Q28E - University Physics with Modern Physics

Question

A 22-caliber riffle bullet traveling at $\mathbf{350}\mathbf{m}\mathbf{/}\mathbf{s}$ strikes a large tree and penetrates it to a depth of $\mathbf{0}\mathbf{.}\mathbf{130}\mathbf{m}$. The mass of the bullet is $\mathbf{1}\mathbf{.}\mathbf{80}\mathbf{g}$. Assume a constant retarding force. (a) How much time is required for the bullet to stop? (b) What force, in Newton, does the tree exert on the bullet?

Step-by-Step Solution

Answer

(a) The required time for the bullet to stop is $7.43 \mathrm{x} {10}^{-4 }\mathrm{s}$.

(b) The magnitude of the force, does the tree exerts on the bullet is $8.48 \mathrm{x} {10}^{2 }\mathrm{N}$.

Step 1: Significance of Newton’s second law of motion:

The concept of expression of linear motion helps to determine the required time interval for the bullet to stop, whereas the concept of Newton’s second law helps to evaluate the force exerted by the tree on the bullet.

Step 2: Identification of the given data:

The given data can be listed below as,

The initial speed of a bullet is, $\mathrm{u}=350 \mathrm{m}/\mathrm{s}$.

The penetration depth in a tree is, $\mathrm{d}=0.130 \mathrm{m}$.

The mass of the bullet is, $\mathrm{m}=1.80 \mathrm{g}$.

Step 3: (a) Determine the required time for the bullet to stop:

The expression to calculate the acceleration of the bullet is expressed as follows:

$$\begin{gathered} {\mathrm{v}}^2={\mathrm{u}}^2+2\mathrm{ad} \\ \mathrm{a}=\left[\frac{{\mathrm{v}}^2-{\mathrm{u}}^2}{2\mathrm{d}}\right] \end{gathered}$$

Here, $\mathrm{v}$ is the final speed of the bullet whose value would be $0$ and $\mathrm{a}$ is the acceleration of the bullet.

Substituting the values in the above expression, we get:

$$\begin{gathered} \mathrm{a}=\left[\frac{{\left(0\right)}^2-{\left(350\mathrm{m}/\mathrm{s}\right)}^2}{2\left(0.130\mathrm{m}\right)}\right] \\ =-4.71 \mathrm{x} {10}^5 \mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

The negative sign indicates that the bullet decelerates during penetration in the tree.

The expression to calculate the required time for the bullet to stop is expressed as follows:

$$\begin{gathered} \mathrm{v}=\mathrm{u}+\mathrm{at} \\ \mathrm{t}=\left(\frac{\mathrm{v}-\mathrm{u}}{\mathrm{a}}\right) \end{gathered}$$

Here, $\mathrm{t}$ is the required time for the bullet to stop.

Substituting the values in the above expression, and you get:

$$\begin{gathered} \mathrm{t}=\left(\frac{0-350\mathrm{m}/\mathrm{s}}{-4.71\mathrm{x}{10}^5 \mathrm{m}/\mathrm{s}}\right) \\ =7.43\mathrm{x}{10}^{-4} \mathrm{s} \end{gathered}$$

Thus, the required time for the bullet to stop is $7.43\mathrm{x}{10}^{-4} \mathrm{s}$.

Step 4: (b) Determination of the magnitude of the force does the tree exerts on the bullet

The expression to calculate the magnitude of the force does the tree exerts on the bullet is expressed as follows:

$\mathrm{F}=\mathrm{ma}$

Here, $\mathrm{F}$ is the magnitude of the force does the tree exerts on the bullet.

Substituting the values in the above expression, and you have

$$\begin{gathered} \mathrm{F}=\left[\left(1.80\mathrm{g}\right)\mathrm{x}\left(\frac{{10}^{-3}\mathrm{Kg}}{1\mathrm{g}}\right)\left(4.71 \mathrm{x} {10}^5 \mathrm{m}/{\mathrm{s}}^2\right)\right] \\ =8.48 \mathrm{x} \mathrm{x}{10}^2\mathrm{Kg} \mathrm{m}/{\mathrm{s}}^2 \\ =\left(8.48 \mathrm{x} \mathrm{x}{10}^2\mathrm{Kg} \mathrm{m}/{\mathrm{s}}^2\right)\mathrm{x}\left(\frac{1\mathrm{N}}{1\mathrm{Kg}.\mathrm{m}/{\mathrm{s}}^2}\right) \\ =8.48 \mathrm{x} {10}^2 \mathrm{N} \end{gathered}$$

Thus, the magnitude of the force does the tree exerts on the bullet is$8.48 \mathrm{x} {10}^2\mathrm{N}$