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A \(1500-\mathrm{kg}\) blue convertible is traveling south, and a \(2000-\mathrm{kg}\) red SUV is traveling west. If the total momentum of the system consisting of the two cars is 8000 \(\mathrm{kg} \cdot \mathrm{m} / \mathrm{s}\) directed at \(60.0^{\circ}\) west of south, what is the speed of each vehicle?

Short Answer

Expert verified
The speeds are 2.67 m/s for the convertible and 3.46 m/s for the SUV.

Step by step solution

01

Understand the Momentum Components

Given the total momentum and its direction, decompose it into its southward and westward components. The momentum vector has a magnitude of 8000 kg·m/s directed at 60° west of south. Use trigonometry to find components:- Southward component: \( p_s = 8000 \cdot \cos(60°) \)- Westward component: \( p_w = 8000 \cdot \sin(60°) \).
02

Compute the Component Values

Calculate the southward and westward components:- \( p_s = 8000 \cdot \cos(60°) = 8000 \cdot 0.5 = 4000 \text{ kg·m/s} \)- \( p_w = 8000 \cdot \sin(60°) = 8000 \cdot \frac{\sqrt{3}}{2} \approx 6928 \text{ kg·m/s} \)
03

Apply Momentum Formula for Each Vehicle

Momentum is given by \( p = mv \). For the blue convertible traveling south:\[ 1500 \cdot v_b = 4000 \]Solve for \( v_b \). For the red SUV traveling west:\[ 2000 \cdot v_r = 6928 \]Solve for \( v_r \).
04

Solve for Each Speed

Solving the equations from Step 3:- For the blue convertible:\[ v_b = \frac{4000}{1500} = 2.67 \text{ m/s} \]- For the red SUV:\[ v_r = \frac{6928}{2000} \approx 3.46 \text{ m/s} \]
05

Verify Dimensions and Units

Ensure that the units used for masses and velocities (in kg, m/s) ensure correct units for momentum (kg·m/s) and that all calculations align with expected units.

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Key Concepts

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

Vector Decomposition
Vector decomposition is a fundamental concept in physics and mathematics, particularly when dealing with forces, velocities, or momentum that have different directional components. Imagine a force being applied in multiple directions simultaneously. To simplify problems, we often break this multi-directional force into its simpler components using vector decomposition.
To visualize this, consider the momentum of a vehicle traveling, not in a straight line but at an angle. The vehicle's total momentum can be broken into parts: one that aligns purely south and one that aligns purely west, depending on the direction it travels.
  • The southward momentum component can be found using the cosine function, calculated as: \( p_s = 8000 \cdot \cos(60°) \).
  • The westward momentum component can be found using the sine function, calculated as: \( p_w = 8000 \cdot \sin(60°) \).
  • This process helps us analyze each direction's contribution separately.
By understanding vector decomposition, you can tackle complex motion situations and break them into simpler, more manageable problems.
Trigonometry in Physics
Trigonometry plays a crucial role in physics, especially when dealing with vectors and directions. Understanding angles and their sine and cosine functions help us resolve vectors into components. This is essential in problems that require finding the effect of forces, momentum, or velocity in different directions.
When a force or momentum is given at an angle, like in this problem (60° west of south), we utilize cosine and sine to find the specific north-south and east-west components.
  • \( \cos(60°) \) helps find the component parallel to the reference line (south, in this example).
  • \( \sin(60°) \) determines the perpendicular component (west, here).
  • Trigonometric functions allow us to precisely calculate each component of the vector.
Mastering trigonometry in physics enables a better understanding of how forces and movements can be decomposed and analyzed in real-world applications.
Solving Linear Equations
Linear equations are mathematical expressions of a straight line and are crucial in deriving unknown variables in physics. When solving problems involving momentum and velocity, they help to determine speeds or other quantities by establishing relationships between them.
In the given exercise, the momentum of each vehicle is expressed as a linear equation \( p = mv \), where \( p \) is momentum, \( m \) is mass, and \( v \) is velocity.
  • For the blue convertible: \( 1500 \cdot v_b = 4000 \) leads to solving for \( v_b \).
  • For the red SUV: \( 2000 \cdot v_r = 6928 \) solves for \( v_r \).
  • These equations allow us to isolate and calculate the speed (\( v \)) of each moving object.
Understanding how to solve such equations empowers you to approach physics problems analytically, providing clear solutions and insights into physical systems.

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Most popular questions from this chapter

A \(5.00-\mathrm{g}\) bullet is shot through a \(1.00-\mathrm{kg}\) wood block suspended on a string 2.00 \(\mathrm{m}\) long. The center of mass of the block rises a distance of 0.45 \(\mathrm{cm} .\) Find the speed of the bullet as it emerges from the block if its initial speed is 450 \(\mathrm{m} / \mathrm{s}\) .

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Suppose you hold a small ball in contact with, and directly over, the center of a large ball. If you then drop the small ball a short time after dropping the large ball, the small ball rebounds with surprising speed. To show the extreme case, ignore air resistance and suppose the large ball makes an clastic collision with the floor and then rebounds to make an elastic collision with the still- descending small ball. Just before the collision between the two balls, the large ball is moving upward with velocity \(\overrightarrow{\boldsymbol{v}}\) and the small ball has velocity \(-\overrightarrow{\boldsymbol{v}}\) . (Do you see why? Assume the large ball has a much greater mass than the small ball. (a) What is the velocity of the small ball immediately after its collision with the large hall? (b) From the answer to part (a), what is the ratio of the small ball's rebound distance to the distance it fell before the collision?

Just before it is struck by a racket, a tennis ball weighing 0.560 \(\mathrm{N}\) has a velocity of \((20.0 \mathrm{m} / \mathrm{s}) \hat{\imath}-(4.0 \mathrm{m} / \mathrm{s}) \hat{\mathrm{J}}\) . During the 3.00 \(\mathrm{ms}\) that the racket and ball are in contact, the net force on the ball is constant and equal to \(-(380 \mathrm{N}) \hat{\imath}+(110 \mathrm{N}) \mathrm{J}\) . (a) What are the \(x\) -and \(y\) components of the impulse of the net force applied to the ball? \((b)\) What are the \(x-\) and \(y\) -components of the final velocity of the ball?

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