/*! 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 15 I As part of a safety investigat... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 15

I As part of a safety investigation, two \(1400 \mathrm{kg}\) cars traveling at \(20 \mathrm{m} / \mathrm{s}\) are crashed into different barriers. Find the average forces exerted on (a) the car that hits a line of water barrels and takes \(1.5 \mathrm{s}\) to stop, and (b) the car that hits a concrete barrier and takes \(0.10 \mathrm{s}\) to stop.

Short Answer

Expert verified
The average forces exerted on (a) the car hitting the water barrels is \(1400 \times -13.33 \) N, i.e -18662 N (the negative sign shows that the force is acting in the opposite direction to the motion of the car), and (b) on the car hitting the concrete barrier is \(1400 \times -200 \) N, i.e -280000 N.

Step by step solution

01

Identify the Given Information

For both cars: \- Mass (m) of the car = 1400 kg \- Initial speed (u) of the car = 20 m/s \- Final speed (v) of the car = 0 m/s (as the cars finally stop) \For the car crashing into water barrels: \- Time taken to stop (t) = 1.5 s \For the car crashing into a concrete barrier: \- Time taken to stop (t) = 0.1 s
02

Calculate the Acceleration

Acceleration (a) is calculated using the formula, \( a = (v - u) / t \). Apply this formula separately for each scenario to calculate the acceleration.
03

Calculate the Force

The force exerted on the car (F) can be calculated by using the formula, \( F = m \times a \). Apply this formula for each scenario to calculate the force exerted on each car.

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 91影视!

Key Concepts

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

Force Calculation
The concept of force calculation is pivotal in understanding how objects interact with each other, particularly in scenarios like car crashes. According to Newton's Second Law of Motion, the force exerted on an object is the product of its mass and acceleration. The formula is expressed as:\[ F = m \times a \]Where:
  • \(F\) is the force exerted
  • \(m\) is the mass of the object
  • \(a\) is the acceleration

When analyzing a car collision, we first need to identify the change in velocity and the time over which this change occurs. This lets us compute the acceleration, which is simply the rate of change of velocity.
To find the average force exerted on the car, multiply the mass (in this case, 1400 kg) by the determined acceleration. Remember, a higher acceleration results in a greater force. Therefore, understanding how to accurately calculate force is crucial for predicting the impacts during collisions.
Acceleration
Acceleration is the measure of how quickly an object changes its velocity. It is a vector quantity, meaning it has both magnitude and direction. In the context of a car crash, acceleration describes how rapidly a car is brought to a halt when a collision occurs.
The formula to calculate acceleration is:\[ a = \frac{(v - u)}{t} \]Where:
  • \(a\) is the acceleration
  • \(v\) is the final velocity (0 m/s in our crash scenario, since the car stops)
  • \(u\) is the initial velocity (20 m/s here)
  • \(t\) is the time it takes to stop

Different stopping times produce different accelerations. A car that stops quickly (like the one hitting the concrete barrier) experiences greater acceleration compared to one that stops more slowly (like the one hitting water barrels). This directly affects the force exerted on the car and ultimately, the damage it may sustain.
Collision Physics
Collision physics deals with the effects of forces between two bodies during a collision. During a car accident, understanding the physics of collision is vital for analyzing safety features and determining crash outcomes.
When a moving car crashes, it experiences a sudden change in its momentum, directly linked to the forces applied during the impact. The time duration over which the car comes to a stop significantly influences the force experienced by the vehicle.
This explains how in our scenario, despite both cars having the same speed and mass, they experience different forces upon crashing into different barriers. The car hitting the concrete barrier stops in 0.1 seconds, experiencing a much more intense force than the car that takes 1.5 seconds to stop against water barrels.
The key principles at play are those of impulse and momentum, where impulse is the product of force and the time duration during which the force acts. This is why energy-absorbing materials, like water barrels, can help reduce the impact force by prolonging the stopping time, increasing safety during collisions.

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

\(A\) typical raindrop is much more massive than a mosquito and much faster than a mosquito flies. How does a mosquito survive the impact? Recent research has found that the collision of a falling raindrop with a mosquito is a perfectly inelastic collision. That is, the mosquito is "swept up" by the raindrop and ends up traveling along with the raindrop. Once the relative speed between the mosquito and the raindrop is zero, the mosquito is able to detach itself from the drop and fly away. a. A hovering mosquito is hit by a raindrop that is 40 times as massive and falling at \(8.2 \mathrm{m} / \mathrm{s},\) a typical raindrop speed. How fast is the raindrop, with the attached mosquito, falling immediately afterward if the collision is perfectly inelastic? b. Because a raindrop is "soft" and deformable, the collision duration is a relatively long \(8.0 \mathrm{ms}\). What is the mosquito's average acceleration, in \(g\) 's, during the collision? The peak acceleration is roughly twice the value you found, but the mosquito's rigid exoskeleton allows it to survive accelerations of this magnitude. In contrast, humans cannot survive an acceleration of more than about \(10 \mathrm{g}\).

I trap-jaw ant snaps its mandibles shut at very high speed, a good trait for catching small prey. But an ant can also slam its mandibles into the ground; the resulting force can launch the ant into the air for a quick escape. A 12 mg ant hits the ground with an average force of \(47 \mathrm{mN}\) for a time of \(0.13 \mathrm{ms} ;\) these are all typical values. At what speed does it leave the ground?

\(\|\) Ice skaters often end their performances with spin turns, where they spin very fast about their center of mass with their arms folded in and legs together. Upon ending, their arms extend outward, proclaiming their finish. Not quite as noticeably, one leg goes out as well. Suppose that the moment of inertia of a skater with arms out and one leg extended is \(3.2 \mathrm{kg} \cdot \mathrm{m}^{2}\) and for arms and legs in is \(0.80 \mathrm{kg} \cdot \mathrm{m}^{2}\). If she starts out spinning at 5.0 rev/s, what is her angular speed (in rev/s) when her arms and one leg open outward?

A 15 g bullet is fired at \(610 \mathrm{m} / \mathrm{s}\) into a \(4.0 \mathrm{kg}\) block that sits at the edge of a \(75-\mathrm{cm}\) -high table. The bullet embeds itself in the block and carries it off the table. How far from the point directly below the table's edge does the block land?

II A 20 g ball of clay is thrown horizontally at \(30 \mathrm{m} / \mathrm{s}\) toward a \(1.0 \mathrm{kg}\) block sitting at rest on a frictionless surface. The clay hits and sticks to the block. a. What is the speed of the block and clay right after the collision? b. Use the block's initial and final speeds to calculate the impulse the clay exerts on the block. c. Use the clay's initial and final speeds to calculate the impulse the block exerts on the clay. d. Does \(\vec{J}_{\text {blockonclay }}=-\vec{J}_{\text {clayon block }} ?\)
See all solutions

Recommended explanations on Physics Textbooks

Rotational Dynamics

Read Explanation

Modern Physics

Read Explanation

Fields in Physics

Read Explanation

Kinematics Physics

Read Explanation

Quantum Physics

Read Explanation

Energy Physics

Read Explanation
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.