/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 27 \(\bullet\) A hungry 11.5 \(\mat... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

\(\bullet\) A hungry 11.5 \(\mathrm{kg}\) predator fish is coasting from west to east to east at 75.0 \(\mathrm{cm} / \mathrm{s}\) when it suddenly swallows a 1.25 \(\mathrm{kg}\) fish swim- ming from north to south at 3.60 \(\mathrm{m} / \mathrm{s} .\) Find the magnitude and direction of the velocity of the large fish just after it snapped up this meal. Neglect any effects due to the drag of the water.

Short Answer

Expert verified
The final velocity is 0.764 m/s at 27.9° south of east.

Step by step solution

01

Understand Conservation of Momentum

In this problem, we need to apply the principle of conservation of momentum because the predator fish and its prey can be treated as an isolated system (negligible effects of water drag). The total momentum before the predator swallows its prey is equal to the total momentum just after.
02

Calculate Initial Momentum Components

Calculate the momentum of each fish in both the east and south directions. The predator fish is moving east with a velocity of 75.0 cm/s (or 0.75 m/s), so its momentum in the x-direction is:\[ p_{predator,x} = m_{predator} \times v_{predator,x} = 11.5 \, \text{kg} \times 0.75 \, \text{m/s} = 8.625 \, \text{kg m/s} \]The smaller fish is swimming south, so its momentum in the y-direction is:\[ p_{prey,y} = m_{prey} \times v_{prey,y} = 1.25 \, \text{kg} \times 3.60 \, \text{m/s} = 4.5 \, \text{kg m/s} \]
03

Apply Conservation of Momentum to Find Final Velocity Components

After the predator swallows the prey, both will move together. Thus, the final momentum components in the x and y directions are:\[ p_{final,x} = p_{predator,x} = 8.625 \, \text{kg m/s} \]\[ p_{final,y} = p_{prey,y} = 4.5 \, \text{kg m/s} \]The total mass of the combined system is:\[ m_{total} = m_{predator} + m_{prey} = 11.5 \, \text{kg} + 1.25 \, \text{kg} = 12.75 \, \text{kg} \]Thus, the velocity components after the event are given by:\[ v_{final,x} = \frac{p_{final,x}}{m_{total}} = \frac{8.625}{12.75} = 0.6765 \, \text{m/s} \]\[ v_{final,y} = \frac{p_{final,y}}{m_{total}} = \frac{4.5}{12.75} = 0.3529 \, \text{m/s} \]
04

Calculate Magnitude and Direction of Final Velocity

The magnitude of the final velocity can be calculated using Pythagorean theorem:\[ v_{final} = \sqrt{v_{final,x}^2 + v_{final,y}^2} = \sqrt{(0.6765)^2 + (0.3529)^2} = 0.764 \, \text{m/s} \]The direction (angle \( \theta \) from the east towards south) can be calculated by:\[ \theta = \arctan{\left(\frac{v_{final,y}}{v_{final,x}}\right)} = \arctan{\left(\frac{0.3529}{0.6765}\right)} \approx 27.9^\circ \]

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Velocity Calculation
In physics, calculating velocity is essential for understanding motion. Velocity is a vector quantity, which means it has both magnitude and direction.
For scenarios involving multiple objects, like in our fish example, we find individual velocity components first. Velocity components are typically separated into perpendicular directions, such as east-west for the x-axis and north-south for the y-axis.
By computing these components separately, we can easily understand how each object contributes to the overall motion. After finding the components, we use them to determine the resultant velocity using the Pythagorean theorem. This theorem applies because the velocity components form a right-angled triangle.
Thus, the magnitude of velocity is:
  • \( v=\sqrt{v_{x}^2 + v_{y}^2} \)
Momentum Components
Momentum is another vector quantity, similar to velocity. It is derived from the product of an object’s mass and its velocity. In the case of our two fish, each has momentum in their respective moving directions.
Breaking momentum into components makes it easier to handle complex interactions, especially when applied on different paths. For example:
  • Eastward or x-direction: predator's momentum
  • Southward or y-direction: prey's momentum
In the x-direction, the conservation of momentum allows us to retain the total eastward momentum before and after the event. Similarly, in the y-direction, the southward momentum remains constant in an isolated system.
These components are crucial for predicting the system's behavior when combined as a single entity. Calculating these accurately can give insights into resultant effects and help in precise determination of motion.
Isolated System
An isolated system is one where no external forces affect the internal components. In physics, assuming a system is isolated simplifies many calculations because we can ignore factors like friction, drag, or external interventions.
In our fish scenario, since water drag is neglected, the predator and prey fish form an isolated system for the conservation of momentum. This condition holds true because only internal forces (collision and swallowing) are considered, and they cannot change the system’s total momentum independently.
The concept of an isolated system ensures that the momentum observed before an interaction stays constant afterward, aiding in the calculation of final states like velocity and direction.
Momentum Conservation Principle
The principle of momentum conservation is a fundamental law of physics stating that if no external forces act on a system, the system's total momentum remains constant. This principle applies to our fish example because the effect of water resistance is negligible, allowing us to treat the system as closed.
In practice, how does this apply to our fish? Before the hungry predator captures its prey, the total momentum is the sum of their individual momentums in their respective directions. After swallowing, the momentum of the combined system remains unchanged.
Thus, using the equation:
  • Pre-capture momentum = Post-capture momentum
This shows that whatever happens inside this system, the total momentum stays the same. It is a key tool used in predicting outcomes after collisions or mergers, like in our fish example.
Understanding this concept is critical to analyze and predict interactions in similar physical scenarios effectively.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

