Chapter 9, Problem 89AP

(a)

To determine

The final speed of bullet.

Answer to Problem 89AP

The final speed of bullet is $100\text{m/s}$.

Explanation of Solution

In collision of the bullet with the block the momentum of the system remains conserved.

Write the expression for the conservation of momentum.

    $p_i=p_f$                                                                                                          (I)

Here, $p_i$ is the initial momentum and $p_f$ is the final momentum.

Write the expression for the initial momentum of the system.

    $p_i=m{v}_i+M{V}_i$

Here, $m$ is the mass of bullet, $v_i$ is the initial velocity of bullet, $M$ is the mass of the block and $V_i$ is the initial velocity of block.

Substitute $0\text{m/s}$ for $V_i$ in above equation.

    $p_i=m{v}_i$                                                                                                        (II)

Write the expression for the final momentum of the system.

    $p_f=m{v}_f+M{V}_f$                                                                                          (III)

Here, $v_f$ is the final velocity of bullet and $V_f$ is the final velocity of block.

After the collision, the velocity acquired by block is used to compress the spring. Hence the kinetic energy is converted to potential energy.

Write the expression for conservation of energy.

    $K_E=U_S$                                                                                                      (IV)

Here, $K_E$ is the kinetic energy and $U_S$ is the spring potential energy.

Write the expression for the kinetic energy of the block.

    $K_E=\frac{1}{2}M{V}_f^2$                                                                                                 (V)

Write the expression for the spring potential energy.

    $U_S=\frac{1}{2}K{x}^2$                                                                                                 (VI)

Here, $U_S$ is the spring potential energy, $K$ is the spring constant and $x$ is the displacement.

Substitute $\frac{1}{2}M{V}_f^2$ for $K_E$ and $\frac{1}{2}K{x}^2$ for $U_S$ in equation (II).

    $\frac{1}{2}M{V}_f^2=\frac{1}{2}K{x}^2$

Simplify the above expression for value of $V_f$.

    $V_f=\sqrt{\frac{K{x}^2}{M}}$                                                                                                (VII)

Substitute $\sqrt{\frac{K{x}^2}{M}}$ for $V_f$ in equation (III) and Simplify for $p_f$.

    $p_f=m{v}_f+M\left(\sqrt{\frac{K{x}^2}{M}}\right)$                                                                          (VIII)

Substitute $m{v}_f+M\left(\sqrt{\frac{K{x}^2}{M}}\right)$ for $p_f$  and  $m{v}_i$ for $p_i$ in equation (I).

    $m{v}_i=m{v}_f+M\left(\sqrt{\frac{K{x}^2}{M}}\right)$

Simplify the above equation for value of $v_f$.

    $v_f=\frac{\left(m{v}_i-M\left(\sqrt{\frac{K{x}^2}{M}}\right)\right)}{m}$                                                                           (IX)

Conclusion:

Substitute $5\text{g}$ for $m$ , $400\text{m/s}$ for $v_i$ , $1\text{kg}$ for $M$, $900\text{N/m}$ for $K$ and $5\text{cm}$ for $x$ in equation (IX).

    $\begin{array}{c}{v}_f=\frac{\left(\left(5\text{g}\right)\left(400\text{m/s}\right)-\left(1\text{kg}\right)\left(\sqrt{\frac{\left(900\text{N/s}\right){\left(5\text{cm}\right)}^2}{\left(1\text{kg}\right)}}\right)\right)}{5\text{g}}\\ =\frac{\left(\left(5\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\right)\left(400\text{m/s}\right)-\left(1\text{kg}\right)\left(\sqrt{\frac{\left(900\text{N/s}\right){\left(5\text{cm}\left(\frac{1\text{m}}{100\text{cm}}\right)\right)}^2}{\left(1\text{kg}\right)}}\right)\right)}{5\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)}\\ =100\text{m/s}\end{array}$

Thus, the final speed of bullet is $100\text{m/s}$.

(b)

To determine

The amount of initial kinetic energy of bullet converted to internal energy of bullet block system.

