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Mario, a hockey player, is skating due south at a speed of \(7.0 \mathrm{~m} / \mathrm{s}\) relative to the ice. A teammate passes the puck to him. The puck has a speed of \(11.0 \mathrm{~m} / \mathrm{s}\) and is moving in a direction of \(22^{\circ}\) west of south, relative to the ice. What are the magnitude and direction (relative to due south) of the puck's velocity, as observed by Mario?

Short Answer

Expert verified
Puck's speed relative to Mario is approximately 6.9 m/s, moving 43° west of south.

Step by step solution

01

Define the Coordinate System

Assign south as the negative y-direction and west as the negative x-direction. This helps in breaking down the puck's velocity into components relative to Mario.
02

Determine Puck's Velocity Components

The puck's velocity relative to the ice is \(11.0 \ m/s\). Its direction is \(22^\circ\) west of south, meaning \(22^\circ\) from the negative y-axis towards the negative x-axis. Calculate the components:- The southward (y-component): \[ v_{py} = 11.0 \cos(22^\circ) \]- The westward (x-component): \[ v_{px} = 11.0 \sin(22^\circ) \]
03

Calculate Puck's Y-Component Relative to Mario

As Mario moves south at \(7.0 \ m/s\), the puck's southward component relative to Mario is the difference between the puck's and Mario's velocities in the y-direction:\[ v_{py}' = v_{py} - 7.0 \ m/s \]
04

Calculate Puck's Velocity Relative to Mario

The puck's westward component remains the same relative to Mario, as there is no corresponding movement in the x-direction by Mario. Thus, the puck's velocity relative to Mario is:\[ v' = \sqrt{(v_{px})^2 + (v_{py}')^2} \]
05

Determine the Direction of Puck's Velocity Relative to Mario

To find the angle \( \theta' \) with respect to south, use the arctangent function:\[ \theta' = \tan^{-1}\left(\frac{v_{px}}{v_{py}'}\right) \]The angle \( \theta' \) will be west of south.

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Key Concepts

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

Vector components
In physics, understanding vector components is crucial when dealing with forces or velocities, particularly when they're not aligned with a single axis. A vector is a quantity that has both magnitude and direction. For practical calculations, you can break down vectors into their components along the axes of a coordinate system. This process is known as vector resolution.

When we say a vector component, we're referring to the projection of the vector along a certain direction or axis, which simplifies calculations:
  • The southward component in the original exercise is aligned with the negative y-axis.
  • The westward component aligns with the negative x-axis.
By using trigonometric functions, you can determine these components. The y-component of a vector is usually determined by multiplying the vector's magnitude by the cosine of its angle when measured from the axis. Meanwhile, the x-component is determined using the sine function. This breakdown is critical to reshaping the problem into simpler, dimension-independent calculations that can be easily manipulated and recombined.
Coordinate system
A coordinate system in physics provides a framework for describing the position and movement of an object. Choosing an appropriate coordinate system is essential because it allows you to simplify complex physical scenarios into manageable mathematics.

In our specific scenario of the hockey player and the puck, the coordinate system is assigned such that:
  • South is the negative y-direction.
  • West is the negative x-direction.
This assignment allows for clear and unambiguous interpretation of directions relative to each player. By placing southward and westward movements in negative directions, we align the coordinate system with common geographic direction norms. This consistency aids in visualizing components and solving physics problems by maintaining clarity on which directions we consider positive and negative.
Trigonometry in physics
Trigonometry is a powerful tool in physics for resolving vectors and understanding angles relative to specified directions. When dealing with motion or forces at an angle, trigonometric functions like sine, cosine, and tangent are used to decompose complex vectors into simpler parts.

For example, consider a vector at an angle from a reference direction. In the exercise given:
  • The puck's velocity vector makes an angle of \(22^{\circ}\) west of south.
  • The cosine function helps find how much of the puck's speed is directed southward.
  • The sine function finds the westward component.
  • The arctangent function calculates the direction of the resultant velocity relative to Mario's motion.
These trigonometric identities and methods allow us to convert diagonal or angled forces and movements into straight paths that align neatly with the established coordinate system, making complex scenarios more digestible.

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Most popular questions from this chapter

A car drives straight off the edge of a cliff that is \(54 \mathrm{~m}\) high. The police at the scene of the accident observe that the point of impact is \(130 \mathrm{~m}\) from the base of the cliff. How fast was the car traveling when it went over the cliff?

On a spacecraft, two engines are turned on for \(684 \mathrm{~s}\) at a moment when the velocity of the craft has \(x\) and \(y\) components of \(v_{0 x}=4370 \mathrm{~m} / \mathrm{s}\) and \(v_{0 y}=6280 \mathrm{~m} / \mathrm{s} .\) While the engines are firing, the craft undergoes a displacement that has components of \(x=4.11 \times 10^{6} \mathrm{~m}\) and \(y=6.07 \times 10^{6} \mathrm{~m} .\) Find the \(x\) and \(y\) components of the craft's acceleration.

Interactive Solution \(\underline{3.11}\) at presents a model for solving this problem. The earth moves around the sun in a nearly circular orbit of radius \(1.50 \times 10^{11} \mathrm{~m} .\) During the three summer months (an elapsed time of \(7.89 \times 10^{6} \mathrm{~s}\) ), the earth moves one-fourth of the distance around the sun. (a) What is the average speed of the earth? (b) What is the magnitude of the average velocity of the earth during this period?

Concept Simulation 3.2 at reviews the concepts that are important in this problem. A golfer imparts a speed of \(30.3 \mathrm{~m} / \mathrm{s}\) to a ball, and it travels the maximum possible distance before landing on the green. The tee and the green are at the same elevation. (a) How much time does the ball spend in the air? (b) What is the longest "hole in one" that the golfer can make, if the ball does not roll when it hits the green?

As preparation for this problem, review Conceptual Example \(10 .\) The two stones described there have identical initial speeds of \(v_{0}=13.0 \mathrm{~m} / \mathrm{s}\) and are thrown at an angle \(\theta=30.0^{\circ},\) one below the horizontal and one above the horizontal. What is the distance between the points where the stones strike the ground?

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