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Three birds are flying in a compact formation. The first bird, with a mass of \(100 . \mathrm{g}\) is flying \(35.0^{\circ}\) east of north at a speed of \(8.00 \mathrm{~m} / \mathrm{s}\). The second bird, with a mass of \(123 \mathrm{~g}\), is flying \(2.00^{\circ}\) east of north at a speed of \(11.0 \mathrm{~m} / \mathrm{s}\). The third bird, with a mass of \(112 \mathrm{~g}\), is flying \(22.0^{\circ}\) west of north at a speed of \(10.0 \mathrm{~m} / \mathrm{s}\). What is the momentum vector of the formation? What would be the speed and direction of a \(115-\mathrm{g}\) bird with the same momentum?

Short Answer

Expert verified
Answer: The speed and direction of the fourth bird are given by the formula velocity4 and direction4, which are calculated by dividing the magnitude of the total momentum vector by the mass of the fourth bird and using the direction of the total momentum vector, respectively.

Step by step solution

01

Calculate the momentum of each bird individually

First, we'll find the momentum of each bird in the formation using the formula: momentum = mass × velocity. Since each bird has different mass, velocity, and direction, we'll find the respective momentum of each bird along the x and y axis, respectively.
02

Step 1.1: Calculate the x and y components of the momentum for the first bird

The mass of the first bird is \(100\) g. Convert this mass into kg by dividing by 1000: Mass \(= \frac{100}{1000} = 0.1 \,\text{kg}\) Now, find the x and y components of its velocity using trigonometry: v_x1 = v1 * cos(35.0°) v_y1 = v1 * sin(35.0°) Calculate the momentum in the x and y directions: momentum_x1 = mass1 * v_x1 momentum_y1 = mass1 * v_y1
03

Step 1.2: Calculate the x and y components of the momentum for the second bird

The mass of the second bird is \(123\) g. Convert this mass into kg by dividing by 1000: Mass \(= \frac{123}{1000} = 0.123 \,\text{kg}\) Now, find the x and y components of its velocity using trigonometry: v_x2 = v2 * cos(2.0°) v_y2 = v2 * sin(2.0°) Calculate the momentum in the x and y directions: momentum_x2 = mass2 * v_x2 momentum_y2 = mass2 * v_y2
04

Step 1.3: Calculate the x and y components of the momentum for the third bird

The mass of the third bird is \(112\) g. Convert this mass into kg by dividing by 1000: Mass \(= \frac{112}{1000} = 0.112 \,\text{kg}\) Now, find the x and y components of its velocity using trigonometry: v_x3 = v3 * cos(-22.0°) v_y3 = v3 * sin(-22.0°) Calculate the momentum in the x and y directions: momentum_x3 = mass3 * v_x3 momentum_y3 = mass3 * v_y3
05

Calculate the momentum vector of the formation

Now that we have the momentum components of each bird in the x and y directions, we can find the total momentum vector of the formation by summing the individual components: total_momentum_x = momentum_x1 + momentum_x2 + momentum_x3 total_momentum_y = momentum_y1 + momentum_y2 + momentum_y3 Now that we have the total momentum in x and y directions, we can find the magnitude and direction of the momentum vector using Pythagorean theorem and arctan function: magnitude = sqrt(total_momentum_x^2 + total_momentum_y^2) direction = atan(total_momentum_y / total_momentum_x)
06

Calculate the speed and direction of the fourth bird

Given that the fourth bird has a mass of \(115\) g and the same momentum as the formation, we can find its speed and direction. First, convert the mass into kg: mass4 = \(\frac{115}{1000} = 0.115 \,\text{kg}\) Now, use the momentum magnitude we found in step 2 to find the fourth bird's velocity: velocity4 = magnitude / mass4 Use the direction of the momentum vector to find the direction in which the fourth bird should fly: direction4 = direction Now we have the speed and direction of the fourth bird that has the same momentum as the formation.

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Key Concepts

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

Physics Momentum
When we talk about momentum in physics, we're discussing a measurement of the quantity of motion an object possesses. This concept is vital in understanding both motion and collisions. The momentum of an object is determined by two factors: its mass and its velocity. In mathematical terms, momentum (\textbf{p}) is expressed as the product of an object's mass (m) and its velocity (\textbf{v}), so the formula is \[ \textbf{p} = m \times \textbf{v} \].

In a scenario with multiple objects, like our birds in formation, the total momentum of the system is the vector sum of all individual momenta. To compute this in a problem involving different directions, we need to consider the components of the momentum vector.

In the exercise provided, as each bird flies at a different speed and direction, their individual momenta contribute uniquely to the formation's overall momentum. This accumulation is essential in determining the behavior of the formation if any external forces act upon it. Moreover, in a closed system, the law of conservation of momentum states that the total momentum remains constant, unless acted upon by an external force.
Vector Components
Vector components are a way to break down a vector into its parts along the axes of the coordinate system, usually along the x (horizontal) and y (vertical) axes. This breakdown is immensely helpful in simplifying calculations and understanding the motion in each direction independently.

