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Bird Defense. To protect their young in the nest, 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 \(\mathrm{m} / \mathrm{s}\) hit a 1.50 -kg raven flying at 9.0 \(\mathrm{m} / \mathrm{s} .\) The falcon hit the raven at right angles to its original path and bounced back at 5.0 \(\mathrm{m} / \mathrm{s}\) . (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 approximately 42° and moves at 13.45 m/s after the collision.

Step by step solution

01

Understanding the Problem

We have a situation involving a falcon and a raven in a collision. The problem asks for two things: (a) the angle at which the raven's direction of motion changes after the collision and (b) the raven's speed right after the collision.
02

Conservation of Momentum in Two Dimensions

Since the collision happens at right angles, we apply the conservation of momentum separately in the x and y directions. Before the collision, the falcon has momentum only in the x-direction, and the raven has momentum in the y-direction.
03

Define Known Quantities and Directions

The falcon's initial velocity is 20 m/s in the x-direction, and the raven's is 9 m/s in the y-direction. After the collision, the falcon rebounds at -5 m/s in the x-direction (since it bounces back). The raven's final velocities in x and y directions are unknown.
04

Apply Conservation of Momentum in x-direction

Initial momentum in x-direction: \( p_{x,falcon} = m_{falcon} \cdot v_{x,falcon} = 0.6 \cdot 20 \) kg·m/s. After collision: \( p_{x,falcon}' = m_{falcon} \cdot v_{x,falcon}' = 0.6 \cdot (-5) \) kg·m/s and \( p_{x,raven}' = m_{raven} \cdot v_{x,raven} \). \( 12 = -3 + 1.5 \cdot v_{x,raven} \); solve for \(v_{x,raven}\).
05

Apply Conservation of Momentum in y-direction

Initial momentum in y-direction: \( p_{y,raven} = m_{raven} \cdot v_{y,raven} = 1.5 \cdot 9 \) kg·m/s. After collision: \( p_{y,raven}' = m_{raven} \cdot v_{y,raven}' \). \( 13.5 = 1.5 \cdot v_{y,raven} \); solve for \(v_{y,raven}\).
06

Calculate the Final Velocity Components of the Raven

From Step 4, we have \( 1.5 \cdot v_{x,raven} = 15 \), so \( v_{x,raven} = 10 \) m/s. From Step 5, \( v_{y,raven} = 9 \) m/s. The final components are \( v_{x,raven} = 10 \) m/s and \( v_{y,raven} = 9 \) m/s.
07

Calculate the Raven's Speed After Collision

The speed of the raven can be found using the Pythagorean theorem: \( v_{raven} = \sqrt{v_{x,raven}^2 + v_{y,raven}^2} = \sqrt{10^2 + 9^2} = \sqrt{181} \approx 13.45 \) m/s.
08

Angle of the Raven's New Direction

The angle \( \theta \) is calculated using tangent: \( \tan{\theta} = \frac{v_{y,raven}}{v_{x,raven}} = \frac{9}{10} \). Thus, \( \theta = \tan^{-1}\left(\frac{9}{10}\right) \approx 42.0^\circ \).

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Key Concepts

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

Conservation of Momentum
In the world of physics, the conservation of momentum is a fundamental principle. It tells us that the total momentum of an isolated system remains constant if it is not subject to external forces. When it comes to collisions, this principle plays a crucial role in understanding the aftermath of the impact.

When two objects collide, their overall momentum before the collision is equal to their overall momentum after the collision. This holds true even in a two-dimensional space, as we will explore in the next sections. To apply this, we look at each direction (horizontal and vertical) separately. By doing so, we can find out how each component of velocity changes after the collision.
  • The momentum in the x-direction before and after the collision must balance out.
  • The same balance applies for the momentum in the y-direction.
For instance, in the falcon and raven scenario, the conservation of momentum helps us determine the unknown velocity components of the raven after it gets hit.
Two-Dimensional Motion
In physics problems, objects often move in two dimensions. This means we have to account for both horizontal and vertical motions separately. In the case of a collision, it's important to consider these dimensions to understand how each object moves afterward.

The falcon initially moves horizontally, while the raven moves vertically. After they collide, their motion is influenced by their velocities in both dimensions. The concept of two-dimensional motion allows us to split movement into:
  • x-direction: representing horizontal movement
  • y-direction: representing vertical movement
By using the conservation of momentum separately in these two directions, we can find out the velocity components of the raven after the collision. This helps us paint a clearer picture of the motion post-impact.
Impact Angles
Impact angles describe how the direction of an object's motion changes due to a collision. In our example, the falcon changes the raven's direction of motion when it strikes.

