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Chapter 6: Problem 19

The front \(1.20 \mathrm{~m}\) of a \(1400-\mathrm{kg}\) car is designed as a "crumple zone" that collapses to absorb the shock of a collision. If a car traveling \(25.0 \mathrm{~m} / \mathrm{s}\) stops uniformly in \(1.20 \mathrm{~m}\), (a) how long does the collision last, (b) what is the magnitude of the average force on the car, and (c) what is the acceleration of the car? Express the acceleration as a multiple of the acceleration of gravity.

Short Answer

Expert verified
a) The collision lasts 0.048 seconds. b) The average force on the car is -728000 Newtons. c) The car's acceleration is -520 m/s^2 or -53 times the acceleration due to gravity.

Step by step solution

01

Calculate Time

Given the initial velocity \(v_0 = 25.0 \, \mathrm{m/s}\), final velocity \(v = 0 \, \mathrm{m/s}\), and displacement \(d=1.2 \, \mathrm{m}\), the time can be calculated using the equation of motion \(d = v_0 t + \frac{1}{2}a t^2\). However, since the final velocity is 0, the equation modifies to \(d = v_0 t\). Therefore, the time \(t\) the collision lasts is then calculated as \(t = \frac{d}{v_0} = \frac{1.2 \, \mathrm{m}}{25.0 \, \mathrm{m/s}} = 0.048 \, \mathrm{s}\).
02

Calculate Acceleration

The acceleration could be calculated using the formula \(a = \frac{\Delta v}{\Delta t}\). The velocity difference is \(\Delta v = v - v_0 = -25.0 \, \mathrm{m/s}\) and the time difference is \(\Delta t = 0.048 \, \mathrm{s}\). Thus, calculating the acceleration we get \(a = \frac{-25.0 \, \mathrm{m/s}}{0.048 \, \mathrm{s}} = -520 \, \mathrm{m/s^2}\). Note that acceleration is negative because the car is slowing down.
03

Calculate Force

The force exerted on the car could be found using Newton鈥檚 second law \(F = ma\), where \(m\) is the mass and \(a\) is the acceleration. Given that \(m = 1400 \, \mathrm{kg}\) and \(a = -520 \, \mathrm{m/s^2}\), the force \(F\) is then \(F = 1400 \, \mathrm{kg} \times -520 \, \mathrm{m/s^2} = -728000 \, \mathrm{N}\). The negative sign indicates the force is directed in the opposite direction of motion.
04

Express Acceleration in Terms of Gravity

Finally, to express acceleration as a multiple of the acceleration due to gravity (g), find the ratio of \(a\) to \(g\). Considering the gravitational constant, \(g = 9.81 \, \mathrm{m/s^2}\), the acceleration \(a = -520 \, \mathrm{m/s^2}\) is then given as \( \frac{a}{g} = \frac{-520 \, \mathrm{m/s^2}}{9.81 \, \mathrm{m/s^2}} = -53 \, g\). Thus, the car decelerates at a rate 53 times greater than the acceleration due to gravity.

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Key Concepts

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

Crumple Zone
A crumple zone is a crucial safety feature in modern vehicles, designed to absorb the energy from a collision. It works by crumpling upon impact, which reduces the force transmitted to the occupants. This design is primarily aimed at minimizing injuries during a crash.

Key aspects include:
  • Energy Absorption: By deforming, the crumple zone absorbs a significant portion of the impact energy, reducing the force that reaches the passengers.
  • Controlled Collapse: The structure is engineered to collapse in a controlled manner to maximize protection.
  • Impact Force Reduction: By extending the time of the collision, it helps reduce the overall force, as force is a product of change in momentum over time.
The crumple zone in the given problem measures 1.20 meters and is highly effective at stopping the car safely by absorbing the kinetic energy. This ensures that the force felt by the occupants is far lesser than what it would be without such a mechanism.
Uniform Motion
Uniform motion refers to movement where the object covers equal distances in equal intervals of time. In essence, the speed remains constant throughout without any change in velocity.

However, in the scenario described, the problem involves a car that comes to a stop uniformly. This implies that while the initial design of the problem hints at uniform motion, in reality, the motion involves deceleration. Uniform deceleration can be understood as the car slowing down at a constant rate until it completely stops.
  • Constant Deceleration: The car exhibits a uniform deceleration (negative acceleration) as it travels through the crumple zone.
  • Equation of Motion: The equation used, \( v = u + at \), highlights how final velocity \( v \) reaches zero through uniform negative acceleration \( a \), with an initial velocity \( u \).
Understanding uniform motion and its variants, like uniform deceleration, is essential for solving kinematics problems accurately.
Newton's Second Law
Newton's second law of motion is central to the understanding of how objects move under the influence of forces. The law states that the force applied on an object is equal to the mass of the object times its acceleration: \( F = ma \).

