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Chapter 4: Problem 5

A 15 -g bullet is fired from a rifle. It takes \(2.50 \times 10^{-3} \mathrm{~s} \mathrm{~s}\) for the bullet to travel the length of the barrel, and it exits the barrel with a speed of \(715 \mathrm{~m} / \mathrm{s}\). Assuming that the acceleration of the bullet is constant, find the average net force exerted on the bullet.

Short Answer

Expert verified
The average net force exerted on the bullet is 4290 N.

Step by step solution

01

Identify Given Data

We have the following information:- Mass of the bullet: 15 g = 0.015 kg- Time taken: \( t = 2.50 \times 10^{-3} \mathrm{~s} \)- Final velocity of the bullet: \( v = 715 \mathrm{~m/s} \)- Initial velocity \( u = 0 \mathrm{~m/s} \) (since the bullet starts from rest).We are required to find the average net force exerted on the bullet.
02

Calculate Acceleration

Since we have constant acceleration, we use the formula for acceleration, \( a \), which is\[a = \frac{v - u}{t}\].Substitute the known values:\[a = \frac{715 \mathrm{~m/s} - 0 \mathrm{~m/s}}{2.50 \times 10^{-3} \mathrm{~s}} = \frac{715}{2.50 \times 10^{-3}} \approx 286000 \mathrm{~m/s^2}.\]
03

Calculate Net Force Using Newton's Second Law

The average net force \( F \) exerted on the bullet can be found using Newton's second law, \( F = ma \).Substitute the known values:\[F = 0.015 \mathrm{~kg} \times 286000 \mathrm{~m/s^2} = 4290 \mathrm{~N}.\]
04

Verify Units and Calculation

Ensure the units are consistent and the calculations are correct:- Mass is in kilograms (kg),- Acceleration is in meters per second squared (m/s虏),- Force is in newtons (N).The calculations should result in \( F = 4290 \mathrm{~N} \) which are correct when recalculated.

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Key Concepts

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

Constant Acceleration
In physics, constant acceleration means that the speed of an object changes by the same amount every second. This concept is crucial when analyzing any motion that doesn't change its velocity abruptly.
In the case of the bullet, we consider its movement from rest to a very high speed over a short period. Since its acceleration is constant, we accurately use the formula for acceleration \[ \text{Acceleration } (a) = \frac{\text{Final Velocity } (v) - \text{Initial Velocity } (u)}{\text{Time } (t)} \].
Understanding this is vital when using the formula to ensure the calculations for acceleration鈥攁nd consequently the force鈥攁re correct.
Average Net Force
The average net force refers to the total force exerted on an object over a period. This force causes an object to accelerate, based on Newton's Second Law of Motion.
For the bullet, we calculated the net force exerted during its travel through the rifle barrel. Newton's second law helps, stating that force equals mass times acceleration \[F = ma \].
By substituting the values, we found the bullet undergoes an average net force of 4290 N. This force is what propels the bullet to its high speed so quickly.
Bullet Motion
Bullet motion involves complex dynamics, but it primarily deals with the motion from within the barrel to the moment it exits.
Initially stationary, a bullet rapidly accelerates through the barrel due to the force from the explosion behind it.
Its motion is dominated by the conversion of chemical energy into kinetic energy, which occurs so swiftly. This motion underlines topics of both speed and acceleration, and understanding these helps optimize factors such as firing accuracy and impact force.
Kinematics
Kinematics is the branch of physics that studies the motion of objects without considering the forces that cause the motion.
In our exercise, we focus on the bullet's motion: starting at rest and reaching a top speed of 715 m/s.
By breaking down the bullet's movement into initial velocity, final velocity, time, and acceleration, kinematics offers a simplified way to predict the effects of forces on an object's trajectory over time.
Dynamics
Dynamics, unlike kinematics, focuses on forces and their effects on motion. It's the bigger picture where not only the movement but also the 'why' behind that movement is explored.
In our scenario, dynamics explains not just the bullet鈥檚 final speed but how forces, such as gunpowder鈥檚 explosive force, come into play.
Newton's second law fits into dynamics by directly relating these forces to changes in motion, offering insights into both motion and cause.

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Most popular questions from this chapter

When a parachute opens, the air exerts a large drag force on it. This upward force is initially greater than the weight of the sky diver and, thus, slows him down. Suppose the weight of the sky diver is \(915 \mathrm{~N}\) and the drag force has a magnitude of \(1027 \mathrm{~N}\). The mass of the sky diver is \(93.4 \mathrm{~kg} .\) What are the magnitude and direction of his acceleration?

Two blocks are sliding to the right across a horizontal surface, as the drawing shows. In Case A the masses of both blocks are \(3.0 \mathrm{~kg} .\) In Case \(\mathrm{B}\) the mass of block \(1,\) the block behind, is \(6.0 \mathrm{~kg},\) and the mass of block 2 is \(3.0 \mathrm{~kg} .\) No frictional force acts on block 1 in either Case \(A\) or Case B. However, a kinetic frictional force does act on block 2 in both cases and opposes the motion. (a) Identify the forces that contribute to the horizontal net force acting on block \(1 .\) (b) Identify the forces that contribute to the horizontal net force acting on block 2 . (c) In which case, if either, do the blocks push against each other with greater forces? Explain. (d) Are the blocks accelerating or decelerating, and in which case, if either, is the magnitude of the acceleration greater? Problem The magnitude of the kinetic frictional force acting on block 2 in the drawing is \(5.8 \mathrm{~N}\). For both Case \(\mathrm{A}\) and Case \(\mathrm{B}\) deterine (a) the magnitude of the forces with which the blocks push against each other and (b) the acceleration of the blocks. Check to see that your answers are consistent with your answers to the Concept Questions.

A spacecraft is on a journey to the moon. At what point, as measured from the center of the earth, does the gravitational force exerted on the spacecraft by the earth balance that exerted by the moon? This point lies on a line between the centers of the earth and the moon. The distance between the earth and the moon is \(3.85 \times 10^{8} \mathrm{~m}\), and the mass of the earth is 81.4 times as great as that of the moon.

A train consists of 50 cars, each of which has a mass of \(6.8 \times 10^{3} \mathrm{~kg}\). The train has an acceleration of \(+8.0 \times 10^{-2} \mathrm{~m} / \mathrm{s}^{2}\). Ignore friction and determine the tension in the coupling (a) between the 30 th and 31 st cars and (b) between the 49 th and 50 th cars.

A rocket blasts off from rest and attains a speed of \(45 \mathrm{~m} / \mathrm{s}\) in \(15 \mathrm{~s}\). An astronaut has a mass of \(57 \mathrm{~kg}\). What is the astronaut's apparent weight during takeoff?
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