To calculate how much the direction changes, we can use trigonometry. Specifically, the tangent function relates the two velocity components (horizontal and vertical) to the angle of movement. In the falcon and raven situation, the angle \( \theta \) that measures the new direction of the raven compared to its original path is given by:
  • \( \tan{\theta} = \frac{v_{y,raven}}{v_{x,raven}} \)
  • Solving gives \( \theta = \tan^{-1}\left(\frac{9}{10}\right) \approx 42.0^\circ \)
This angle tells us how much the raven's flight path has changed due to the falcon's impact.
Velocity Components
Understanding velocity components is essential in analyzing collision problems. Velocity components break down an object's velocity into two perpendicular directions: x (horizontal) and y (vertical). This allows us to assess the object's motion in each direction separately.

In the falcon and raven example, we calculated the velocity components of the raven post-collision:
  • The x-component \( v_{x,raven} \) tells us how fast the raven moves horizontally.
  • The y-component \( v_{y,raven} \) describes its vertical motion.
After pinpointing these components, we combined them using the Pythagorean theorem to determine the overall speed of the raven. This gives the actual velocity magnitude:
  • Overall speed: \( \sqrt{v_{x,raven}^2 + v_{y,raven}^2} \approx 13.45 \text{ m/s} \)
Both components are necessary to find not just the speed but also the direction of motion post-collision.

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Most popular questions from this chapter

Block \(A\) in Fig. E8.24 has mass \(1.00 \mathrm{kg},\) and block \(B\) has mass 3.00 \(\mathrm{kg}\) . The blocks are forced together, compressing a spring \(S\) between them; then the system is released from rest on a level, frictionless surface. The spring, which has negligible mass, is not fastened to either block and drops to the surface after it has expanded. Block \(B\) acquires a speed of 1.20 \(\mathrm{m} / \mathrm{s}\) . (a) What is the final speed of block \(A\) ? (b) How much potential energy was stored in the compressed spring?

On a very muddy football field, a 110 -kg linebacker tackles an \(85-\) kg halfback. Immediately before the collision, the line-backer is slipping with a velocity of 8.8 \(\mathrm{m} / \mathrm{s}\) north and the halfback is sliding with a velocity of 7.2 \(\mathrm{m} / \mathrm{s}\) east. What is the velocity (magnitude and direction) at which the two players move together immediately after the collision?

A machine part consists of a thin, uniform \(4.00-\mathrm{kg}\) bar that is 1.50 \(\mathrm{m}\) long, hinged perpendicular to a similar vertical bar of mass 3.00 \(\mathrm{kg}\) and length 1.80 \(\mathrm{m} .\) The longer bar has a small but dense \(2.00-\mathrm{kg}\) ball at one end (Fig. E8.55). By what distance will the center of mass of this part move horizontally and vertically if the vertical bar is pivoted counterclockwise through \(90^{\circ}\) to make the entire part horizontal?

A neutron with mass \(m\) makes a head-on, elastic collision with a nucleus of mass \(M,\) which is initially at rest. (a) Show that if the neutron's initial kinetic energy is \(K_{0}\) , the kinetic energy that it loses during the collision is 4\(m M K_{0} /(M+m)^{2}\) . (b) For what value of \(M\) does the incident neutron lose the most energy? (c) When \(M\) has the value calculated in part (b), what is the speed of the neutron after the collision?

A 10.0 -g marble slides to the left with a velocity of magnitude 0.400 \(\mathrm{m} / \mathrm{s}\) on the frictionless, horizontal surface of an icy New York side- walk and has a head- on, elastic collision with a larger 30.0 -g marble sliding to the right with a velocity of magnitude 0.200 \(\mathrm{m} / \mathrm{s}(\mathrm{Fig} . \mathrm{E} 8.48) .\) (a) Find the velocity of each marble (magnitude and direction) after the collision. (Since the collision is head-on, all the motion is along a line.) (b) Calculate the change in momentum (that is, the momentum after the collision minus the momentum before the collision for each marble. Compare the values you get for each marble. (c) Calculate the change in kinetic energy (that is, the kinetic energy after the collision minus the kinetic energy before the collision) for each marble. Compare the values you get for each marble.

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