\(\cdot\) Three odd-shaped blocks of chocolate have the following masses and center-of-mass coordinates: \((1) \quad 0.300\) kg, \((0.200 \mathrm{m}, 0.300 \mathrm{m}) ;\) (2) \(0.400 \mathrm{kg}, \quad(0.100 \mathrm{m},-0.400 \mathrm{m})\) (3) \(0.200 \mathrm{kg},(-0.300 \mathrm{m}, 0.600 \mathrm{m}) .\) Find the coordinates of the center of mass of the system of three chocolate blocks.

Accident analysis. Two cars collide at an intersection. Car \(A\), with a mass of \(2000 \mathrm{~kg}\), is going from west to east, while car \(B\), of mass \(1500 \mathrm{~kg}\), is going from north to south at \(15 \mathrm{~m} / \mathrm{s}\). As a result of this collision, the two cars become enmeshed and move as one afterward. In your role as an expert witness, you inspect the scene and determine that, after the collision, the enmeshed cars moved at an angle of \(65^{\circ}\) south of east from the point of impact. (a) How fast were the enmeshed cars moving just after the collision? (b) How fast was car \(A\) going just before the collision?

. Changing your center of mass. To keep the calculations fairly simple, but still reasonable, we shall model a human leg that is 92.0 \(\mathrm{cm}\) long (measured from the hip joint) by assuming that he upper leg and the lower leg (which includes the foot) have equal lengths and that each of them is uniform. For a 70.0 kg per- son, the mass of the upper leg would be 8.60 \(\mathrm{kg}\) , while that of the lower leg (including the foot) would be 5.25 \(\mathrm{kg}\) . Find the location of the center of mass of this leg, relative to the hip joint, if it is (a) stretched out horizontally and (b) bent at the knee to form a right angle with the upper leg remaining horizontal.

\(\bullet\) A \(70-\mathrm{kg}\) astronaut floating in space in a \(110-\mathrm{kg}\) MMU (manned maneuvering unit) experiences an acceleration of 0.029 \(\mathrm{m} / \mathrm{s}^{2}\) when he fires one of the MMU's thrusters. (a) If the speed of the escaping \(\mathrm{N}_{2}\) gas relative to the astronaut is 490 \(\mathrm{m} / \mathrm{s}\) , how much gas is used by the thruster in 5.0 \(\mathrm{s} \%\) (b) What is the thrust of the thruster?

\(\bullet\) Jonathan and Jane are sitting in a sleigh that is at rest on frictionless ice. Jonathan's weight is \(800 \mathrm{N},\) Jane's weight is \(600 \mathrm{N},\) and that of the sleigh is 1000 \(\mathrm{N} .\) They see a poisonous spider on the floor of the sleigh and immediately jump off. Jonathan jumps to the left with a velocity (relative to the ice) of 5.00 \(\mathrm{m} / \mathrm{s}\) at \(30.0^{\circ}\) above the horizontal, and Jane jumps to the right at 7.00 \(\mathrm{m} / \mathrm{s}\) at \(36.9^{\circ}\) above the horizontal (relative to the ice). Calculate the sleigh's horizontal velocity (magnitude and direction) after they jump out.

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.