Question

a 22 caliber rifle bullet traveling at 350 m/s, strikes a large tree and penetrates it to a depth of 0.130 m. the mass of the bullet is 1.80 g. assume a constant retarding force. how much time is required for the bullet to stop? express your answer with the appropriate units.

Explanation:

Step1: Identify the motion type

The bullet undergoes uniformly decelerated motion (constant retarding force implies constant acceleration, here deceleration). We can use the kinematic equation for motion with constant acceleration: $v = v_0+at$ (final velocity $v = 0$), and also the equation for displacement: $x=v_0t+\frac{1}{2}at^2$. Alternatively, we can use the average velocity formula. The average velocity $\bar{v}=\frac{v_0 + v}{2}$, since $v = 0$, $\bar{v}=\frac{v_0}{2}$. And displacement $x=\bar{v}t$.

Step2: Rearrange the average velocity formula for time

From $x=\frac{v_0}{2}t$, we can solve for $t$: $t=\frac{2x}{v_0}$.

Step3: Plug in the values

Given $v_0 = 350\space m/s$, $x = 0.130\space m$.
Substitute into the formula: $t=\frac{2\times0.130}{350}$.

Step4: Calculate the result

$t=\frac{0.26}{350}\approx7.43\times 10^{-4}\space s$ or convert to milliseconds: $7.43\times 10^{-4}\space s=0.743\space ms$ (or microseconds: $743\space \mu s$)

Answer:

Value: $7.43\times 10^{-4}$ (or $0.743$ if in ms, or $743$ if in $\mu s$), Units: seconds (or milliseconds, or microseconds). The most appropriate unit here is seconds, so $t = \boldsymbol{7.43\times 10^{-4}\space s}$ (or $743\space \mu s$ as $1\space s = 10^6\space \mu s$, so $7.43\times 10^{-4}\space s=7.43\times 10^{-4}\times 10^6\space \mu s = 743\space \mu s$)