Q76P - University Physics with Modern Physics

Question

The spring of a spring gun has force constant k = 400 N/m and negligible mass. The spring is compressed 6.00 cm , and a ball with mass 0.0300 kg is placed in the horizontal barrel against the compressed spring. The spring is then released, and the ball is propelled out the barrel of the gun. The barrel is 6.00 cm long, so the ball leaves the barrel at the same point that it loses contact with the spring. The gun is held so that the barrel is horizontal. (a)Calculate the speed with which the ball leaves the barrel if you canignore friction. (b) Calculate the speed of the ball as it leaves the barrel if a constant resisting force of 6.00 N acts on the ball as it moves along the barrel. (c) For the situation in part (b), at what position along the barrel does the ball have the greatest speed, and what is that speed? (In this case, the maximum speed does not occur at the end of the barrel. 

Step-by-Step Solution

Answer

1. The speed of the ball at the time of leaving is $6.93 \mathrm{m}/\mathrm{s}$ .
2. The speed of the ball at the time of leaving is $4.90 \mathrm{m}/\mathrm{s}$ .
3. The maximum speed of the ball is $5.20 \mathrm{m}/\mathrm{s}$ .

Step 1: Energy stored in the spring

The amount of energy stored in the spring can be expressed in terms of,

${\mathbf{E}}_{\mathbf{R}}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}{\mathbf{kx}}^{\mathbf{2}}$ 

Where k and x are spring constant and deformation in the spring, respectively.

Step 2: Given data

Force constant, $\mathrm{k}=400 \mathrm{N}/\mathrm{m}$

Compressed distance, $\mathrm{x}=6.0 \mathrm{cm}\left(\frac{{10}^{-2} \mathrm{m}}{1\mathrm{cm}}\right)=0.06 \mathrm{m}$ 

Mass of the ball is, $\mathrm{m}=0.03 \mathrm{kg}$ 

The length of the barrel is, $\mathrm{L}=6.0 \mathrm{cm}\left(\frac{{10}^{-2} \mathrm{m}}{1\mathrm{cm}}\right)=0.06 \mathrm{m}$ .  

Step 3: (a) Find the speed of the ball at the time of leaving

From the law of conservation,

 

 $$\begin{gathered} \frac{1}{2}{\mathrm{kx}}^2=\frac{1}{2}{\mathrm{mv}}^2 \\ \left(400 \mathrm{N}/\mathrm{m}\right){\left(0.06 \mathrm{m}\right)}^2=\left(0.03 \mathrm{kg}\right){\mathrm{v}}^2 \\ \mathrm{v}=6.93 \mathrm{m}/\mathrm{s} \end{gathered}$$

 

Hence, the speed of the ball at the time of leaving is  $6.93 \mathrm{m}/\mathrm{s}$.

Step 4: (b) Find thespeed of the ball at the time of leaving

The magnitude of resisting force is, $F_r=6.0 N$

Then, again by using the law of conservation of energy we can determine the speed of the ball,

 

$$\begin{gathered} \frac{1}{2}{\mathrm{kx}}^2-{\mathrm{F}}_{\mathrm{r}} \mathrm{x}=\frac{1}{2}{\mathrm{mv}}^2 \\ \left(400 \mathrm{N}/\mathrm{m}\right){\left(0.06 \mathrm{m}\right)}^2-\left(6.0 \mathrm{N}\right)\left(0.06 \mathrm{m}\right)=\left(0.03 \mathrm{kg}\right){\mathrm{v}}^2 \\ \mathrm{v}=4.90 \mathrm{m}/\mathrm{s} \end{gathered}$$ 

 

Hence, the speed of the ball at the time of leaving is $4.90 \mathrm{m}/\mathrm{s}$ .

Step 5: (c) Find the maximum speed of the ball

Let the maximum speed occurs at x'  , 

And at that point, the net force is zero 

The distance moved by the ball is given as,

 

$$\begin{gathered} \mathrm{kx}\text{'}={\mathrm{F}}_{\mathrm{r}} \\ \mathrm{x}\text{'}=\frac{6.0 \mathrm{N}}{400 \mathrm{N}/\mathrm{m}} \\ \mathrm{x}\text{'}=0.0150\mathrm{m} \end{gathered}$$ 

 

The maximum speed is calculated by calculating the total work done,

 

$$\begin{gathered} \frac{1}{2}{\mathrm{mv}}_{\max }^2+{\mathrm{F}}_{\mathrm{r}}\left(\mathrm{x}-\mathrm{x}\text{'}\right)=\frac{1}{2}\mathrm{k}\left({\mathrm{x}}^2-\mathrm{x}{\text{'}}^2\right) \\ \frac{1}{2}\left(0.03 \mathrm{kg}\right){\mathrm{v}}_{\max }^2+\left(6.0 \mathrm{N}\right)\left(0.06 \mathrm{m}-0.015\mathrm{m}\right) \\ =\frac{1}{2}\left(400\mathrm{N}/\mathrm{m}\right)\left[{\left(0.06 \mathrm{m}\right)}^2-{\left(0.015 \mathrm{m}\right)}^2\right] \\ {\mathrm{v}}_{\max }=5.20 \mathrm{m}/\mathrm{s} \end{gathered}$$ 

 

Hence,the maximum speed of the ball is $5.20 \mathrm{m}/\mathrm{s}$.