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A 2646 lb car is moving on the freeway at 68 mph. (a) Find the magnitude of its momentum and its kinetic energy in SI units. (b) At what speed, in \(\mathrm{m} / \mathrm{s}\), will the car's momentum be half of what it is in part (a)? (c) At what speed, in \(\mathrm{m} / \mathrm{s}\), will the car's kinetic energy be half of what it is in part (a)?

Short Answer

Expert verified
Momentum: 36486.5 kg·m/s, KE: 554316 J. New speeds: (b) 15.201 m/s; (c) 15.188 m/s.

Step by step solution

01

Convert Units

First, convert the car's weight from pounds (lb) to kilograms (kg) and its speed from miles per hour (mph) to meters per second (m/s). You need these conversions:- 1 lb = 0.453592 kg.- 1 mph = 0.44704 m/s.For the car's mass: \[ 2646 \text{ lb} \times 0.453592 \approx 1200.38 \text{ kg} \]For the car's speed: \[ 68 \text{ mph} \times 0.44704 \approx 30.403 \text{ m/s} \]
02

Calculate Initial Momentum

Momentum is calculated using the formula: \[ p = mv \]Where \( m \) is mass and \( v \) is velocity. Substitute the converted values:\[ p = 1200.38 \times 30.403 \approx 36486.5 \, \text{kg} \cdot \text{m/s} \]
03

Calculate Initial Kinetic Energy

Kinetic Energy (KE) is calculated using the formula:\[ KE = \frac{1}{2}mv^2 \]Substitute the mass and velocity:\[ KE = \frac{1}{2} \times 1200.38 \times (30.403)^2 \approx 554316.0 \, \text{J} \]
04

Calculate New Speed for Half Momentum

If the car's momentum is half, let its new speed be \( v_1 \).\[ \frac{1}{2} \times 36486.5 = 1200.38 \times v_1 \]Solving for \( v_1 \):\[ v_1 = \frac{18243.25}{1200.38} \approx 15.201 \, \text{m/s} \]
05

Calculate New Speed for Half Kinetic Energy

If the car's kinetic energy is half, let its new speed be \( v_2 \).\[ \frac{1}{2} \times 554316.0 = \frac{1}{2} \times 1200.38 \times v_2^2 \]Cancel the \( \frac{1}{2} \) and solve for \( v_2 \):\[ 277158.0 = 1200.38 \times v_2^2 \]\[ v_2^2 = \frac{277158.0}{1200.38} \]\[ v_2 \approx \sqrt{230.77} \approx 15.188 \, \text{m/s} \]

