Chapter 8, Problem 21P

(a)

To determine

 Speed of the projectile when it leaves the barrel.

Answer to Problem 21P

Speed of the projectile when it leaves the barrel is $1.40\text{m}/\text{s}$.

Explanation of Solution

In the system elastic potential energy is due to the spring and as friction force is acting on the ball total energy of the system is due to kinetic energy, eleastic potential energy and energy due to frictional force.

When the cannon is fired initial speed of the ball is zero as initially ball is at rest so, initial kinetic energy of the system is zero and spring finally returns to equilibriam so final elastic potential energy is also zero.

Write the expression for change in kinetic energy of the system.   

  $\Delta K=\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2$                                                                                       (I)

Here, $\Delta K$ is change in kinetic energy, $v_f$ is final velocity, $v_i$ is initial velocity and $m$ is mass.

Write the expression for change inelastic potential energy.

  $\Delta U=\frac{1}{2}k{x}_f^2-\frac{1}{2}k{x}_i^2$                                                                                       (II)

Here, $\Delta U$ is change inelastic potential energy, $k$ is force constant, $x_i$ is initial stretch of the spring and $x_f$ is final stretch of the spring.

Write the expression for total internal energy.

    $\Delta E_{\mathrm{int}}=f_{k}d$                                                                                                  (III)

Here, $\Delta E_{\mathrm{int}}$ total internal energy stored in the system, $f_k$ is frictional force and $d$ is the distance covered by ball when it comes out of cannon.

Total energy of the system remains conserved therefore sum of all energies is equal to zero.

Write the expression for change conservation of energy for the system.

    $\Delta K+\Delta U+\Delta E_{\mathrm{int}}=0$                                                                                   (IV)

Substitute  $\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2$ for $\Delta K$, $\frac{1}{2}k{x}_f^2-\frac{1}{2}k{x}_i^2$ for $\Delta U$ and $f_{k}d$ for $\Delta E_{\mathrm{int}}$ in equation (IV).

    $\left(\frac{1}{2}m{v}_f^2-\frac{1}{2}m{v}_i^2\right)+\left(\frac{1}{2}k{x}_f^2-\frac{1}{2}k{x}_i^2\right)+\left(f_{k}d\right)=0$                                           (V)

Substitute  $0$ for $\frac{1}{2}m{v}_i^2$ and $0$ for  $\frac{1}{2}k{x}_f^2$ in equation (V).      

  $\left(\frac{1}{2}m{v}_f^2\right)-\left(\frac{1}{2}k{x}_i^2\right)+\left(f_{k}d\right)=0$

Multiply above equation by $\left(2\right)$.

    $\left(m{v}_f^2\right)-\left(k{x}_i^2\right)+\left(2f_{k}d\right)=0$

Rearrange the above equation for $v_f$.

    $v_f=\sqrt{\frac{\left(k{x}_i^2\right)-\left(2f_{k}d\right)}{m}}$                                                                                 (VI)

Conclusion:

Substitute $\left(5.30\times {10}^{-3}\text{kg}\right)$ for $m$, $\left(5.00\times {10}^{-2}\text{m}\right)$ for $x_i$, $\left(8.00\text{N}/\text{m}\right)$ for $k$, $\left(\text{15}\times {\text{10}}^{-2}\text{m}\right)$ for $d$ and $\left(\text{0}\text{.0320}\text{N}\right)$ for $f_k$ in equation  (VI).

    $\begin{array}{c}{v}_f=\sqrt{\frac{\left(\left(8.00\text{N}/\text{m}\right){\left(5.00\times {10}^{-2}\text{m}\right)}^2\right)-\left(2\left(\text{0}\text{.0320}\text{N}\right)\left(\text{15}\times {\text{10}}^{-2}\text{m}\right)\right)}{\left(5.30\times {10}^{-3}\text{kg}\right)}}\\ =\sqrt{\frac{0.0104\text{N}\cdot \text{m}}{\left(5.30\times {10}^{-3}\text{kg}\right)}}\\ =1.40\text{m}/\text{s}\end{array}$

Thus, Speed of the projectile when it leaves the barrel is $1.40\text{m}/\text{s}$.

(b)

To determine

The distance at which the ball has maximum speed.

Answer to Problem 21P

The distance at which the ball has maximum speed is $4.60\text{cm}$ .

Explanation of Solution

The ball has maximum speed when there is no further acceleration. It happens when the force due to spring acting on the ball is equal to frictional force.

    $F_s=f_k$                                                                                                      (VII)

Here, $F_s$ is force due to spring and $f_k$ is frictional force.

Write the expression for force due to spring on the ball.

