A paddle ball toy consists of a flat wooden paddle and asmall rubber ball that are attached to each other by an elastic band (Figure 2.61). You have a paddle ball toy for which the mass of the ball is 0.015 kg, the stiffness of the elastic band is 0.9 N/m, and the relaxed length of the elastic band is 0.30 m. You are holding the paddle so the ball hangs suspended under it, when your cat comes along and bats the ball around, setting it in motion. At a particular instant the momentum of the ball is$\mathbf{⟨}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{02}\mathbf{,}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{01}\mathbf{,}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{02}\mathbf{⟩}\mathbf{kg}\mathbf{\cdot }\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{}$ , and the moving ball is at location $\mathbf{⟨}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{,}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{61}\mathbf{,}\mathbf{0}\mathbf{⟩}\mathbf{m}\mathbf{}\mathbf{}$ relative to an origin located at the point where the elastic band is attached to the paddle.

(a) Determine the position of the ball 0.1 s later, $\mathbf{}\mathit{\Delta }\mathit{t}\mathbf{}$using of 0.1 s.

(b) Starting with the same initial position

$\mathbf{(}\mathbf{⟨}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{2}\mathbf{,}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{61}\mathbf{,}\mathbf{0}\mathbf{⟩}\mathbf{m}\mathbf{)}$and momentum$\mathbf{(}\mathbf{⟨}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{02}\mathbf{,}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{01}\mathbf{,}\mathbf{-}\mathbf{0}\mathbf{.}\mathbf{02}\mathbf{⟩}\mathbf{}\mathbf{kg}\mathbf{\cdot }\mathbf{m}\mathbf{/}\mathbf{s}\mathbf{)}$ , determine the position of the ball 0.1 s later, using a of 0.05 s. (c) If your answers are different, explain why.

Mass of the ball is, $\mathrm{m}=0.15\mathrm{kg}$

Stiffness of elastic band,${\mathrm{k}}_{\mathrm{s}}=0.9\mathrm{N}/\mathrm{m}$

Relaxed strength,${\mathrm{L}}_0=0.30\mathrm{m}$

The initial Moment of inertia of the ball$\left(\overrightarrow{{\mathrm{p}}_{\mathrm{i}}}\right)=⟨-0.02,-0.01,-0.02⟩\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}$

The initial location of the ball is given${\mathrm{r}}_{\mathrm{i}}=⟨-0.2,-0.61,0⟩\mathrm{m}$

Gravitational force

$\overrightarrow{\mathbf{F}}\mathbf{}\mathbf{=}\overrightarrow{\mathbf{m}\mathbf{g}\mathbf{}}$

Force by a spring

$\overrightarrow{\mathbf{F}}\mathbf{=}\mathbf{-}{\mathit{k}}_{\mathbf{s}}\mathit{s}\hat{\mathbf{L}}$

The s is stretched, and the vector$\hat{\mathbf{L}}$extends from the point of attachment of the

spring to the movable end. ${\mathbf{k}}_{\mathbf{s}}$ is the “spring stiffness” (also called “spring constant”).

The forces acting on the ball are gravitational force and the force of spring

Gravitational force can be given as

Spring forces

We know that the position of the ball can be described by the equation

$r_f=r_1+\frac{\overrightarrow{p_i}}{m}\Delta t+\frac{1}{2}\frac{\overrightarrow{F_{net}}}{m}\Delta t^2$

Where $r_f$final position,$r_j$ initial position, the$p_i$ initial momentum$\overrightarrow{F_{net}}$ is net force and

$∆t$is time step.

Now substituting the values into the equation

$$\begin{gathered} {\mathrm{r}}_{\mathrm{f}}=⟨-0.2,-0.61,0⟩\mathrm{m}+\frac{\left(⟨-0.02,-0.01,-0.02⟩\mathrm{kg}\cdot \mathrm{m}/\mathrm{s}\right)}{(0.015\mathrm{kg})}(0.1\mathrm{s})+\frac{1}{2}\frac{⟨0.095,1.44,0⟩\mathrm{N}}{(0.015\mathrm{kg})}(0.1\mathrm{s}{)}^2 \\ {\mathrm{r}}_{\mathrm{f}}=⟨-0.2,-0.61,0⟩\mathrm{m}+⟨-0.133,-0.067,-0.133⟩\mathrm{m}+⟨0.032,0.048,0⟩\mathrm{m} \\ {\mathrm{r}}_{\mathrm{f}}=⟨-0.301,-0.629,-0.133⟩\mathrm{m} \end{gathered}$$

The position of the ball 0.1 s later, using $∆t$of 0.1 s, is $⟨-0.301,-0.629,-0.133⟩\mathrm{m}$

From part (a)

We know,

${\overrightarrow{F}}_{net}=⟨0.095,1.44,0⟩N$

Again, using the equation

${\mathrm{r}}_{\mathrm{f}}={\mathrm{r}}_1+\frac{\overrightarrow{{\mathrm{p}}_{\mathrm{i}}}}{\mathrm{m}}\mathrm{\Delta t}+\frac{1}{2}\frac{{\mathrm{F}}_{\mathrm{net}}}{\mathrm{m}}{\mathrm{\Delta t}}^2$

Where$r_f$ final position,$r_i$ initial position, the$p_j$ initial momentum$\overrightarrow{F_{net}}$ is net force and

$∆t$is time interval

Now substituting the values into the equation

The position of the ball 0.1 s later, using $∆t$of 0.5 s, is${\mathrm{r}}_{\mathrm{f}}=⟨-0.259,-0.631,-0.067⟩\mathrm{m}$

Yes, in both the cases we are getting the different answers, with of 0.1 s the final position is $⟨-0.301,-0.629,-0.133⟩\mathrm{m}.$and for$∆t$ of 0.5 s, the final position is ${\mathrm{r}}_{\mathrm{f}}=⟨-0.259,-0.631,-0.067⟩\mathrm{m}.$ This is because of using the different time steps. When we use smaller time steps, we will get a more accurate result.