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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.

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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.

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Most popular questions from this chapter

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?

A 110 kg linebacker running at \(2.0 \mathrm{m} / \mathrm{s}\) and an \(82 \mathrm{kg}\) quarterback running at \(3.0 \mathrm{m} / \mathrm{s}\) have a head- on collision in midair. The linebacker grabs and holds onto the quarterback. Who ends up moving forward after they hit?

A 20 g ball of clay traveling east at \(3.0 \mathrm{m} / \mathrm{s}\) collides with a 30 g ball of clay traveling north at \(2.0 \mathrm{m} / \mathrm{s}\). What are the speed and the direction of the resulting 50 g blob of clay?

A \(20 \mathrm{g}\) ball of clay traveling east at \(2.0 \mathrm{m} / \mathrm{s}\) collides with a \(30 \mathrm{g}\) ball of clay traveling \(30^{\circ}\) south of west at \(1.0 \mathrm{m} / \mathrm{s}\). What are the speed and direction of the resulting \(50 \mathrm{g}\) blob of clay?

A 3.0 -m-diameter merry-go-round with a mass of \(250 \mathrm{kg}\) is spinning at 20 rpm. John runs around the merry-go-round at \(5.0 \mathrm{m} / \mathrm{s},\) in the same direction that it is turning, and jumps onto the outer edge. John's mass is 30 kg. What is the merry-goround's angular speed, in rpm, after John jumps on?

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