Chapter 3, Problem 82P

(a)

To determine

The acceleration of the bullet.

Answer to Problem 82P

The bullet has an acceleration of $3.20\times {10}^4{\text{m/s}}^{\text{2}}$.

Explanation of Solution

Write the kinematic equation for the motion of the bullet.

    $v=v_0+a_b\Delta t$        (I)

Here, $v$ is the final velocity of the bullet, $v_0$ is the initial velocity of the bullet, $a_b$ is the acceleration of the bullet and $\Delta t$ is the time taken by the bullet to leave the barrel.

Rewrite equation (I) to obtain an equation for $a_b$.

    $a_b=\frac{v-v_0}{\Delta t}$

Conclusion:

Substitute $320\text{m/s}$ for $v$, $0\text{m/s}$ for $v_0$ and $0.01\text{s}$ for $\Delta t$.

    $\begin{array}{c}{a}_b=\frac{320\text{m/s}-0\text{m/s}}{0.01\text{s}}\\ =3.20\times {10}^4{\text{m/s}}^{\text{2}}\end{array}$

Therefore, the acceleration of the bullet is $3.20\times {10}^4{\text{m/s}}^{\text{2}}$.

(b)

To determine

The force on the bullet.

Answer to Problem 82P

The force on the bullet is $96.0\text{N}$.

Explanation of Solution

Write the equation for the force on the bullet using Newton’s second law.

    $F_b=m{a}_b$

Here, $F_b$ is the force on the bullet, $m$ is the mass of the bullet and $a_b$ is the acceleration of the bullet.

Conclusion:

Substitute $3.00\times {10}^{-3}\text{kg}$ for $m$ and $3.20\times {10}^4{\text{m/s}}^{\text{2}}$ for $a_b$.

    $\begin{array}{c}{F}_b=\left(3.00\times {10}^{-3}\text{kg}\right)\left(3.20\times {10}^4{\text{m/s}}^{\text{2}}\right)\\ =96.0\text{N}\end{array}$

Therefore, the force on the bullet is $96.0\text{N}$.

(c)

To determine

The magnitude of force exerted on the gun.

Answer to Problem 82P

The magnitude of force exerted on the gun is $96.0\text{N}$.

Explanation of Solution

According to Newton’s third law of motion, for every action, there is an equal and opposite reaction. Hence write the equation for the force exerted on the gun.

    $F_b=-F_r$

Here, $F_b$ is the force on the bullet and $F_r$ is the reverse force exerted by the bullet on the gun.

Conclusion:

Substitute $96.0\text{N}$ for $F_b$.

    $F_r=-96.0\text{N}$

Therefore, the magnitude of force exerted on the gun is $96.0\text{N}$. The negative sign in the answer indicates that the reaction is in the opposite direction.

(d)

To determine

The acceleration experienced by the gun

Answer to Problem 82P

The gun experiences an acceleration of $-18.5{\text{m/s}}^{\text{2}}$.

Explanation of Solution

Write the equation for the force on the gun using Newton’s second law.

    $F_r=M{a}_r$

Here, $F_b$ is the force on the gun, $M$ is the mass of the gun and $a_r$ is the acceleration of the gun.

Conclusion:

Substitute $-96.0\text{N}$ for $F_r$ and $5.20\text{kg}$ for $M$ to solve for $a_r$.

    $\begin{array}{c}-96.0\text{N}=\left(5.20\text{kg}\right)a_r\\ a_r=\frac{-96.0\text{N}}{5.20\text{kg}}\\ =-18.5{\text{m/s}}^{\text{2}}\end{array}$

Therefore, the acceleration experienced by the gun is $-18.5{\text{m/s}}^{\text{2}}$.

(e)

To determine

The ratio of mass $m$ and $M$ to that of the acceleration.

Answer to Problem 82P

The acceleration ration is the inverse mass ratio.

Explanation of Solution

Write the ratio for mass $m$ to $M$ and substitute the values.

    $\begin{array}{c}\frac{M}{m}=\frac{5.20\text{kg}}{3.00\times {10}^{-3}\text{kg}}\\ =1.73\times {10}^3\end{array}$        (II)

Write the ratio of the acceleration and substitute.

    $\begin{array}{c}\frac{a_b}{a_r}=\frac{3.20\times {10}^4{\text{m/s}}^{\text{2}}}{-18.5{\text{m/s}}^{\text{2}}}\\ =-1.73\times {10}^3\end{array}$        (III)

Conclusion:

From equation (II) and (III), the acceleration ration is the inverse ratio of the mass.