Chapter 3, Problem 95P

(a)

To determine

Acceleration of the bullet.

Answer to Problem 95P

Acceleration is $\underset{\_}{-5.6\times {10}^5\text{m}/{\text{s}}^{\text{2}}}$.

Explanation of Solution

Then using the equation of motion

    $v^2=v_0^2+2aS$        (I)

Here $v$ is the final velocity, $v_0$ is the initial velocity, $S$ is the distance covered and $a$ is the acceleration

The final velocity is zero. Then rewrite (I) in terms of $a$

    $a=\frac{-v_0{}^2}{2S}$        (II)

Conclusion:

Substitute $300\text{m}/\text{s}$ for $v_0$, $8.1\text{cm}$ for $S$ in (II)

    $\begin{array}{c}a=\frac{-{\left(300\text{m}/\text{s}\right)}^2}{2\left(8.1\text{cm}\left(\frac{{10}^{-2}\text{m}}{1\text{cm}}\right)\right)}\\ =-5.6\times {10}^5\text{m}/{\text{s}}^{\text{2}}\end{array}$

Acceleration is $\underset{\_}{-5.6\times {10}^5\text{m}/{\text{s}}^{\text{2}}}$

(b)

To determine

Time taken by the bullet to come to rest.

Answer to Problem 95P

The time is $\underset{\_}{\text{5}\text{.4}\times {\text{10}}^{-\text{4}}\text{s}}$.

Explanation of Solution

Write the equation of motion

    $v=v_0 +at$        (III)

Substitute zero for final velocity and rearrange (III) in terms of $t$

    $t=\frac{-v_0}{a}$        (IV)

Conclusion:

Substitute $300\text{m}/\text{s}$ for $v_0$, $-5.6\times {10}^5\text{m}/{\text{s}}^{\text{2}}$ for $a$ in (IV)

    $\begin{array}{c}t=\frac{-\left(300\text{m}/\text{s}\right)}{\left(-5.6\times {10}^5\text{m}/{\text{s}}^{\text{2}}\right)}\\ =5.4\times {10}^{-4}\text{s}\end{array}$

The time is $\underset{\_}{\text{5}\text{.4}\times {\text{10}}^{-\text{4}}\text{s}}$.

(c)

To determine

Force exerted by the wood on the bullet

Answer to Problem 95P

Force exerted by the wood on the bullet is $\underset{\_}{\text{5}\text{.6}\times {\text{10}}^{\text{3}}\text{N}}$.

Explanation of Solution

The only horizontal force acting on the bullet as it decelerated was a force of kinetic friction.

The net force is given by

    $\begin{array}{c}\Sigma F=F_{frictional}\\ =ma\end{array}$        (V)

Here $m$ is the mass of the bullet

Conclusion:

Substitute $10\text{g}$ for $m$ and $-5.6\times {10}^5\text{m}/{\text{s}}^{\text{2}}$ for $a$ in (V)

    $\begin{array}{c}{F}_{frictional}=\left(10\text{g}\left(\frac{{10}^{-3}\text{kg}}{1\text{g}}\right)\right)\left(-5.6\times {10}^5\text{m}/{\text{s}}^{\text{2}}\right)\\ =5.6\times {10}^3\text{N}\end{array}$

Force exerted by the wood on the bullet is $\underset{\_}{\text{5}\text{.6}\times {\text{10}}^{\text{3}}\text{N}}$.