Answer:
The second dart leaves the gun two times as faster than the first one.
Explanation:
Assuming no energy loss during the spring-dart energy transfer, we have by the conservation of energy principle
[tex] U_s = K_d \\ \frac{1}{2} kx^2 = \frac{1}{2}mv^2 \\ v = \sqrt{\frac{k}{m}x^2}. [/tex]
Given an arbitrary [tex] x [/tex] and its double, [tex] 2x [/tex], launch velocities are
[tex] v_1 = \sqrt{\frac{k}{m}x^2} \text{ and} \\ v_2 = \sqrt{\frac{k}{m}\left(2x\right)^2} = \sqrt{\frac{k}{m}4x^2} = 2\sqrt{\frac{k}{m}x^2} = \mathbf{2v_1}. [/tex]
