Question

Two children are playing a game in which they try to hit a small box on the floor with a marble fired from a spring-loaded gun that is mounted on a table. The target box is horizontal distance D = 2.20 mfrom the edge of the table; see Figure. Bobby compresses the spring 1.10cm, but the center of the marble falls 27.0 cmshort of the center of the box. How far should Rhoda compress the spring to score a direct hit? Assume that neither the spring nor the ball encounters friction in the gun.

Short answer

The compression of the spring in the second shot is $x_2=1.25m$

Step-by-step solution

Step 1: Given

1. The horizontal distance of the target box from the edge of the table is, D = 2.20 m
2. The compression of the spring is x = 1.10 cm = 0.011 m
3. In the first shot, the horizontal distance covered by the marble is, d = 27.0 cm = 0.27 m.

Determining the concept

Use the concept of the energy conservation law and elastic potential energy of the spring. Find the horizontal distance covered by the marble in the first and second shot by using kinematical equations. According to the law of energy conservation, energy can neither be created, nor be destroyed.

Formulae:

$$\begin{gathered} x=v_{0}t+\frac{1}{2}a{t}^2 \\ U=mgh \\ U\left(x\right)=\frac{1}{2}k{x}^2 \\ K=\frac{1}{2}m{v}^2 \end{gathered}$$

where, K is kinetic energy, Uis potential energy, m is mass, v is velocity, g is an acceleration due to gravity, x is displacement,a is an acceleration, k is spring constant, t is time and h is height.

Determining thecompression of the spring in the second shot

According to the figure, the marble has horizontal as well as vertical motion. The horizontal distance covered by the marble is,

$x=v_{0}t$ (i)

For the vertical motion, the initial vertical velocity of the marble is zero. According to the second kinematical equation, the vertical distance covered by the marble is,

$$\begin{gathered} h=v_{0}t+\frac{1}{2}g{t}^2 \\ h=\frac{1}{2}g{t}^2 \\ t=\sqrt{\frac{2h}{g}} \end{gathered}$$

Equation (i) becomes,

$x=v_0\sqrt{\frac{2h}{g}}$ (ii)

The horizontal distance covered by the marble is directly proportional to the initial velocity of the marble.

Let, $v_{01}$and $v_{02}$be the initial speed of the first and second shot of the marble respectively and $D_1$and D be the horizontal distances covered by the marble respectively.

$$\begin{gathered} D_1=D-d \\ {D}_1=2.20\mathrm{m}-0.27\mathrm{m} \\ {D}_1=1.93\mathrm{m} \end{gathered}$$

From equation (ii), as,

$D_1=v_{01}$ (iii)

$D=v_{02}$ (iv)

Dividing equation (iii) by (iv),

$\frac{D_1}{D}=\frac{v_{01}}{v_{02}}$ (v)

The spring is compressed by the marble.Hence, it has elastic potential energy. According to the energy conservation law, the elastic potential energy of the spring is converted into the kinetic energy of the marble,

$\frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2$

The compression of the spring is directly proportional to the initial velocity of the marble. Hence,

$$\begin{gathered} x_1=v_{01} \\ {x}_2=v_{02} \end{gathered}$$

The equation (v) becomes,

$$\begin{gathered} \frac{D_1}{D}=\frac{x_1}{x_2} \\ {x}_2=\frac{D}{D_1}{x}_1 \\ {x}_2=\frac{2.2.\mathrm{m}\times 0.011\mathrm{m}}{1.93\mathrm{m}} \\ {x}_2=1.25\mathrm{m} \end{gathered}$$

Hence, the compression of the spring in the second shot is$x_2=1.25\mathrm{m}$

Therefore,the compression of the spring in the second shot can be found by using the concept of conservation of energy and the elastic potential energy.