Question

The high-speed winds around a tornado can drive projectiles into trees, building walls, and even metal traffic signs. In a laboratory simulation, a standard wood toothpick was shot by pneumatic gun into an oak branch. The toothpick’s mass was $\mathbf{0}\mathbf{.}\mathbf{13}\mathbf{g}$, its speed before entering the branch was$\mathbf{220}\mathbf{}\mathbf{m}\mathbf{/}\mathbf{s}$ , and its penetration depth was role="math" localid="1657007959021" $\mathbf{15}\mathbf{}\mathbf{mm}$ . If its speed was decreased at a uniform rate, what was the magnitude of the force of the branch on the toothpick?

Short answer

Force of branch on the toothpick is $2.1\times {10}^2\mathrm{N}$.

Step-by-step solution

Given information

1. Mass of the toothpick is$\mathrm{m}=0.13\mathrm{g}$ ,
2. Initial speed is ${\mathrm{v}}_0=220\mathrm{m}/\mathrm{s}$
3. Penetrating distance is $∆\times =15\mathrm{mm}\left(\frac{{10}^{-3}\mathrm{m}}{1\mathrm{mm}}\right)=0.015\mathrm{m}$

Understanding the concept of stopping force

Newton’s second law of motion states that the force acting on the object is equal to the product of mass and acceleration. The acceleration is equal to the time rate of change of velocity. When the final velocity is smaller than the initial velocity, then it is called deceleration. It means the force is acting to stop the moving object. This force is also referred to as stopping force.

Force acting on a toothpick is stopping force. Therefore, by using initial velocity and final velocity (considered as zero) we can calculate its acceleration. In addition, by using that acceleration and mass we can conclude its force.

Formulae:

${\mathbf{v}}_{\mathbf{f}}^{\mathbf{2}}\mathbf{}\mathbf{=}{\mathbf{v}}_{\mathbf{0}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathbf{a}\mathbf{∆}\mathbf{\times }$ (1)

localid="1657008578053" $\mathbf{F}\mathbf{=}\mathbf{ma}$ (2)

Calculation of the acceleration of the toothpick 

We know the initial velocity, final velocity, and distance travel, so we can use the kinematic equation given by equation (i) to calculate the acceleration,

$$\begin{gathered} {\mathrm{v}}_{\mathrm{f}}^2={\mathrm{v}}_0^2+2\mathrm{a}∆\times \\ \mathrm{a}=\frac{{\mathrm{v}}_{\mathrm{f}}^2-{\mathrm{v}}_0^2}{2∆\times } \end{gathered}$$

Substitute the values, and we get

a =role="math" localid="1657009447305" $$\begin{gathered} \mathrm{a}=\frac{{\left(0\mathrm{m}/\mathrm{s}\right)}^2-{\left(220\mathrm{m}/\mathrm{s}\right)}^2}{2\times \left(0.015\mathrm{m}\right)} \\ =-1.61\times {10}^6\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Hence, the acceleration of the toothpick is $-1.61\times {10}^6\mathrm{m}/{\mathrm{s}}^2$.

Calculate the magnitude of the force

From equation (ii), we can find the value of force by substituting the known values

$\mathrm{F}=\mathrm{ma}$$=\left(1.3\times {10}^{-4}\mathrm{kg}\right)\times \left(1.61\times {10}^6\mathrm{m}/\mathrm{s}\right)$

$=2.1\times {10}^2\mathrm{N}$

Hence, the magnitude of the force of the branch on the toothpick is $2.1\times {10}^2\mathrm{N}$.