Chapter 10, Problem 70PQ

A ballistic pendulum is used to measure the speed of bullets. It comprises a heavy block of wood of mass M suspended by two long cords. A bullet of mass m is fired into the block horizontally. The block, with the bullet embedded in it, swings upward (Fig. P10.70). The center of mass of the combination rises through a vertical distance h before coming to rest momentarily. In a particular experiment, a bullet of mass 40.0 g is fired into a wooden block of mass 10.0 kg. The block–bullet combination is observed to rise to a maximum height of 20.0 cm above the block’s initial height. a. What is the initial speed of the bullet? b. What is the fraction of initial kinetic energy lost after the bullet is embedded in the block?

FIGURE P10.70

To determine

The initial speed of the bullet.

Answer to Problem 70PQ

The initial speed of the bullet is $497\text{m}/\text{s}$ .

Explanation of Solution

Write the expression of the conservation of linear momentum before and after collision.

  $m_{\text{bullet}}{\overrightarrow{\text{v}}}_{\text{bullet}}=\left(m_{\text{bullet}}+M_{\text{block}}\right)\overrightarrow{\text{V}}$

Here, $m_{\text{bullet}}$ is the mass of bullet, $M_{\text{block}}$ is the mass of the block, ${\overrightarrow{\text{v}}}_{\text{bullet}}$ is the velocity of the bullet before collision and $\overrightarrow{\text{V}}$ is the final velocity combined block-bullet system.

Rearrange above equation to get ${\overrightarrow{\text{v}}}_{\text{bullet}}$.

  ${\overrightarrow{\text{v}}}_{\text{bullet}}=\frac{\left(m_{\text{bullet}}+M_{\text{block}}\right)\overrightarrow{\text{V}}}{m_{\text{bullet}}}$                                                                                         (I)

According to conservation of mechanical energy, kinetic energy of the bullet-block system immediately after collision is equal to gravitational potential energy of the bullet-block system at maximum displacement.

Write the mathematical expression for conservation of energy.

  $K_i=U_f$                                                                                                                  (II)

Here, $K_i$ is the kinetic energy of the bullet-block system immediately after the collision and $U_f$ is the gravitational potential energy of the bullet-block system at maximum displacement.

Write the expression for $K_i$ .

  $K_i=\frac{1}{2}\left(m_{\text{bullet}}+M_{\text{block}}\right)V^2$                                                                                     (III)

Write the expression for $U_f$ .

  $U_f=\left(m_{\text{bullet}}+M_{\text{block}}\right)gh$                                                                                       (IV)

Here, $h$ is the maximum height that system can attain.

Put equations (III) and (IV) in equation (II) and rearrange it to get $V$ .

  $\begin{array}{c}\frac{1}{2}\left(m_{\text{bullet}}+M_{\text{block}}\right)V^2=\left(m_{\text{bullet}}+M_{\text{block}}\right)gh\\ V=\sqrt{2gh}\end{array}$

Substitute $\sqrt{2gh}$ for $V$ in (I) to get $v_{\text{bullet}}$ .

  $v_{\text{bullet}}=\frac{\left(m_{\text{bullet}}+M_{\text{block}}\right)\left(\sqrt{2gh}\right)}{m_{\text{bullet}}}$                                                                             (V)

Conclusion:

Substitute $40.0\text{g}$ for $m_{\text{bullet}}$, $10.0\text{kg}$ for $m_{\text{block}}$, $9.81{\text{m}/\text{s}}^{\text{2}}$ for $g$ and $20.0\text{cm}$ for $h$ in equation (V) to get $v_{\text{bullet}}$ .

  $\begin{array}{c}{v}_{\text{bullet}}=\frac{\left(40.0\text{g}\times \frac{1\text{kg}}{1000\text{g}}+10.0\text{kg}\right)\left(\sqrt{2\left(9.81{\text{m}/\text{s}}^{\text{2}}\right)\left(20.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)}\right)}{40.0\text{g}\times \frac{1\text{kg}}{1000\text{g}}}\\ =497\text{m}/\text{s}\end{array}$

Therefore, the initial speed of the bullet is $497\text{m}/\text{s}$ .

To determine

The fraction of initial kinetic energy lost after the bullet is embedded in the block.

Answer to Problem 70PQ

The initial kinetic energy of the bullet is lost by $99.6\%$ .

Explanation of Solution

The collision of bullet with block results in loss of some initial kinetic energy so that final kinetic energy after impact might be less than initial kinetic energy.

Initial kinetic energy of the system is equal to kinetic energy of the bullet before collision.

Write the expression for the initial kinetic energy.

  $K_i=\frac{1}{2}{m}_{\text{bullet}}{v}_{\text{bullet}}^2$                                                                                                    (VI)

Lose of kinetic energy is equal to difference between the final kinetic energy after the impact and initial kinetic energy of the bullet.

Final kinetic energy after the impact is equal to final potential energy of the block-bullet system at maximum displacement position.

Write the expression for the final kinetic energy.

  $K_f=U_f$

Substitute $\left(m_{\text{bullet}}+M_{\text{block}}\right)gh$ for $U_f$ in above equation to get $K_f$ .

  $K_f=\left(m_{\text{bullet}}+M_{\text{block}}\right)gh$                                                                                       (VII)

Write the expression for the percentage change in kinetic energy.

  $\frac{\Delta K}{K_i}\times 100=\frac{K_i-K_f}{K_i}\times 100$                                                                                  (VIII)

Conclusion:

Substitute $40.0\text{g}$ for $m_{\text{bullet}}$ and $497\text{m}/\text{s}$ for $v_{\text{bullet}}$ in equation (VI) to get $K_i$ .

  $\begin{array}{c}{K}_i=\frac{1}{2}\left(40.0\text{g}\times \frac{1\text{kg}}{1000\text{g}}\right){\left(497\text{m}/\text{s}\right)}^2\\ =4.94\times {10}^3\text{J}\end{array}$

Substitute $40.0\text{g}$ for $m_{\text{bullet}}$, $10.0\text{kg}$ for $m_{\text{block}}$, $9.81{\text{m}/\text{s}}^{\text{2}}$ for $g$ and $20.0\text{cm}$ for $h$ in equation (VII) to get $K_f$ .

  $\begin{array}{c}{K}_f=\left(40.0\text{g}\frac{1\text{ kg}}{1000\text{ g}}+10.0\text{kg}\right)\left(9.81{\text{m}/\text{s}}^{\text{2}}\right)\left(20.0\text{cm}\times \frac{1\text{m}}{100\text{cm}}\right)\\ =19.7\text{J}\end{array}$

Substitute $4.94\times {10}^3\text{J}$ for $K_i$ and $19.7\text{J}$ for $K_f$ in equation (VIII) to get the percentage of loss.

  $\begin{array}{c}\frac{\Delta K}{K_i}\times 100=\frac{4.94\times {10}^3\text{J}-19.7\text{J}}{4.94\times {10}^3\text{J}}\times 100\\ =0.996\times 100\\ =99.6\%\end{array}$

Therefore, the initial kinetic energy of the bullet is lost by $99.6\%$ .