How far up does the dart go this time (neglect friction and assume the spring obeys Hooke's law)?


The Attempt at a Solution



I know I have to use 1/2(k)x^2 , but I am not quite sure how to set this problem up.
 
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All of your elastic potential energy is going to be transferred to gravitational potential energy; GPE = mgh. Your elastic potential energy, like you mentioned, is going to be 0.5kx^2. Look at the ratios that form; if the spring is compressed only half way, the potential energy is 1/4 of what it was when pulled back fully - which would proportionally change the height.

That makes sense in my head...it's hard to form into words, so I hope that somewhat makes sense!
 
So it would only travel 6 meters?
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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