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Chapter 8: Problem 40

To protect their young in the nest, peregrine falcons will fly into birds of prey (such as ravens) at high speed. In one such episode, a 600-g falcon flying at 20.0 m/s hit a 1.50-kg raven flying at 9.0 m/s. The falcon hit the raven at right angles to its original path and bounced back at 5.0 m/s. (These figures were estimated by the author as he watched this attack occur in northern New Mexico.) (a) By what angle did the falcon change the raven's direction of motion? (b) What was the raven's speed right after the collision?

Short Answer

Expert verified
The raven changes direction by an angle \( \theta \), and its speed increases to \( v' \) after the collision.

Step by step solution

01

Analyze the Collision

The falcon and raven collide in a two-dimensional motion problem. We'll use the conservation of momentum to solve this. The collision involves two perpendicular components: the direction of the falcon and raven before the collision.
02

Define the Coordinate System

Let the initial velocity of the falcon along the x-axis be +20.0 m/s. The raven travels along the y-axis with a velocity of +9.0 m/s.
03

Apply Conservation of Momentum in X-direction

In the x-direction, the initial momentum is the falcon's momentum: \( p_{fx} = m_f v_{fx} \). After the collision, the falcon has momentum: \( -m_f v_{f'x} \). Therefore, \( m_f v_{fx} = m_f v_{f'x} + m_r v_{rx} \).
04

Apply Conservation of Momentum in Y-direction

Initially, only the raven has momentum in the y-direction: \( m_r v_{ry} \). After the collision, the momentum is \( m_f v_{f'y} + m_r v_{ry} \). Thus, \(m_r v_{ry} = m_f v_{f'y} + m_r v'_{ry} \).
05

Calculate Resultant Raven Velocity

The momentum equations are solved for the new velocity components of the raven. From conservation laws:\[ m_f v_{f'x} = m_r v_{rx} \text{ and } m_r v_{ry} = m_f v_{f'y} + m_r v'_{ry} \] We solve these for the new velocities:
06

Determine the Angle of Deflection

To find the angle, use \( \tan(\theta) = \frac{p_{ry}}{p_{rx}} \). Find the components from steps above: \( \theta = \tan^{-1} \left( \frac{v'_{ry}}{v_{rx}} \right) \). \( \theta \) is the angle of deflection.
07

Calculate Final Velocity Magnitude of Raven

Combine \( v_{rx} \) and \( v'_{ry} \) as squares to obtain the magnitude: \( v' = \sqrt{v_{rx}^2 + v_{ry}^2} \). Use Pythagorean theorem for this calculation.

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Key Concepts

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

Two-dimensional collisions
In physics, two-dimensional collisions involve objects moving in a plane, typically described in terms of x and y components. When dealing with collisions in this way, we pay attention to how the objects move horizontally and vertically. Before a collision occurs, each object has a specific momentum in these directions.
  • The horizontal (x-direction) momentum depends on the object's mass and horizontal velocity.
  • The vertical (y-direction) momentum depends on the object's mass and vertical velocity.

The total momentum in each direction gets conserved; meaning the total momentum (sum of all objects') before the collision equals the total momentum after. This lets us solve many physics problems related to movement and collisions, like when two vehicles collide at an intersection.
Peregrine falcons physics problem
The peregrine falcon problem is an excellent example of two-dimensional collision principles in action. Peregrine falcons are known for their swift, predatory skills in which they strike birds of prey, such as ravens, to protect their territory. This real-world scenario provided insights into conservation of momentum.
When the falcon strikes a raven at right angles, a two-dimensional collision occurs. Here, both horizontal and vertical components of momentum must be considered to determine the motion of the falcon and raven post-collision.
  • The falcon's motion primarily influences the x-direction momentum.
  • The raven's initial flight affects the y-direction momentum.

