A spring-loaded toy gun is used to shoot a ball straight up in the air. The ball reaches a...


A spring-loaded toy gun is used to shoot a ball straight up in the air. The ball reaches a maximum height H, measured from the equilibrium position of the spring. The same ball is shot straight up a second time from the same gun, but this time the spring is compressed only half as far before firing. How far up does the ball go this time? Neglect friction. Assume that the spring is ideal and that the distance by which the spring is compressed is negligible compared to H.

Application of Conservation of Energy:

According to the conservation of energy, energy can not be created or destroyed, it can only be converted from one energy to another energy. When there is no supply or dissipation of energy then the total energy of the system remains constant.

Answer and Explanation: 1

When the spring is compressed then it stores energy and when it releases then it transferred its stored energy to the ball as kinetic energy. And when the ball rises to some height then the kinetic energy of the ball is converted to potential energy.

When a spring is compressed by a distance {eq}k {/eq} then the potential energy stored in the spring is {eq}P=\dfrac12kx^2 {/eq}, where {eq}k {/eq} is the spring constant.

Let this storing of energy ultimately led the ball to rise a height of {eq}h {/eq}. Then according to the conservation of energy,

$$\begin{align} mgh=\dfrac12kx^2 \end{align} $$

Here, {eq}m {/eq} is the mass of the ball.

Hence we can find the relation:

$$\begin{align} mgh&=\dfrac12kx^2\\[.3 cm] h&=\dfrac{k}{2gh}x^2\\[.3 cm] \color{blue}{h}&\color{blue}{\propto x^2}\\[.3 cm] \Rightarrow \dfrac{h_2}{h_1}&=\left(\dfrac{x_2}{x_1}\right)^2 \end{align} $$


  • {eq}x_2=\dfrac{x_1}2 {/eq}
  • {eq}h_1=H {/eq}


  • {eq}h_2=? {/eq}

Using the formula we get:

$$\begin{align} \dfrac{h_2}{h_1}&=\left(\dfrac{x_2}{x_1}\right)^2\\[.3 cm] \dfrac{h_2}{H}&=\left(\dfrac{\frac{x_1}2}{x_1}\right)^2\\[.3 cm] \dfrac{h_2}{H}&=\dfrac14\\[.3 cm] h_2&=\dfrac H4 \end{align} $$

Hence, in the 2nd case, the ball will reach one fourth of the initial height.

Learn more about this topic:

Practice Applying Spring Constant Formulas


Chapter 17 / Lesson 11

In this lesson, you'll have the chance to practice using the spring constant formula. The lesson includes four problems of medium difficulty involving a variety of real-life applications.

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