Math

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A ball of mass mm is loaded into a launcher with a spring of spring constant kk. The ball is pushed down until it is at a vertical position y=0y=0, and the spring is compressed a distance ΔL\Delta L, as shown. The ball is then released from rest. Immediately after leaving the launcher, the xx - and yy-components of the ball's velocity are vxv_{x} and vyv_{y}, respectively. The ball reaches a maximum height of ymax y_{\text {max }} and lands a horizontal distance Δx\Delta x away from its initial position. Energy losses due to friction are negligible. Which of the following is a correct conservation of energy equation that compares the total mechanical energy of the ball-spring-Earth system immediately before the ball is launched to the total mechanical energy of the ball-spring-Earth system the moment the ball reaches its maximum height?

Studdy Solution
According to the conservation of energy, the total mechanical energy before the ball is launched is equal to the total mechanical energy at the maximum height:
12k(ΔL)2=mgymax \frac{1}{2} k (\Delta L)^2 = mgy_{\text{max}}
This equation represents the conservation of energy for the ball-spring-Earth system.
The correct conservation of energy equation is:
12k(ΔL)2=mgymax \boxed{\frac{1}{2} k (\Delta L)^2 = mgy_{\text{max}}}

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