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In a performance test two cars take the same time to accelerate from rest up to the same speed. Car A has a mass of \(1400 \mathrm{~kg}\), and car \(\mathrm{B}\) has a mass of \(1900 \mathrm{~kg}\). During the test, which car (a) has the greater change in momentum, (b) experiences the greater impulse, and (c) is acted upon by the greater net average force? In each case, give your reasoning.

Short Answer

Expert verified
(a) Car B; (b) Car B; (c) Car B.

Step by step solution

01

Understanding Momentum

Momentum is calculated as the product of an object's mass and its velocity. Mathematically, it's expressed as \( p = mv \), where \( p \) is the momentum, \( m \) is mass, and \( v \) is velocity. Since both cars reach the same speed from rest, the change in velocity is the same for both. Therefore, car A will have a change in momentum by \( \Delta p_A = 1400 \, \text{kg} \cdot v \), and car B by \( \Delta p_B = 1900 \, \text{kg} \cdot v \).
02

Comparing Changes in Momentum

The change in momentum for each car can be compared directly using their mass. Since \( 1900 \, \text{kg} \cdot v > 1400 \, \text{kg} \cdot v \) for the same \( v \), Car B has a greater change in momentum.
03

Understanding Impulse

Impulse is defined as the change in momentum of an object when a force is applied over a period of time. It's given by the equation \( J = \Delta p \). Since impulse is equal to the change in momentum, Car B experiences a greater impulse than Car A.
04

Understanding Net Average Force

From Newton's second law, the net force is related to the rate of change of momentum, which can be expressed as \( F = \frac{\Delta p}{\Delta t} \). If both cars take the same time to reach the final speed, then \( F_A = \frac{1400v}{t} \) and \( F_B = \frac{1900v}{t} \). Since \( 1900v > 1400v \), Car B experiences a greater net average force.

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Key Concepts

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

Impulse
Impulse is a fascinating concept in physics that describes the effect of a force applied over a period of time. It's calculated using the formula \( J = \Delta p \), where \( J \) represents impulse and \( \Delta p \) signifies the change in momentum. When you apply a force to an object, the object's momentum changes, and this change is captured in the term impulse.

For our exercise, both cars start from rest and accelerate to the same speed. This means they both experience a force over the same time period. However, because impulse is directly tied to the change in momentum, and we know from the solution that Car B, with a mass of 1900 kg, has a greater change in momentum than Car A, it logically follows that Car B experiences a greater impulse.

Some quick points to remember about impulse:
  • Impulse is essentially the amount of momentum change an object undergoes.
  • If the same force acts over different durations, a longer time results in greater impulse.
  • The unit for impulse is the same as momentum: kilogram meter per second (\( \text{kg} \cdot \text{m/s} \)).
Net Average Force
The net average force is a superbly intuitive concept when you start breaking it down. According to Newton's Second Law, force is the cause of change in momentum. The net average force exerted on an object can be calculated using the formula \( F = \frac{\Delta p}{\Delta t} \). Here, \( F \) stands for net force, \( \Delta p \) is the change in momentum, and \( \Delta t \) is the time period over which the force acts.

In the case of our two cars, since they both accelerate for the same duration to reach the same final speed, the difference in net average force arises solely because of the difference in their masses. For Car B, which is heavier, the change in momentum is larger. Thus, when we divide this larger \( \Delta p \) by the same \( \Delta t \) for both cars, the net average force on Car B is greater.

Quick highlights about net average force:
  • Net average force is responsible for accelerating an object to its new speed.
  • It is dependent on the rate at which momentum changes in a given time.
  • The unit for force is Newtons (\( \text{N} \)).
Change in Momentum
The change in momentum is a central concept in classical mechanics and a foundational element in understanding motion. Momentum is the product of mass and velocity (\( p = mv \)). A change in momentum occurs when either the mass or velocity of an object changes. Since the exercise specifies constant mass for each car and a change from rest to the same final speed, the change in momentum depends on the mass.

For our two cars, Car A and Car B, they both begin at rest and increase speed, meaning their change in velocity is identical. However, due to Car B's higher mass (1900 kg compared to Car A's 1400 kg), Car B experiences a larger change in momentum than Car A. This can be mathematically represented as \( \Delta p_B = 1900 \cdot v \) and \( \Delta p_A = 1400 \cdot v \), showing \( \Delta p_B > \Delta p_A \).

To sum up:
  • A higher mass or greater change in velocity results in a greater change in momentum.
  • Momentum is conserved in an isolated system unless acted upon by an external force.
  • The units of momentum are the same as impulse: kilogram meter per second (\( \text{kg} \cdot \text{m/s} \)).

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Most popular questions from this chapter

Part \(a\) of the drawing shows a bullet approaching two blocks resting on a horizontal frictionless surface. Air resistance is negligible. The bullet passes completely through the first block (an inelastic collision) and embeds itself in the second one, as indicated in part \(b\). Note that both blocks are moving after the collision with the bullet. (a) Can the conservation of linear momentum be applied to this three-object system, even though the second collision occurs a bit later than the first one? Justify your answer. Neglect any mass removed from the first block by the bullet. (b) Is the total kinetic energy of this three-body system conserved? If not, would the total kinetic energy after the collisions be greater than or smaller than that before the collisions? Justify your answer.

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