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Chapter 4: Problem 28

A .22 caliber rifle bullet traveling at \(350 \mathrm{~m} / \mathrm{s}\) strikes a large tree and penetrates it 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
To calculate the time it takes for the bullet to stop, use the equations of motion. Solving the quadratic equation gives the time value. The force exerted by the tree can be calculated using Newton's second law and the deceleration calculated in step 1.

Step by step solution

01

Calculate the time it takes for the bullet to stop

Firstly we set v, the final velocity to 0 (since the bullet comes to a stop), and u, the initial velocity to 350 m/s (bullet's speed). Using the equation of motion \( v = u + at \), where a is deceleration, we solve the equation for 'a' first before substituting it into the second equation of motion to find 't'. So, rearranging the equation gives us \( a = (v-u)/t \). Substituting the known values gives \( a = (0 - 350) m/s / t \) or \( a = -350m/s / t \).
02

Use the second equation of motion

We now substitute 'a' into the second equation of motion \( s = ut + 0.5at^2 \) to solve for 't'. Substituting 'a' from step 1 and 's' from the problem (s equals 0.130m), the equation becomes \( 0.130 \mathrm{~m} = 350 \mathrm{~m/s} \cdot t + 0.5(-350m/s / t) \cdot t^2 \). Solving for 't' gives a quadratic equation \( -175t^2 + 350t - 0.130 = 0 \), which we can solve to get the value for 't'.
03

Calculate the retarding force exerted by the tree

The force can be calculated using Newton's second law of motion \( f = m \cdot a \), where 'f' is force, 'm' is mass and 'a' is acceleration (or deceleration in this case). Substituting the value of 'a' from Step 1 and the mass of the bullet 'm' (1.80g or 0.0018kg) gives \( f = ma = 0.0018 \mathrm{~kg} \cdot 350 \mathrm{~m/s^2} \cdot t \). Substituting the value of 't' calculated from step 2 will give the force in newtons.

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Key Concepts

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

Equations of Motion
Understanding the equations of motion is essential when dealing with kinematics problems like the one involving a rifle bullet. Equations of motion provide the tools needed to describe the movement of objects using terms like initial velocity, final velocity, acceleration, and displacement.

Here's a breakdown:
  • Initial velocity (\( u \)): This is the speed at which an object starts moving. For the bullet, this was 350 m/s.
  • Final velocity (\( v \)): The speed at which an object finishes its motion. Here, the bullet's final velocity is 0 m/s because it stops.
  • Acceleration (\( a \)): This can be positive (acceleration) or negative (deceleration) and indicates the rate of change of velocity.
  • Displacement (\( s \)): The overall change in position. In this case, the bullet moved into the tree by 0.130 meters.
Using these values, we first use the equation \[ v = u + at \] to find \( a \), and then use \[ s = ut + 0.5at^2 \] to find the time \( t \) it takes for the bullet to come to a stop. Solving these equations helps us deeply understand how the bullet's motion is affected by external forces.
Newton's Second Law
Newton's second law is key when calculating the force exerted on an object. It is expressed in the equation \( F = ma \), where \( F \) is the force, \( m \) is the mass of the object, and \( a \) is the acceleration.

For the bullet striking the tree, the mass is small, 1.80 g or 0.0018 kg, but the deceleration is substantial, as found in solving the equations of motion.

By inputting these values into Newton's second law, we can discern the magnitude of the force the tree exerts on the bullet, aiding us in understanding the impact dynamics in such situations. The calculated force is vital for evaluating how different materials can absorb and dissipate the kinetic energy of moving objects. Newton's second law provides a tangible way to analyze these effects practically.
Deceleration
Deceleration is simply negative acceleration, occurring when an object slows down. In this scenario, deceleration is crucial as the bullet experiences a significant slowdown due to the tree's resistance.

The magnitude of deceleration can be calculated using motion equations, which were introduced earlier. It's vital here because it influences not only the time it takes for the bullet to come to a stop but also the force exerted, as described in Newton's second law.

Generally, the greater the deceleration, the quicker an object will come to a complete stop. Understanding deceleration helps explain why different materials can stop a projectile differently. Trees can exert substantial deceleration forces, effectively stopping bullets within short distances.

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Most popular questions from this chapter

World-class sprinters can accelerate out of the starting blocks with an acceleration that is nearly horizontal and has magnitude \(15 \mathrm{~m} / \mathrm{s}^{2} .\) How much horizontal force must a \(55 \mathrm{~kg}\) sprinter exert on the starting blocks to produce this acceleration? Which object exerts the force that propels the sprinter: the blocks or the sprinter herself?

A student of mass \(45 \mathrm{~kg}\) jumps off a high diving board. What is the acceleration of the earth toward her as she accelerates toward the earth with an acceleration of \(9.8 \mathrm{~m} / \mathrm{s}^{2}\) ? Use \(6.0 \times 10^{24} \mathrm{~kg}\) for the mass of the earth, and assume that the net force on the earth is the force of gravity she exerts on it.

Two blocks connected by a light horizontal rope sit at rest on a horizontal, frictionless surface. Block \(A\) has mass \(15.0 \mathrm{~kg},\) and block \(B\) has mass \(m\). A constant horizontal force \(F=60.0 \mathrm{~N}\) is applied to block \(A\) (Fig. \(\mathbf{P 4 . 3 8}\) ). In the first 5.00 s after the force is applied, block \(A\) moves \(18.0 \mathrm{~m}\) to the right. (a) While the blocks are moving, what is the tension \(T\) in the rope that connects the two blocks? (b) What is the mass of block \(B ?\)

After an annual checkup, you leave your physician's office, where you weighed \(683 \mathrm{~N}\). You then get into an elevator that, conveniently, has a scale. Find the magnitude and direction of the elevator's acceleration if the scale reads (a) \(725 \mathrm{~N}\) and (b) \(595 \mathrm{~N}\).

A small car of mass \(380 \mathrm{~kg}\) is pushing a large truck of mass 900 \(\mathrm{kg}\) due east on a level road. The car exerts a horizontal force of \(1600 \mathrm{~N}\) on the truck. What is the magnitude of the force that the truck exerts on the car?
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