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Chapter 9: Problem 7

Truck Traveling North A \(2100 \mathrm{~kg}\) truck traveling north at \(41 \mathrm{~km} / \mathrm{h}\) turns east and accelerates to \(51 \mathrm{~km} / \mathrm{h}\). (a) What is the change in the kinetic energy of the truck? What are the (b) magnitude and (c) direction of the change in the translational momentum of the truck?

Short Answer

Expert verified
Change in kinetic energy: 75,244.77 J. Magnitude of change in momentum: 37,947 kg m/s. Direction: 38.66掳 south of east.

Step by step solution

01

Convert speeds to meters per second

First, convert the initial and final speeds from km/h to m/s using the conversion factor \ 1 \text{ km/h} = \frac{1}{3.6} \text{ m/s}: \( v_i = 41 \text{ km/h} \times \frac{1}{3.6} = 11.39 \text{ m/s}\) and \( v_f = 51 \text{ km/h} \times \frac{1}{3.6} = 14.17 \text{ m/s}\)
02

Calculate initial and final kinetic energy

Use the kinetic energy formula \ \( KE = \frac{1}{2}mv^2 \): Initial kinetic energy: \( KE_i = \frac{1}{2} \times 2100 \text{ kg} \times (11.39 \text{ m/s})^2 = 136,052.595 \text{ J}\) and Final kinetic energy: \( KE_f = \frac{1}{2} \times 2100 \text{ kg} \times (14.17 \text{ m/s})^2 = 211,297.365 \text{ J}\)
03

Determine the change in kinetic energy

Subtract the initial kinetic energy from the final kinetic energy: \( \Delta KE = KE_f - KE_i = 211,297.365 \text{ J} - 136,052.595 \text{ J} = 75,244.77 \text{ J}\)
04

Calculate initial and final momentum components

Find the momentum components using \ \( p = mv \): Initially, \( p_{i,x} = 0 \text{ kg}\text{ m/s}\) (north) and \( p_{i,y} = 2100 \text{ kg} \times 11.39 \text{ m/s} = 23,919 \text{ kg}\text{ m/s}\) Finally, \( p_{f,x} = 2100 \text{ kg} \times 14.17 \text{ m/s} = 29,757 \text{ kg}\text{ m/s}\) (east) and \( p_{f,y} = 0 \text{ kg}\text{ m/s}\)
05

Compute change in momentum components

Calculate the change in momentum for each component: \( \Delta p_x = p_{f,x} - p_{i,x} = 29,757 \text{ kg m/s} - 0 = 29,757 \text{ kg m/s}\) and \( \Delta p_y = p_{f,y} - p_{i,y} = 0 - 23,919 \text{ kg m/s} = -23,919 \text{ kg m/s} \ (south)\)
06

Determine magnitude of the change in momentum

Use the Pythagorean theorem: \ \( \Delta p = \sqrt{(\Delta p_x)^2 + (\Delta p_y)^2} = \sqrt{(29,757)^2 + (-23,919)^2} = 37,947 \text{ kg}\text{ m/s}\)
07

Find the direction of the change in momentum

Calculate the angle using \ \( \theta = \tan^{-1} \left( \frac{\Delta p_y}{\Delta p_x} \right) \): \( \theta = \tan^{-1} \left( \frac{-23,919}{29,757} \right) = -38.66^\text{\circ}\) The direction is 38.66掳 south of east.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy an object has due to its motion. The formula to calculate kinetic energy (\text{KE}) is given by: \[ KE = \frac{1}{2} mv^2 \] where 'm' denotes mass and 'v' is the velocity. In the example of the truck, the calculations involve converting the truck's speeds from km/h to m/s before applying the kinetic energy formula. This conversion is key because the SI unit for velocity in the kinetic energy equation is meters per second (m/s). The kinetic energy of the truck increases from an initial value of 136,052.595 Joules to a final value of 211,297.365 Joules as it changes direction from north to east and accelerates.
Translational Momentum
Translational momentum, also known simply as momentum, measures the quantity of motion an object has. It is a vector quantity, meaning it has both magnitude and direction, and is calculated using: \[ p = mv \] Here, 'p' stands for momentum, 'm' is the mass, and 'v' is the velocity. In the truck example, the initial and final momentum vectors must be calculated separately to determine the change in momentum. The truck initially has a momentum component only in the north direction, and after turning, it has a component in the east direction. The change in momentum involves a vector subtraction of these components.
Pythagorean Theorem
The Pythagorean theorem is essential for determining the magnitude of the change in momentum, which is a vector. The theorem states: \[ a^2 + b^2 = c^2 \] For the truck, we use the theorem to find the resultant momentum vector from its components. We calculate the northward component (negative due to southward change) and the eastward component. The magnitude of the total change in momentum is \[ \Delta p = \sqrt{(\Delta p_x)^2 + (\Delta p_y)^2} \]. This gives a value that combines both directional changes into a single measure.
Tangent Function
The tangent function helps derive the direction of the change in momentum. The angle (\theta) relative to the east can be calculated using the inverse tangent (\tan^{-1}) function: \[ \theta = \tan^{-1} \left( \frac{\Delta p_y}{\Delta p_x} \right) \]. In this case, \( \Delta p_y \) is the southward momentum (negative value), and \( \Delta p_x \) is the eastward momentum. The resulting angle tells us how far south of east the momentum has shifted. Basic trigonometry allows students to relate sides of a right-angled triangle with its angles.
Conversion of Units
Conversion of units is crucial for consistency, especially in physics problems where different units are used. In the exercise, speeds were converted from km/h to m/s using the conversion factor, \(1 \text{ km/h} = \frac{1}{3.6} \text{ m/s} \). This conversion ensures the calculated kinetic energy and momentum values are correct. Standardized units simplify calculations and reduce errors. Always check that your units match the formula requirements before solving a problem.

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Most popular questions from this chapter

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