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Chapter 4: Problem 34

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?

Short Answer

Expert verified
(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 by step solution

01

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 \( m = 1.80 \, \mathrm{g} = 0.0018 \, \mathrm{kg} \).
02

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:\[ 0 = 350^2 + 2a(0.130) \]\[ 0 = 122500 + 0.26a \]\[ a = -\frac{122500}{0.26} = -471153.846 \, \mathrm{m/s^2} \]
03

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:\[ 0 = 350 + (-471153.846)t \]\[ t = \frac{-350}{-471153.846} \approx 0.000742 \mathrm{\, s} \]
04

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.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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鈥攖he 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.

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Most popular questions from this chapter

If a net horizontal force of 132 \(\mathrm{N}\) is applied to a person with mass 60 \(\mathrm{kg}\) who is resting on the edge of a swimming pool, what horizontal acceleration is produced?

If we know \(F(t),\) the force as a function of time, for straight-line motion, Newton's second law gives us \(a(t),\) the acceleration as a function of time. We can then integrate \(a(t)\) to find \(v(t)\) and \(x(t)\) . However, suppose we know \(F(v)\) instead. (a) The net force on a body moving along the \(x\) -axis equals \(-C v^{2} .\) Use Newton's second law written as \(\Sigma F=m d v / d t\) and two integrations to show that \(x-x_{0}=(m / C) \ln \left(v_{0} / v\right) .\) (b) Show that Newton's second law can be written as \(\Sigma F=m v d v / d x .\) Derive the same expression as in part (a) using this form of the second law and one integration.

A ball is hanging from a long string that is tied to the ceiling of a train car traveling castward on horizontal tracks. An observer inside the train car sees the ball hang motionless. Draw a clearly labeled free-body diagram for the ball if (a) the train has a uniform velocity, and (b) the train is speeding up uniformly. Is the net force on the ball zero in either case? Explain.

What magnitude of net force is required to give a \(135-\mathrm{kg}\) refrigerator an acceleration of magnitude 1.40 \(\mathrm{m} / \mathrm{s}^{2}\) ?

A box rests on a frozen pond, which serves as a frictionless horizontal surface. If a fisherman applies a horizontal force with magnitude 48.0 \(\mathrm{N}\) to the box and produces an acceleration of magnitude \(3.00 \mathrm{m} / \mathrm{s}^{2},\) what is the mass of the box?
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