Q93CP - University Physics with Modern Physics

Question

A Spring with Mass. We usually ignore the kinetic energy of the moving coils of a spring, but let’s try to get a reasonable approximation to this. Consider a spring of mass M, equilibrium length ${\mathit{L}}_{\mathbf{0}}$, and force constant k. The work done to stretch or compress the spring by a distance L is $\frac{\mathbf{1}}{\mathbf{2}}\mathit{k}{\mathit{X}}^{\mathbf{2}}$, where $\mathit{X}\mathbf{=}\mathit{L}\mathbf{-}{\mathit{L}}_{\mathbf{0}}$. Consider a spring, as described above, that has one end fixed and the other end moving with speed v. Assume that the speed of points along the length of the spring varies linearly with distance l from the fixed end. Assume also that the mass M of the spring is distributed uniformly along the length of the spring. (a) Calculate the kinetic energy of the spring in terms of M and v. (Hint: Divide the spring into pieces of length dl; find the speed of each piece in terms of l, v, and L; find the mass of each piece in terms of dl, M, and L; and integrate from 0 to L. The result is not $\frac{\mathbf{1}}{\mathbf{2}}\mathit{M}{\mathit{v}}^{\mathbf{2}}$, since not all of the spring moves with the same speed.) In a spring gun, a spring of mass $0.243-\mathrm{kg}$ and force constant $3200 \mathrm{N}/\mathrm{m}$ is compressed $2.50 \mathrm{cm}$ from its upstretched length. When the trigger is pulled, the spring pushes horizontally on a $0.053-\mathrm{kg}$ ball. The work done by friction is negligible. Calculate the ball’s speed when the spring reaches its uncompressed length (b) ignoring the mass of the spring and (c) including, using the results of part (a), the mass of the spring. (d) In part (c), what is the final kinetic energy of the ball and of the spring?

Step-by-Step Solution

Answer

1. The kinetic energy is $K=\frac{1}{6}M{v}^2$.
2. The required velocity is 6.1 m/s .
3. The required velocity is 3.9 m/s .
4. The final kinetic energy of the ball and of the spring are 0.40 J and 0.60 J, respectively.

Step 1: Identification of given data

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 $x=2.5\times {10}^{-2} \mathrm{m}$.

Step 2: Concept/Significance of kinetic energy

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.

Step 3: Determine the kinetic energy of the spring in terms of M and v

(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,

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

 

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

Step 4: Determine the ball’s speed when the spring reaches its uncompressed length and ignore the mass of the spring

(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 $2.5\times {10}^{-2} \mathrm{m}$ for x in equation (1).

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

 

Therefore, the required velocity is 6.1 m/s.

Step 5: Determine the ball’s speed when the spring reaches its uncompressed length and including the mass of the spring

(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 $2.5\times {10}^{-2} \mathrm{m}$ for $x$ in equation (2).

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

 

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

Step 6: Determine the final kinetic energy of the ball and of the spring

(d)

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

 

Find the final kinetic energy of the ball as follows.

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

 

Find the final kinetic energy of the spring as follows.

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

 

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