Understanding these interactions helps us see how natural events adhere to principles of physics.
Angle of deflection calculation
In this scenario, calculating the angle of deflection involves understanding how the raven's direction changes after the collision. This angle tells us the extent to which the raven's path has been altered. To find this, we use the tangent function, \[\tan(\theta) = \frac{v'_{ry}}{v_{rx}}\]
where \(v'_{ry}\) and \(v_{rx}\) are the new velocity components in the y and x directions, respectively. By computing the arctangent of these velocity components, we solve for the angle \(\theta\).
This angle helps us understand the dynamics of the interaction, specifically how the raven鈥檚 flight path was significantly altered by the falcon's strike.
Momentum conservation in x and y directions
The concept of momentum conservation is key to solving two-dimensional collision problems. It states that unless external forces act, the momentum of a system remains constant. For this scenario, we tackled each direction separately鈥攂ecause they act independently.
  • **X-direction**: The falcon had initial momentum due to its speed and mass. After colliding, the raven gained part of this momentum.
    Using the conservation equation:\[ m_f v_{fx} = m_f v_{f'x} + m_r v_{rx} \]
  • **Y-direction**: While initially only the raven had momentum, both the falcon and raven had y-direction components post-collision.
    Calculating with:\[m_r v_{ry} = m_f v_{f'y} + m_r v'_{ry} \]

This breakup in separable components simplifies analyzing results, assuring precise calculations for post-collision speeds and directions.

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Most popular questions from this chapter

A fireworks rocket is fired vertically upward. At its maximum height of 80.0 m, it explodes and breaks into two pieces: one with mass 1.40 kg and the other with mass 0.28 kg. In the explosion, 860 J of chemical energy is converted to kinetic energy of the two fragments. (a) What is the speed of each fragment just after the explosion? (b) It is observed that the two fragments hit the ground at the same time. What is the distance between the points on the ground where they land? Assume that the ground is level and air resistance can be ignored.

A 4.00-g bullet, traveling horizontally with a velocity of magnitude 400 m/s, is fired into a wooden block with mass 0.800 kg, initially at rest on a level surface. The bullet passes through the block and emerges with its speed reduced to 190 m/s. The block slides a distance of 72.0 cm along the surface from its initial position. (a) What is the coefficient of kinetic friction between block and surface? (b) What is the decrease in kinetic energy of the bullet? (c) What is the kinetic energy of the block at the instant after the bullet passes through it?

A 10.0-g marble slides to the left at a speed of 0.400 m/s on the frictionless, horizontal surface of an icy New York sidewalk and has a head- on, elastic collision with a larger 30.0-g marble sliding to the right at a speed of 0.200 m/s (\(\textbf{Fig. E8.48}\)). (a) Find the velocity of each marble (magnitude and direction) after the collision. (Since the collision is head-on, all motion is along a line.) (b) Calculate the \(change\) \(in\) \(momentum\) (the momentum after the collision minus the momentum before the collision) for each marble. Compare your values for each marble. (c) Calculate the \(change\) \(in\) \(kinetic\) \(energy\) (the kinetic energy after the collision minus the kinetic energy before the collision) for each marble. Compare your values for each marble.

You (mass 55 kg) are riding a frictionless skateboard (mass 5.0 kg) in a straight line at a speed of 4.5 m/s. A friend standing on a balcony above you drops a 2.5-kg sack of flour straight down into your arms. (a) What is your new speed while you hold the sack? (b) Since the sack was dropped vertically, how can it affect your \(horizontal\) motion? Explain. (c) Now you try to rid yourself of the extra weight by throwing the sack straight up. What will be your speed while the sack is in the air? Explain.

Block \(B\) (mass 4.00 kg) is at rest at the edge of a smooth platform, 2.60 m above the floor. Block \(A\) (mass 2.00 kg) is sliding with a speed of 8.00 m/s along the platform toward block \(B\). \(A\) strikes \(B\) and rebounds with a speed of 2.00 m/s. The collision projects \(B\) horizontally off the platform. What is the speed of \(B\) just before it strikes the floor?
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