/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 52 In \(1.00 \mathrm{~s}\), an auto... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

In \(1.00 \mathrm{~s}\), an automatic paint ball gun can fire 15 balls, each with a mass of \(0.113 \mathrm{~g}\), at a muzzle velocity of \(88.5 \mathrm{~m} / \mathrm{s}\). Calculate the average recoil force experienced by the player who's holding the gun.

Short Answer

Expert verified
The average recoil force is approximately 0.15 N.

Step by step solution

01

Convert Mass to Kilograms

First, convert the mass of one paintball from grams to kilograms to use it in further calculations. Since there are 1000 grams in a kilogram, we have:\[m = 0.113 \text{ g} = 0.113 \times 10^{-3} \text{ kg} = 0.000113 \text{ kg}\]
02

Calculate the Momentum of One Paintball

The momentum of an object can be calculated using the formula \( p = mv \), where \( m \) is the mass and \( v \) is the velocity. For one paintball:\[p = 0.000113 \text{ kg} \times 88.5 \text{ m/s} = 0.0099855 \text{ kg} \cdot \text{m/s}\]
03

Calculate Total Momentum for 15 Paintballs

In one second, 15 paintballs are fired. The total momentum for these paintballs is the sum of the momentum of each individual paintball:\[P_{\text{total}} = 15 \times 0.0099855 \text{ kg} \cdot \text{m/s} = 0.1497825 \text{ kg} \cdot \text{m/s}\]
04

Use Newton's Second Law to Find Average Recoil Force

According to Newton's second law, force is the change in momentum over time. Since the change in momentum is equal to the total momentum over one second, the average recoil force \( F \) is:\[F = \frac{\Delta p}{\Delta t} = \frac{0.1497825 \text{ kg} \cdot \text{m/s}}{1 \text{ s}} = 0.1497825 \text{ N}\]
05

Conclusion

The average recoil force experienced by the player holding the paintball gun is approximately 0.15 N.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Momentum
Momentum is a key concept in physics that describes the quantity of motion an object has. It's determined by both the object's mass and velocity. To calculate momentum, we use the formula:
  • \( p = mv \) where \( p \) is the momentum, \( m \) is the mass, and \( v \) is the velocity.
In our paintball problem, we determined the momentum of a single paintball by multiplying its mass (which was converted to kilograms) and its velocity, resulting in \( p = 0.0099855 \text{ kg} \cdot \text{m/s} \).

It's essential to note that more mass or higher velocity will increase the momentum of an object, meaning they will have a stronger influence when they collide with something. Here, even though each paintball has a tiny mass, their high velocity contributes a substantial momentum to each shot.
Newton's Second Law
Newton's Second Law of Motion forms the backbone of understanding how forces and motions interact. It is usually expressed as:
  • \( F = ma \), which means force equals mass times acceleration.
  • Or, it's often used in the context of changing momentum as: \( F = \frac{\Delta p}{\Delta t} \), where \( \Delta p \) is the change in momentum and \( \Delta t \) is the change in time.
In the case of the paintball gun, each time a paintball is fired, there's a change in momentum that occurs instantaneously as the paintball is pushed out of the gun's barrel.

Because this momentum change happens over the duration of one second for 15 paintballs, we use the principle of Newton's second law to find the average recoil force. This step is crucial because it accounts for the rapid firing, encompassing the fast-paced ejection of paintballs.
Muzzle Velocity
Muzzle velocity is the speed a projectile has when it exits the barrel of a firearm or similar device. It's a fundamental parameter because it not only affects the momentum calculation but also other characteristics like range or impact force.

In the provided exercise, the muzzle velocity is given as 88.5 meters per second. This high velocity signifies that the paintballs are shot out swiftly, which significantly contributes to their momentum.
  • A higher muzzle velocity increases the paintball's momentum proportionally, making the firing experience more intense.
  • This high exit speed also evokes a noticeable recoil force, which is felt as the backward shove when the player fires the gun.
Understanding how muzzle velocity impacts the system allows one to adjust or evaluate the paintball gun's performance effectively, whether for improved accuracy or force management.
Mass Conversion
Mass conversion is a routine yet vital step in physics calculations. This involves converting units from one system to another, ensuring smooth calculations and meaningful interpretations.

In the problem at hand, converting the paintball's mass from grams to kilograms was necessary:
  • One gram equals \(0.001\) kilograms, so multiplying by this factor converts the given mass correctly.
  • For example, \(0.113\) grams converts to \(0.000113\) kilograms.
Performing this conversion is crucial because standard physics formulas often use SI units, like kilograms for mass. Utilizing the correct units ensures precision and accuracy in momentum calculations. Each step where mass and force are involved benefits from correctly formatted and converted units to facilitate reliable results.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

The speed of the fastest-pitched baseball was \(45 \mathrm{~m} / \mathrm{s},\) and the ball's mass was \(145 \mathrm{~g}\). (a) What was the magnitude of the momentum of this ball, and how many joules of kinetic energy did it have? (b) How fast would a 57 gram ball have to travel to have the same amount of (i) kinetic energy, and (ii) momentum?

On a frictionless, horizontal air table, puck \(A\) (with mass \(0.250 \mathrm{~kg}\) ) is moving to the right toward puck \(B\) (with mass \(0.350 \mathrm{~kg}\) ), which is initially at rest. After the collision, puck \(A\) has a velocity of \(0.120 \mathrm{~m} / \mathrm{s}\) to the left, and puck \(B\) has a velocity of \(0.650 \mathrm{~m} / \mathrm{s}\) to the right. (a) What was the speed of puck \(A\) before the collision? (b) Calculate the change in the total kinetic energy of the system that occurs during the collision.

A \(750 \mathrm{~kg}\) car is stalled on an icy road during a snowstorm. A \(1000 \mathrm{~kg}\) car traveling eastbound at \(10 \mathrm{~m} / \mathrm{s}\) collides with the rear of the stalled car. After being hit, the \(750 \mathrm{~kg}\) car slides on the ice at \(4 \mathrm{~m} / \mathrm{s}\) in a direction \(30^{\circ}\) north of east. (a) What are the magnitude and direction of the velocity of the \(1000 \mathrm{~kg}\) car after the collision? (b) Calculate the ratio of the kinetic energy of the two cars just after the collision to that just before the collision. (You may ignore the effects of friction during the collision.)

Combining conservation laws. A \(5.00 \mathrm{~kg}\) chunk of ice is sliding at \(12.0 \mathrm{~m} / \mathrm{s}\) on the floor of an ice-covered valley when it collides with and sticks to another \(5.00 \mathrm{~kg}\) chunk of ice that is initially at rest. (See Figure \(8.39 .\) ) Since the valley is icy, there is no friction. After the collision, the blocks slide partially up a hillside and then slide back down. How fast are they moving when they reach the valley floor again? (Hint: Break this problem into two parts - the collision and the behavior after the collision - and apply the appropriate conservation law to each part.)

A \(0.300 \mathrm{~kg}\) glider is moving to the right on a frictionless, horizontal air track with a speed of \(0.80 \mathrm{~m} / \mathrm{s}\) when it makes a head-on collision with a stationary \(0.150 \mathrm{~kg}\) glider. (a) Find the magnitude and direction of the final velocity of each glider if the collision is elastic. (b) Find the final kinetic energy of each glider.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.