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. part b what force, in newtons, does the tree exert on the bullet? express your answer in newtons.

Explanation:

Step1: Convert mass to kg

The mass of the bullet is \( m = 1.80\ \text{g} = 1.80\times 10^{- 3}\ \text{kg} \), initial velocity \( v_0=350\ \text{m/s} \), final velocity \( v = 0\ \text{m/s} \), and displacement \( s=0.130\ \text{m} \). We use the kinematic equation \( v^{2}=v_{0}^{2}+2as \) to find the acceleration \( a \), then use Newton's second law \( F = ma \) to find the force.

First, from \( v^{2}=v_{0}^{2}+2as \), we solve for \( a \):
\( a=\frac{v^{2}-v_{0}^{2}}{2s} \)
Substitute \( v = 0 \), \( v_0 = 350\ \text{m/s} \), \( s=0.130\ \text{m} \) into the formula:
\( a=\frac{0-(350)^{2}}{2\times0.130}=\frac{- 350^{2}}{0.260}\ \text{m/s}^2 \)

Step2: Calculate acceleration

Calculate \( 350^{2}=122500 \), then \( a=\frac{- 122500}{0.260}\approx - 471153.85\ \text{m/s}^2 \) (the negative sign indicates deceleration)

Step3: Calculate force using Newton's second law

Newton's second law is \( F=ma \), substitute \( m = 1.80\times 10^{-3}\ \text{kg} \) and \( a\approx - 471153.85\ \text{m/s}^2 \)
\( F=1.80\times 10^{-3}\ \text{kg}\times(- 471153.85\ \text{m/s}^2)\approx - 848.08\ \text{N} \)
The negative sign indicates the force is opposite to the direction of motion. The magnitude of the force exerted by the tree on the bullet is approximately \( 848\ \text{N} \) (we can also calculate directly using the work - energy theorem \( W=\Delta K=F\cdot s\cos\theta \), since the force and displacement are in opposite directions, \( \theta = 180^{\circ} \), \( \cos\theta=- 1 \), and \( \Delta K=\frac{1}{2}mv^{2}-\frac{1}{2}mv_{0}^{2}=-\frac{1}{2}mv_{0}^{2} \), so \( F\cdot s\times(- 1)=-\frac{1}{2}mv_{0}^{2} \), then \( F=\frac{mv_{0}^{2}}{2s} \))

Using the work - energy theorem formula \( F=\frac{mv_{0}^{2}}{2s} \)
Substitute \( m = 1.80\times 10^{-3}\ \text{kg} \), \( v_0 = 350\ \text{m/s} \), \( s = 0.130\ \text{m} \)
\( F=\frac{1.80\times 10^{-3}\times350^{2}}{2\times0.130}=\frac{1.80\times 10^{-3}\times122500}{0.260}=\frac{220.5}{0.260}\approx848\ \text{N} \)

Answer:

\(\boxed{848}\) (The force exerted by the tree on the bullet is approximately 848 newtons. The negative sign in the acceleration - based calculation indicates the direction of the force is opposite to the bullet's motion, and we take the magnitude here as the question asks for the force exerted by the tree on the bullet.)