    $F_s=kx\text{'}$

Here, $F_s$ is force due to spring, $k$ is force constant and $x\text{'}$ is distance at which ball has no further acceleration.

Substitute $kx\text{'}$ for $F_s$ in equation (VII)

    $kx\text{'}=f_k$

Rearrange the above equation for $x\text{'}$ .

    $x\text{'}=\frac{f_k}{k}$                                                                                                    (VIII)

Write the expression for distance where the speed is maximum.

    $x_{\max }=x-x\text{'}$                                                                                               (IX)

Here, $x_{\max }$ is distance where speed of the ball is maximum, $x$ is initial compression of the string.

Conclusion:

Substitute $\left(8.00\text{N}/\text{m}\right)$ for $k$ and $\left(\text{0}\text{.0320}\text{N}\right)$ for $f_k$ in equation (VIII).

    $\begin{array}{c}x\text{'}=\frac{\left(\text{0}\text{.0320}\text{N}\right)}{\left(8.00\text{N}/\text{m}\right)}\\ =0.004\text{m}\end{array}$

Substitute $0.004\text{m}$ for $x\text{'}$ and $\left(\text{0}\text{.05m}\right)$ for $x$ in equation (IX).

    $\begin{array}{c}{x}_{\max }=0.05\text{m}\text{- 0}\text{.004}\text{m}\\ \text{=0}\text{.046 m}\left(\frac{100\text{ cm}}{1\text{ m}}\right)\\ \text{=4}\text{.60 cm}\end{array}$

Thus, the distance at which the ball has maximum speed is $4.60\text{cm}$ .

(c)

To determine

Maximum speed at the point asked in part (b).

Answer to Problem 21P

Maximum speed at the point asked in part (b) is $1.79\text{m}/\text{s}$.

Explanation of Solution

At this point initial speed of the ball is zero as initially ball is at rest so, initial kinetic energy of the system is zero .

Substitute  $0$ for $\frac{1}{2}m{v}_i^2$, $x\text{'}$ for $x_f$ and $x_{\max }$ for  $d$ in equation (V).      

  $\left(\frac{1}{2}m{v}_f^2\right)+\left(\frac{1}{2}kx{\text{'}}^2-\frac{1}{2}k{x}_i^2\right)+\left(f_k{x}_{\max }\right)=0$

Rearrange above equation.

    $\frac{1}{2}m{v}_f^2=\frac{1}{2}k{x}_i^2-\frac{1}{2}kx{\text{'}}^2+f_k{x}_{\max }$

Multiply above equation by $\left(2\right)$ .

    $m{v}_f^2=k{x}_i^2-kx{\text{'}}^2+2f_k{x}_{\max }$

Rearrange above equation for $v_f$ .

    $v_f=\sqrt{\frac{k{x}_i^2-kx{\text{'}}^2+2f_k{x}_{\max }}{m}}$                                                                           (X)

Conclusion:

Substitute $\left(5.30\times {10}^{-3}\text{kg}\right)$ for $m$, $\left(5.00\times {10}^{-2}\text{m}\right)$ for $x_i$, $\left(4.0\times {10}^{-3}\text{m}\right)$ for $x\text{'}$, $\left(8.00\text{N}/\text{m}\right)$ for $k$, $\left(4.6\times {\text{10}}^{-2}\text{m}\right)$ for $x_{\max }$ and $\left(\text{0}\text{.0320}\text{N}\right)$ for $f_k$ in equation  (X).

    $\begin{array}{c}{v}_f=\sqrt{\frac{\left[\begin{array}{l}\left(\left(8.00\text{N}/\text{m}\right){\left(5.00\times {10}^{-2}\text{m}\right)}^2\right)-\left(\left(8.00\text{N}/\text{m}\right){\left(4.0\times {10}^{-3}\text{m}\right)}^2\right)\\ -\left(2\left(\text{0}\text{.0320}\text{N}\right)\left(4.60\times {\text{10}}^{-2}\text{m}\right)\right)\end{array}\right]}{\left(5.30\times {10}^{-3}\text{kg}\right)}}\\ =\sqrt{\frac{\left(0.02\text{N}\cdot \text{m}\right)-\left(1.280\times {10}^{-4}\text{N}\cdot \text{m}\right)-\left(2.94\times {10}^{-3}\text{N}\cdot \text{m}\right)}{\left(5.30\times {10}^{-3}\text{kg}\right)}}\\ =1.787\text{m}/\text{s}\\ \simeq 1.79\text{m}/\text{s}\end{array}$

Thus, the maximum speed at the point asked in part (b) is $1.79\text{m}/\text{s}$.