Problem 34 - University Physics with Modern Physics

Question

A. 22 rifle bullet, traveling at 350 \(\mathrm{m} / \mathrm{s}\) , strikes a large tree, which it penetrates to a depth of 0.130 \(\mathrm{m}\) . The mass of the bullet is 1.80 \(\mathrm{g}\) . Assume a constant retarding force. (a) How much time is required for the bullet to stop? (b) What force, in newtons, does the tree exert on the bullet?

Step-by-Step Solution

Answer

(a) The time required for the bullet to stop is approximately 0.000742 seconds. (b) The force exerted by the tree is approximately 847.08 N.

Step 1: Identify Given Values

First, let's list the known values from the problem. The initial velocity of the bullet \( v_i = 350 \, \mathrm{m/s} \). The final velocity \( v_f = 0 \, \mathrm{m/s} \) since the bullet comes to a stop. The distance it penetrates is \( d = 0.130 \, \mathrm{m} \). The mass of the bullet is \( m = 1.80 \, \mathrm{g} = 0.0018 \, \mathrm{kg} \).

Step 2: Use Kinematic Equation to Find Acceleration

We can use the kinematic equation \( v_f^2 = v_i^2 + 2ad \) to find the acceleration \( a \). Plugging in the values we have:\[ 0 = 350^2 + 2a(0.130) \]\[ 0 = 122500 + 0.26a \]\[ a = -\frac{122500}{0.26} = -471153.846 \, \mathrm{m/s^2} \]

Step 3: Calculate Time Using Acceleration

Now use the equation \( v_f = v_i + at \) to find the time \( t \) it takes for the bullet to stop:\[ 0 = 350 + (-471153.846)t \]\[ t = \frac{-350}{-471153.846} \approx 0.000742 \mathrm{\, s} \]

Step 4: Calculate Force Exerted by the Tree

The force exerted by the tree can be found using Newton's second law \( F = ma \). Substituting the known values:\[ F = 0.0018 \times (-471153.846) \approx -847.0769 \, \mathrm{N} \]The force is negative, indicating it is in the direction opposite to the bullet's motion.

Key Concepts

Kinematic Equations

In physics, kinematic equations are used to describe the motion of objects. These equations relate the five parameters: initial velocity \( v_i \), final velocity \( v_f \), acceleration \( a \), displacement \( d \), and time \( t \). Understanding these equations allows us to solve various motion problems by plugging in the known values to find the unknown ones.

For the exercise involving a bullet penetrating a tree, one kinematic equation used is \( v_f^2 = v_i^2 + 2ad \). Here, this formula helps us derive the bullet's acceleration as it penetrates the tree. The equation rearranges the terms in such a way that by knowing the displacement and velocities, we can solve for acceleration:

• Final velocity \( v_f \) is 0 because the bullet stops.
• Initial velocity \( v_i \) is given as 350 m/s.
• Displacement \( d \) is 0.130 m—the distance the bullet travels inside the tree.

From this, we learn how rapidly the bullet slows down. Practicing kinematic equations is essential for solving similar physics problems effectively.

Newton's Second Law

Newton's Second Law of Motion states that the force acting on an object is equal to the mass of the object times its acceleration: \( F = ma \). This principle is fundamental in understanding how motion is influenced by forces, which is crucial, especially when analyzing how an object like a bullet interacts with other materials.

In the provided bullet penetration problem, we determined the retarding force exerted by the tree using this law. Once the bullet's acceleration due to the tree is calculated, Newton's Second Law helps us translate this acceleration into a force:

• The bullet's mass is 0.0018 kg.
• The negative acceleration (opposing the motion) was found to be \(-471153.846 \, \mathrm{m/s^2}\).

Applying \( F = ma \), we find the magnitude of the force. It illustrates how even a small object, moving at high speeds, can experience significant forces over short distances.

Bullet Penetration Problem

The bullet penetration problem is a classic example that combines multiple physics principles to understand materials' resistance and motion's transformation into a stopping force. This problem offers insight into real-world dynamics where high-speed projectiles interact with solid objects, such as wood.

Here's how the concepts come together in this scenario:

• The initial speed and stopping distance illustrate how quickly the bullet stops once it encounters resisting material.
• Kinematic equations help calculate details such as acceleration and penetration time.
• With Newton's Second Law, the interaction is framed in terms of forces, helping to picture what kind of impact occurs.

Understanding this problem not only provides computational practice but also enhances intuition on how phased interactions between fast-moving entities and stationary bodies work in everyday life.