In the given exercise:
  • Mass of Car: The crumple zone involves a 1400 kg car.
  • Acceleration: Calculated as \( -520 \, \mathrm{m/s^2} \), indicating a rapid deceleration.
  • Force Calculation: Substituting these values into Newton's second law gives the exerted force as \( -728000 \, \mathrm{N} \).
This force is negative, reflecting the opposition to the direction of travel due to the deceleration in the crumple zone. It's vital to note that the force required to stop the car quickly is far greater than what one might encounter under normal driving conditions. Newton's second law provides a profound understanding of how safety features like crumple zones are designed to manage collision forces effectively.

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Most popular questions from this chapter

GP A space probe, initially at rest, undergoes an internal mechanical malfunction and breaks into three pieces. One piece of mass \(m_{1}=48.0 \mathrm{~kg}\) travels in the positive \(x\) -direction at \(12.0 \mathrm{~m} / \mathrm{s}\), and a second piece of mass \(m_{2}=62.0 \mathrm{~kg}\) travels in the \(x y\) -plane at an angle of \(105^{\circ}\) at \(15.0 \mathrm{~m} / \mathrm{s}\). The third piece has mass \(m_{\mathrm{g}}=112 \mathrm{~kg}\). (a) Sketch a diagram of the situation, labeling the different masses and their velocities. (b) Write the general expression for conservation of momentum in the \(x\) - and \(y\) -directions in terms of \(m_{1}, m_{2}, m_{3}, v_{1}, v_{2}\), and \(v_{3}\) and the sines and cosines of the angles, taking \(\theta\) to be the unknown angle. (c) Calculate the final \(x\) -components of the momenta of \(m_{1}\) and \(m_{2}\). (d) Calculate the final \(y\) -components of the momenta of \(m_{1}\) and \(m_{2} .\) (e) Substitute the known momentum components into the general equations of momentum for the \(x\) - and \(y\) -directions, along with the known mass \(m_{3} .\) (f) Solve the two momentum equations for \(v_{3} \cos \theta\) and \(v_{3} \sin \theta\), respectively, and use the identity \(\cos ^{2} \theta+\sin ^{2} \theta=1\) to obtain \(v_{3}\). (g) Divide the equation for \(v_{3} \sin \theta\) by that for \(v_{3} \cos \theta\) to obtain \(\tan \theta\), then obtain the angle by taking the inverse tangent of both sides. (h) In general, would three such pieces necessarily have to move in the same plane? Why?

A \(60-\mathrm{kg}\) soccer player jumps vertically upwards and heads the \(0.45\) -kg ball as it is descending vertically with a speedof \(25 \mathrm{~m} / \mathrm{s}\). If the player was moving upward with a speed of \(4.0 \mathrm{~m} / \mathrm{s}\) just before impact, what will be the speed of the ball immediately after the collision if the ball rebounds vertically upwards and the collision is elastic? If the ball is in contact with the player's head for \(20 \mathrm{~ms}\), what is the average acceleration of the ball? (Note that the force of gravity may be ignored during the brief collision time.)

A \(1200-\mathrm{kg}\) car traveling initially with a speed of \(25.0 \mathrm{~m} / \mathrm{s}\) in an easterly direction crashes into the rear end of a 9000 -kg truck moving in the same direction at \(20.0 \mathrm{~m} / \mathrm{s}\) (Fig. \(\mathrm{P} 6.42\) ). The velocity of the car right after the collision is \(18.0 \mathrm{~m} / \mathrm{s}\) to the east. (a) What is the velocity of the truck right after the collision? (b) How much mechanical energy is lost in the collision? Account for this loss in energy.

\mathrm{\\{} E C P ~ A ~ \(90.0\) -kg fullback running east with a speed of \(5.00 \mathrm{~m} / \mathrm{s}\) is tackled by a \(95.0\) -kg opponent running north with a speed of \(3.00 \mathrm{~m} / \mathrm{s}\). (a) Why does the tackle constitute a perfectly inelastic collision? (b) Calculate the velocity of the players immediately after the tackle and (c) determine the mechanical energy that is lost as a result of the collision. Where did the lost energy go?

ecp Two blocks of masses \(m_{1}\) and \(m_{2}\) approach each other on a horizontal table with the same constant speed, \(v_{0}\), as measured by a laboratory observer. The blocks undergo a perfectly elastic collision, and it is observed that \(m_{1}\) stops but \(m_{2}\) moves opposite its original motion with some constant speed, \(v\). (a) Determine the ratio of the two masses, \(m_{1} / m_{2} .\) (b) What is the ratio of their speeds, \(v / v_{0}\) ?
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