Problem 75 - University Physics with Modern Physics

Question

Two soccer players, Mia and Alice, are running as Alice passes the ball to Mia. Mia is running due north with a speed of 6.00 m/s. The velocity of the ball relative to Mia is 5.00 m/s in a direction 30.0\(^\circ\) east of south. What are the magnitude and direction of the velocity of the ball relative to the ground?

Step-by-Step Solution

Answer

The ball's velocity relative to the ground is 3.01 m/s at 56.3° east of north.

Step 1: Understand the Problem

We need to find the velocity of the ball relative to the ground. We have two velocities: Mia's velocity (6.00 m/s north) and the ball's velocity relative to Mia (5.00 m/s 30.0° east of south). We want to determine the actual or absolute velocity of the ball as seen from the ground.

Step 2: Resolve Ball's Relative Velocity into Components

The velocity of the ball relative to Mia needs to be split into its velocity components in the north-south and east-west directions. Since it's 30.0° east of south:\[ v_{b/M} = 5.00\, \text{m/s} \]- Southward component: \( v_{b/M, s} = 5.00 \times \cos(30.0°) = 5.00 \times \frac{\sqrt{3}}{2} = 4.33 \text{ m/s south} \)- Eastward component: \( v_{b/M, e} = 5.00 \times \sin(30.0°) = 5.00 \times 0.5 = 2.50 \text{ m/s east} \)

Step 3: Combine Mia's Velocity with Ball's Components

Mia is moving north with a velocity of 6.00 m/s; therefore:- The northward component of the ball is Mia's northward velocity minus the southward component from the relative velocity:\[ v_{b, n} = 6.00 - 4.33 = 1.67 \text{ m/s north} \]- The eastward component from the ball relative to Mia remains as it is:\[ v_{b, e} = 2.50 \text{ m/s east} \]

Step 4: Find the Resultant Velocity of the Ball

Now we have the components of the ball's velocity relative to the ground:- Northward component of 1.67 m/s- Eastward component of 2.50 m/sWe use the Pythagorean theorem to find the magnitude of the resultant velocity:\[ v_b = \sqrt{(1.67)^2 + (2.50)^2} \approx \sqrt{2.79 + 6.25} = \sqrt{9.04} \approx 3.01\, \text{m/s} \]

Step 5: Calculate the Direction of the Ball's Velocity

To find the direction, we use the tangent function for the components:\[ \theta = \arctan\left(\frac{2.50}{1.67}\right) \approx \arctan(1.497) = 56.3° \]This angle is measured east of north (since the north component is the base and east is positive). Hence, the direction is 56.3° east of north.

Key Concepts

Vector Components

When analyzing movement, it's crucial to break down velocities into vector components. Think of a vector as an arrow that shows both direction and magnitude (size). For Mia and the ball, we need to decompose the ball's velocity into two directions: north-south (vertical) and east-west (horizontal). In physics, vector components are like the legs of a right triangle, showing how a velocity splits into two perpendicular parts.

• Example: If a ball's speed is 5 m/s, and it travels 30° east of south, the vector components tell us exactly how much of that speed is directed south and how much is east.
• Think of using a compass to map out the direction: one hand points south, and the other east.
• These directions help you understand the movement more precisely, converting a complex direction into simple vertical and horizontal speeds.

Understanding vector components lets you manage these directions mathematically by projecting them onto axes, making complex motion easier to handle.

Pythagorean Theorem

Once we have vector components, the Pythagorean Theorem is our go-to tool for recombining them to find the complete picture: the actual speed and path of the object. This theorem relates the sides of a right triangle, providing a straightforward method to calculate the hypotenuse (the longest side).

• A right triangle has one 90-degree angle, which makes applying Pythagoras easy and powerful.
• The formula is \( c = \sqrt{a^2 + b^2} \) where \( c \) is the hypotenuse, and \( a \) and \( b \) are the triangle's legs.
• In our soccer exercise, the northward and eastward velocity components form a right triangle for the ball.

By using the Pythagorean Theorem, we find the magnitude (speed) of the ball's velocity relative to the ground as the hypotenuse, locking down its real movement path & speed.

Trigonometry in Physics

Trigonometry often steps in for final adjustments, especially for finding angles in physics problems. It uses ratios like sine, cosine, and tangent to relate the angles and sides of triangles, particularly right triangles. For these soccer players, we're interested in the tangent function to finalize the direction the ball moves relative to true north.

• The tangent of an angle in a right triangle is the ratio of opposite side over adjacent side (\( \tan(\theta) = \frac{opposite}{adjacent} \)).
• To pinpoint the ball’s direction, calculate the angle between the direction north and the actual path.
• This process is critical when you need more than just speed—understanding direction gives a full picture of motion.

So when the math says the ball moves at 56.3° east of north, it's showing how trigonometry helps us project real-world movements onto a directional map, combining all components into a single, meaningful result.