91影视

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$\left(\stackrel{鈫�}{{\mathrm{p}}_{\mathrm{i}}}\right)=鉄�-0.02,-0.01,-0.02鉄�\mathrm{kg}鈰�\mathrm{m}/\mathrm{s}$

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

Gravitational force

$\stackrel{\mathbf{鈫�}}{\mathbf{F}}\mathbf{}\mathbf{=}\stackrel{\mathbf{鈫�}}{\mathbf{m}\mathbf{g}\mathbf{}}$

Force by a spring

$\stackrel{\mathbf{鈫�}}{\mathbf{F}}\mathbf{=}\mathbf{-}{\mathit{k}}_{\mathbf{s}}\mathit{s}\hat{\mathbf{L}}$

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 鈥渟pring stiffness鈥� (also called 鈥渟pring constant鈥�).

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

$r_f=r_1+\frac{\stackrel{鈫�}{p_i}}{m}螖t+\frac{1}{2}\frac{\stackrel{鈫�}{F_{net}}}{m}螖t^2$

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

$鈭�t$is time step.

Now substituting the values into the equation

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

The position of the ball 0.1 s later, using $鈭�t$of 0.1 s, is $鉄�-0.301,-0.629,-0.133鉄�\mathrm{m}$

From part (a)

We know,

${\stackrel{鈫�}{F}}_{net}=鉄�0.095,1.44,0鉄�N$

Again, using the equation

${\mathrm{r}}_{\mathrm{f}}={\mathrm{r}}_1+\frac{\stackrel{鈫�}{{\mathrm{p}}_{\mathrm{i}}}}{\mathrm{m}}\mathrm{螖迟}+\frac{1}{2}\frac{{\mathrm{F}}_{\mathrm{net}}}{\mathrm{m}}{\mathrm{螖迟}}^2$

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

$鈭�t$is time interval

Now substituting the values into the equation

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

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