91Ó°ÊÓ

The mass is uniformly distributed.

Velocity is proportional to the length.

v is the end velocity.

L is the total length.

M is the total mass.

The spring constant is $k=3200\mathrm{N}/\mathrm{m}$.

The mass of the bullet is $m=0.053\mathrm{kg}$.

The compression $x=2.5×{10}^{-2}\mathrm{m}$.

The expression of kinetic energy is given by,

$\mathit{K}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}$

Here, m is the mass of the body, and v is velocity of the body.

(a)

Assume that at the fixed point $I=0$, and at the moving end of the spring,$I=L$.

When the spring is moving, $I$will be the function of time, so the velocity of the point corresponding to$I$will be $u$. This velocity is also an implicit function of time.

$u\left(I\right)=\frac{vI}{L}$

The mass of the spring$\left(M\right)$is uniformly distributed along the length of the spring.

Consider a small piece of the spring with length $dI$. The mass is given by,

$dm=\frac{M}{L}dI$

The kinetic energy of the piece is given by,

$$\begin{gathered} dk=\frac{1}{2}dm·u^2 \\ =\frac{1}{2}\left(\frac{M}{L}dI\right)\left(\frac{vI}{L}\right) \\ =\frac{1}{2}\frac{m{v}^2}{L^3}{I}^{2}dI \end{gathered}$$

The kinetic energy of the total spring is given by,

$$\begin{gathered} K={∫}_0^{L}dK \\ ={∫}_0^L\frac{1}{2}\frac{m{v}^2}{L^3}{I}^{2}dI \\ =\frac{1}{2}\frac{m{v}^2}{L^3}{\left(\frac{I^3}{3}\right)}_0^L \\ =\frac{1}{6}M{v}^2 \end{gathered}$$

Hence, the kinetic energy is $K=\frac{1}{6}M{v}^2$.

(b)

It is given that the mass of the spring is negligible.

The kinetic energy of the spring is given by,

$$\begin{gathered} K=\frac{1}{2}m{v}^2=\frac{1}{2}k{x}^2 \\ v=\sqrt{\frac{k}{m}{x}^2}........\left(1\right) \end{gathered}$$

Substitute $0.053\mathrm{kg}$for m, 3200 N/m for k, and $2.5×{10}^{-2}\mathrm{m}$for x in equation (1).

$$\begin{gathered} v=\sqrt{\frac{\left(3200\mathrm{N}/\mathrm{m}\right)}{\left(0.053\mathrm{kg}\right)}{\left(2.5×{10}^{-2}\mathrm{m}\right)}^2} \\ =6.1\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the required velocity is 6.1 m/s.

(c)

The mass of the spring is $M=0.243\mathrm{kg}$.

If the mass of the spring is included, then

$$\begin{gathered} \frac{1}{2}k{x}^2=\frac{1}{2}m{v}^2+\frac{1}{6}M{v}^2 \\ v=\sqrt{\frac{k{x}^2}{\left(m+{\displaystyle \frac{M}{3}}\right)}} \end{gathered}$$ .........(2)

Substitute 0.053 kg for $m$, 3200 N/m for $k$, $0.243\mathrm{kg}$for $M$,and$2.5×{10}^{-2}\mathrm{m}$for$x$in equation (2).

$$\begin{gathered} v=\sqrt{\frac{\left(3200\mathrm{N}/\mathrm{m}\right){\left(2.5×{10}^{-2}\mathrm{m}\right)}^2}{\left(0.053\mathrm{kg}+{\displaystyle \frac{0.243\mathrm{kg}}{3}}\right)}} \\ =3.9\mathrm{m}/\mathrm{s} \end{gathered}$$

Therefore, the required velocity is $3.9\mathrm{m}/\mathrm{s}$.

(d)

The mass of the ball is $m_b=0.053\mathrm{kg}$.

Find the final kinetic energy of the ball as follows.

$$\begin{gathered} K_b=\frac{1}{2}{m}_b{v}^2 \\ =\frac{1}{2}\left(0.053kg\right){\left(3.9m/s\right)}^2 \\ ≈0.40\mathrm{J} \end{gathered}$$

Find the final kinetic energy of the spring as follows.

$$\begin{gathered} K_b=\frac{1}{6}M{v}^2 \\ =\frac{1}{6}\left(0.243\mathrm{kg}\right){\left(3.9\mathrm{m}/\mathrm{s}\right)}^2 \\ ≈0.60\mathrm{J} \end{gathered}$$

Therefore, the final kinetic energy of the ball and of the spring are 0.40 J and 0.60 J, respectively.