Answer to Problem 89AP

The amount of initial kinetic energy of bullet converted to internal energy of bullet block system is $374\text{J}$.

Explanation of Solution

Difference in the final and initial kinetic energy gives the amount of internal energy.

Write the expression for the conservation of energy.

    $\Delta {{\rm K}}_E+\Delta {{\rm E}}_{\mathrm{int}}=0$

Here $\Delta K_E$ is the change in kinetic energy and $\Delta E_{\mathrm{int}}$ is the internal energy.

Simplify the above equation for value of  $\Delta E_{\mathrm{int}}$.

    $\Delta E_{\mathrm{int}}=-\Delta K_E$                                                                                               (X)

Write the expression for the $\Delta K_E$.

    $\Delta K_E=K_{Ef}-K_{Ei}$                                                                                        (XI)

Here $K_{Ef}$ is the final kinetic energy and $K_{Ei}$ is the initial kinetic energy.

Write the expression for the final kinetic energy.

    $K_{Ef}=\frac{1}{2}m{v}_f^2+\frac{1}{2}M{V}_f^2$                                                                               (XII)

Write the expression for initial kinetic energy.

    $K_{Ei}=\frac{1}{2}m{v}_i^2+\frac{1}{2}M{V}_i^2$                                                                               (XIII) 

Substitute $\frac{1}{2}m{v}_i^2+\frac{1}{2}M{V}_i^2$ for $K_{Ei}$ and $\frac{1}{2}m{v}_f^2+\frac{1}{2}M{V}_f^2$ for $K_{Ef}$ in equation (XI).

    $\Delta K_E=\left(\frac{1}{2}m{v}_f^2+\frac{1}{2}M{V}_f^2\right)-\left(\frac{1}{2}m{v}_i^2+\frac{1}{2}M{V}_i^2\right)$                                        (XIV)

Substitute  $\left(\frac{1}{2}m{v}_f^2+\frac{1}{2}M{V}_f^2\right)-\left(\frac{1}{2}m{v}_i^2+\frac{1}{2}M{V}_i^2\right)$ for $\Delta K_E$ in equation (X)

    $\Delta E_{\mathrm{int}}=-\left[\left(\frac{1}{2}m{v}_f^2+\frac{1}{2}M{V}_f^2\right)-\left(\frac{1}{2}m{v}_i^2+\frac{1}{2}M{V}_i^2\right)\right]$                                  (XV)

Conclusion:

Susbtitute $5\text{g}$ for $m$ , $400\text{m/s}$ for $v_i$ , $1\text{kg}$ for $M$, $100\text{m/s}$ for $v_f$, $0\text{m/s}$ for $V_i$ and $1.5\text{m/s}$ in equation (XV).

    $\begin{array}{c}\Delta E_{\mathrm{int}}=-\left[\begin{array}{l}\left(\frac{1}{2}\left(5\text{g}\right){\left(100\text{m/s}\right)}^2+\frac{1}{2}\left(1\text{kg}\right){\left(1.5\text{m/s}\right)}^2\right)\\ -\left(\frac{1}{2}\left(5\text{g}\right){\left(400\text{m/s}\right)}^2+\frac{1}{2}\left(1\text{kg}\right){\left(0\text{m/s}\right)}^2\right)\end{array}\right]\\ =-\frac{1}{2}\left[\begin{array}{l}\left(\left(5\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\right){\left(100\text{m/s}\right)}^2+\left(1\text{kg}\right){\left(1.5\text{m/s}\right)}^2\right)\\ -\left(\left(5\text{g}\left(\frac{1\text{kg}}{1000\text{g}}\right)\right){\left(400\text{m/s}\right)}^2+\left(1\text{kg}\right){\left(0\text{m/s}\right)}^2\right)\end{array}\right]\\ =373.75\text{J}\\ \simeq 374\text{J}\end{array}$

Thus, the amount of initial kinetic energy of bullet converted to internal energy of bullet block system is $374\text{J}$.