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Key Concepts

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

Momentum
Momentum is a fundamental concept in physics, often considered as the "quantity of motion" an object possesses. It is defined as the product of an object's mass and velocity, represented by the formula:\[ p = mv \]Where:
  • \( p \) is the momentum, measured in kilogram meters per second (kg·m/s).
  • \( m \) is the mass of the object in kilograms (kg).
  • \( v \) is the velocity in meters per second (m/s).
Momentum is a vector quantity, which means it has both magnitude and direction. The more mass an object has, or the faster it is going, the greater its momentum. Understanding momentum helps in analyzing the motion of cars, like in our exercise problem, making it easier to predict how they behave in different scenarios. Even when an object changes speed, its momentum calculation guides how force and motion interplay.
Kinetic Energy
Kinetic energy reflects the energy an object possesses due to its motion. Unlike momentum, kinetic energy is a scalar quantity, which means it only has magnitude, not direction. The kinetic energy of an object is calculated using the formula:\[ KE = \frac{1}{2}mv^2 \]Where:
  • \( KE \) stands for kinetic energy, measured in joules (J).
  • \( m \) represents mass in kilograms (kg).
  • \( v \) is velocity in meters per second (m/s).
When calculating kinetic energy, the speed of the object impacts its energy quadratically, meaning if the speed doubles, the kinetic energy increases by four times. This is key in understanding vehicle dynamics and safety. For the car, knowing how speed influences kinetic energy helps in strategizing energy consumption and hazard avoidance.
Unit Conversion
Unit conversion is essential when solving physics problems as it ensures uniformity in calculations. Consistent units prevent errors and allow accurate application of formulas. For our exercise, converting: - Weight from pounds (lb) to kilograms (kg). Use the conversion factor: 1 lb = 0.453592 kg. - Speed from miles per hour (mph) to meters per second (m/s) with: 1 mph = 0.44704 m/s. These conversions are crucial since SI (International System of Units) is the standard in scientific computation, favoring units like kilograms for mass and meters per second for speed. It's like speaking a common language in science. Once the data switches to SI units, the calculations for momentum and kinetic energy become streamlined and accurate, allowing clearer interpretations and comparisons.
Physics Formulas
Physics relies heavily on formulas to describe how the universe works. These formulas are mathematical expressions capturing relationships between physical quantities. In our context, basic formulas like:
  • Momentum: \( p = mv \)
  • Kinetic Energy: \( KE = \frac{1}{2}mv^2 \)
Illuminate the intricacies of motion and energy. They simplify complex real-world phenomena into digestible calculations. Applying these formulas in structured steps solves problems effectively—like finding changes in speed that affect momentum or kinetic energy. Once familiar with these tools, you can bridge mathematical concepts with physical reality, enhancing both problem-solving skills and comprehension of how motion impacts surrounding environments.

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Most popular questions from this chapter

A mass \(m\) is placed at the rim of a frictionless hemispherical bowl with a radius \(R\) and released from rest (Figure 8.49 ). It then slides down and undergoes a perfectly elastic collision with a second mass \(3 m\) sitting at rest at the bottom of the bowl. (a) What are the direction and speed of each mass just after the collision? (b) In terms of \(R\), to what maximum height will each mass travel after the collision?

Tennis players sometimes leap into the air to return a volley. (a) If a \(57 \mathrm{~g}\) tennis ball is traveling horizontally at \(72 \mathrm{~m} / \mathrm{s}\) (which does occur), and a \(61 \mathrm{~kg}\) tennis player leaps vertically upward and hits the ball, causing it to travel at \(45 \mathrm{~m} / \mathrm{s}\) in the reverse direction, how fast will her center of mass be moving horizontally just after hitting the ball? (b) If, as is reasonable, her racket is in contact with the ball for \(30.0 \mathrm{~ms}\), what force does her racket exert on the ball? What force does the ball exert on the racket?

To protect their young in the nest, peregrine falcons will fly into birds of prey (such as ravens) at high speed. In one such episode, a \(600 \mathrm{~g}\) falcon flying at \(20.0 \mathrm{~m} / \mathrm{s}\) flew into a \(1.5 \mathrm{~kg}\) raven flying at \(9.0 \mathrm{~m} / \mathrm{s}\). The falcon hit the raven at right angles to its original path and bounced back with a speed of \(5.0 \mathrm{~m} / \mathrm{s} .\) By what angle did the falcon change the raven's direction of motion?

On a frictionless air track, a \(0.150 \mathrm{~kg}\) glider moving at \(1.20 \mathrm{~m} / \mathrm{s}\) to the right collides with and sticks to a stationary \(0.250 \mathrm{~kg}\) glider. (a) What is the net momentum of this two-glider system before the collision? (b) What must be the net momentum of this system after the collision? Why? (c) Use your answers in parts (a) and (b) to find the speed of the gliders after the collision. (d) Is kinetic energy conserved during the collision?

On a very muddy football field, a \(110 \mathrm{~kg}\) linebacker tackles an \(85 \mathrm{~kg}\) halfback. Immediately before the collision, the linebacker is slipping with a velocity of \(8.8 \mathrm{~m} / \mathrm{s}\) north and the halfback is sliding with a velocity of \(7.2 \mathrm{~m} / \mathrm{s}\) east. What is the velocity (magnitude and direction) at which the two players move together immediately after the collision?

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