Chapter 10, Problem 93E

a)

To determine

What is the angular speed of the board just after the bullet impact.

b)

To determine: What is the maximum height above the equilibrium position does the centre of the board reach before starting to swing down again.

Answer: The maximum height reached by the system is $0.317\text{cm}$.

Explanation:

Given info: The length of the sides of the square target is $0.250\text{m}$. The mass of the target is $0.750\text{kg}$. The mass of the bullet is $1.90\text{g}$. The speed of the bullet is $360\text{m}\cdot {\text{s}}^{\text{-1}}$.

Write the expression for the conservation of energy for collision time and the position where the target reaches its maximum height. Consider the centre of the target to be zero potential level.

$\frac{1}{2}I{\omega }^2=\left(m+M\right)gh$

• $I$ is the moment of inertia of the bullet-target system
• $\omega$ is the angular speed
• $M$ is the mass of the target
• $m$ is the mass of the bullet
• $g$ is the acceleration due to gravity
• $h$ is the maximum height reached by the system before moving down again

Re write the above equation to get an expression for the $h$.

$h=\frac{I{\omega }^2}{2m^{\prime }g}$

Write the expression for the moment of inertia of the system.

$I=\frac{1}{3}M{l}^2+m{\left(\frac{l}{2}\right)}^2$

• $l$ is the length of sides of the target square

Substitute the expression of the $I$ in the expression for $h$.

$h=\frac{{\omega }^2}{2m^{\prime }g}\left(\frac{1}{3}M{l}^2+m{\left(\frac{l}{2}\right)}^2\right)$

The length of the sides of the square target is $0.250\text{m}$. The mass of the target is $0.750\text{kg}$. The mass of the bullet is $1.90\text{g}$. The speed of the bullet is $360\text{m}\cdot {\text{s}}^{\text{-1}}$.

Substitute $0.250\text{m}$ for $l$, $0.750\text{kg}$ for $M$, $1.90\text{g}$ for $m$, $360\text{m}\cdot {\text{s}}^{\text{-1}}$ for $v$, $5.46\text{rad}\cdot {\text{s}}^{\text{-1}}$ for $\omega$, $9.\text{8}\text{m}\cdot {\text{s}}^{\text{-2}}$ for $g$

$\begin{array}{c}h=\frac{{\left(5.46\text{rad}\cdot {\text{s}}^{\text{-1}}\right)}^2}{2\left(1.90\text{g}+0.750\text{kg}\right)\left(9.\text{8}\text{m}\cdot {\text{s}}^{\text{-2}}\right)}\left(\frac{1}{3}\left(0.750\text{kg}\right){\left(0.250\text{m}\right)}^2+\left(1.90\text{g}\right){\left(\frac{0.250\text{m}}{2}\right)}^2\right)\\ =\frac{{\left(5.46\text{rad}\cdot {\text{s}}^{\text{-1}}\right)}^2}{2\left(\left(1.90\text{g}\right)\left({10}^{-3}\text{kg}\cdot {\text{g}}^{\text{-1}}\right)+0.750\text{kg}\right)\left(9.\text{8}\text{m}\cdot {\text{s}}^{\text{-2}}\right)}\left(\frac{1}{3}\left(0.750\text{kg}\right){\left(0.250\text{m}\right)}^2+\left(1.90\text{g}\right)\left({10}^{-3}\text{kg}\cdot {\text{g}}^{\text{-1}}\right){\left(\frac{0.250\text{m}}{2}\right)}^2\right)\\ =0.0317\text{m}\\ =0.317\text{cm}\end{array}$

Conclusion: The maximum height reached by the system is $0.317\text{cm}$.

c)

To determine

What is the minimum bullet speed required for the board to swing all the way over after impact.