For instance, calculating the momentum of each bird means splitting their velocities into horizontal and vertical parts. This is done using trigonometry, specifically the sine and cosine functions, which relate the angle of motion to the respective x and y components.

Once we have calculated the x (horizontal) and y (vertical) components of each bird's momentum, we can find the total momentum for the formation by adding up these components independently. The formula for a momentum component along an axis is \[ \text{momentum}_{\text{axis}} = \text{mass} \times \text{velocity}_{\text{axis}} \].

In the birds' formation, we calculate these components by first converting mass into kilograms and then applying the cosine (for the x-axis) and sine (for the y-axis) to the birds' velocities in conjunction with their flight angles. This operation is key to understanding vector quantities' roles in various physical situations and plays an integral part in problem-solving.
Trigonometry in Physics
Trigonometry is not just a topic in mathematics but also an essential tool in physics, particularly when dealing with vector quantities like velocity, force, and momentum. In our exercise, trigonometry helps determine the direction and magnitude components of vectors.

The two trigonometric functions that come into play with vectors are sine (sin) and cosine (cos). They are used to calculate the x and y components of a vector, given its magnitude and angle of direction. For example, if a bird is flying at an angle \( \theta \) from a reference direction, the velocity components can be found using \[ \text{velocity}_x = \text{velocity} \times \cos(\theta) \] and \[ \text{velocity}_y = \text{velocity} \times \sin(\theta) \].

In the context of our problem, this trigonometric approach allows us to deconstruct each bird's velocity vector into more manageable x and y components, which can be used to find the overall momentum of the bird formation. It's this combination of trigonometry and vector analysis that proves invaluable for solving a wide array of physics problems involving direction and magnitude.

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Most popular questions from this chapter

One of the events in the Scottish Highland Games is the sheaf toss, in which a \(9.09-\mathrm{kg}\) bag of hay is tossed straight up into the air using a pitchfork. During one throw, the sheaf is launched straight up with an initial speed of \(2.7 \mathrm{~m} / \mathrm{s}\). a) What is the impulse exerted on the sheaf by gravity during the upward motion of the sheaf (from launch to maximum height)? b) Neglecting air resistance, what is the impulse exerted by gravity on the sheaf during its downward motion (from maximum height until it hits the ground)? c) Using the total impulse produced by gravity, determine how long the sheaf is airborne.

An 83.0 -kg running back leaps straight ahead toward the end zone with a speed of \(6.50 \mathrm{~m} / \mathrm{s}\). A 115 -kg linebacker, keeping his feet on the ground, catches the running back and applies a force of \(900 . \mathrm{N}\) in the opposite direction for 0.750 s before the running back's feet touch the ground. a) What is the impulse that the linebacker imparts to the running back? b) What change in the running back's momentum does the impulse produce? c) What is the running back's momentum when his feet touch the ground? d) If the linebacker keeps applying the same force after the running back's feet have touched the ground, is this still the only force acting to change the running back's momentum?

A particle \(\left(M_{1}=1.00 \mathrm{~kg}\right)\) moving at \(30.0^{\circ}\) downward from the horizontal with \(v_{1}=2.50 \mathrm{~m} / \mathrm{s}\) hits a second particle \(\left(M_{2}=2.00 \mathrm{~kg}\right),\) which was at rest momentarily. After the collision, the speed of \(M_{1}\) was reduced to \(.500 \mathrm{~m} / \mathrm{s}\), and it was moving at an angle of \(32^{\circ}\) downward with respect to the horizontal. Assume the collision is elastic. a) What is the speed of \(M_{2}\) after the collision? b) What is the angle between the velocity vectors of \(M_{1}\) and \(M_{2}\) after the collision?

When you open the door to an air-conditioned room, you mix hot gas with cool gas. Saying that a gas is hot or cold actually refers to its average energy; that is, the hot gas molecules have a higher kinetic energy than the cold gas molecules. The difference in kinetic energy in the mixed gases decreases over time as a result of elastic collisions between the gas molecules, which redistribute the energy. Consider a two-dimensional collision between two nitrogen molecules \(\left(\mathrm{N}_{2},\right.\) molecular weight \(=28.0 \mathrm{~g} / \mathrm{mol}\) ). One molecule moves at \(30.0^{\circ}\) with respect to the horizontal with a velocity of \(672 \mathrm{~m} / \mathrm{s} .\) This molecule collides with a second molecule moving in the negative horizontal direction at \(246 \mathrm{~m} / \mathrm{s}\). What are the molecules' final velocities if the one that is initially more energetic moves in the vertical direction after the collision?

A bored boy shoots a soft pellet from an air gun at a piece of cheese with mass \(0.25 \mathrm{~kg}\) that sits, keeping cool for dinner guests, on a block of ice. On one particular shot, his 1.2-g pellet gets stuck in the cheese, causing it to slide \(25 \mathrm{~cm}\) before coming to a stop. According to the package the gun came in, the muzzle velocity is \(65 \mathrm{~m} / \mathrm{s}\). What is the coefficient of friction between the cheese and the ice?

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