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One \(110-\mathrm{kg}\) football lineman is running to the right at 2.75 \(\mathrm{m} / \mathrm{s}\) while another \(125-\mathrm{kg}\) lineman is running directly toward him at 2.60 \(\mathrm{m} / \mathrm{s}\) . What are (a) the magnitude and direction of the net momentinm of these two athletes, and (b) their total kinetic energy?

Short Answer

Expert verified
(a) Net momentum: 22.5 kg m/s (direction of second lineman). (b) Total kinetic energy: 838.875 J.

Step by step solution

01

Understand the Problem

We need to find the net momentum and total kinetic energy of two football players moving toward each other. The given data includes their masses and velocities.
02

Calculate Individual Momenta

Momentum is calculated using the formula: \( p = mv \). For the first lineman, \( m = 110 \) kg and \( v = 2.75 \) m/s. Thus, the momentum \( p_1 = 110 \times 2.75 = 302.5 \) kg m/s. For the second lineman, \( m = 125 \) kg and \( v = 2.60 \) m/s (negative because he is running in the opposite direction). Thus, \( p_2 = 125 \times (-2.60) = -325 \) kg m/s.
03

Calculate Net Momentum

The net momentum is the sum of the individual momenta: \( p_{net} = p_1 + p_2 = 302.5 + (-325) = -22.5 \) kg m/s. The minus sign indicates the net momentum is in the direction of the second lineman.
04

Calculate Individual Kinetic Energies

Kinetic energy is calculated using the formula: \( KE = \frac{1}{2}mv^2 \). For the first lineman, \( KE_1 = \frac{1}{2} \times 110 \times (2.75)^2 = 416.375 \) J. For the second lineman, \( KE_2 = \frac{1}{2} \times 125 \times (2.60)^2 = 422.5 \) J.
05

Calculate Total Kinetic Energy

The total kinetic energy is the sum of the individual kinetic energies: \( KE_{total} = KE_1 + KE_2 = 416.375 + 422.5 = 838.875 \) J.

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Key Concepts

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

Kinetic Energy
Kinetic energy is a fascinating concept in physics, describing the energy an object possesses due to its motion. Imagine you're running, skating, or even driving a car—any movement involves kinetic energy. This energy can be calculated using the formula:
\[ KE = \frac{1}{2}mv^2 \]
where \( m \) is the mass of the object and \( v \) is its velocity.
In the context of our football players, kinetic energy helps us understand the power behind their movement. For example, multiplying the half of the mass by the square of velocity gives a tangible measure of their kinetic force. The first lineman, with a mass of 110 kg, and velocity of 2.75 m/s, has a kinetic energy of 416.375 joules. The second lineman, slightly heavier and running at 2.60 m/s, has a kinetic energy of 422.5 joules.
These computations illustrate that even if two athletes run at similar speeds, a slight change in mass or speed can significantly alter their kinetic power. Knowing the total kinetic energy of a system is useful in understanding the energy dynamics involved, especially in sports scenarios where energy transfers and impacts are crucial.
Collisions
Collisions occur when two or more bodies exert forces on each other for a relatively short time. In our scenario, two football players collide, bringing the concept of momentum and forces into play.
During a collision, the key elements to consider are the players' masses and velocities. These parameters contribute to their momentum and subsequently affect the interaction during the collision. It's crucial to understand that momentum, unlike kinetic energy, is a vector quantity, having both magnitude and direction.
Inelastic and elastic collisions are two types commonly discussed. Inelastic collisions involve objects colliding and sticking together, with kinetic energy not conserved. Elastic collisions, on the other hand, conserve both momentum and kinetic energy. The collision between our players can be analyzed for its inelastic nature since the momentum's direction (one player moving towards the other) ultimately impacts the result.
  • The mass of each player affects the collision's outcome significantly.
  • The direction of momentum plays a vital role in understanding the resultant motion.
Analyses and calculations like these offer insights into the physical interactions during sports and can lead to better safety and performance strategies.
Physics Problems
The study of physics problems often involves breaking down complex interactions into simpler components, allowing us to better understand real-world phenomena. In problems like the football players' scenario, identifying and calculating fundamental elements such as momentum and kinetic energy helps in analyzing the situation effectively.
When approaching physics problems, gather all given data, such as masses and velocities for the football scenario, and identify what needs to be calculated. Next, break the problem into manageable steps:
  • Determine relevant formulas, like those for kinetic energy \( KE = \frac{1}{2}mv^2 \) and momentum \( p = mv \).
  • Calculate individual components, like the kinetic energies and momenta of each player.
  • Combine these results to find net values, such as total kinetic energy or net momentum.
These strategies make physics problems more approachable by systematically organizing information and applying it. Such an organized method helps deal not just with abstract equations but also with realistic situations, preparing you to solve problems with confidence.

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Most popular questions from this chapter

You are standing on a sheet of ice that covers the football stadium parking lot in Buffalo; there is negligible friction between your feet and the ice. A friend throws you a \(0.400-\mathrm{kg}\) ball that is traveling horizontally at 10.0 \(\mathrm{m} / \mathrm{s}\) . Your mass is 70.0 \(\mathrm{kg}\) . (a) If you catch the ball, with what speed do you and the ball move after-ward? (b) If the ball hits you and bounces off your chest, so after-ward it is moving horizontally at 8.0 \(\mathrm{m} / \mathrm{s}\) in the opposite direction, what is your speed after the collision?

A \(20.00-\mathrm{kg}\) lead sphere is hanging from a hook by a thin wire 3.50 \(\mathrm{m}\) long, and is free to swing in a complete circle. Suddenly it is struck horizontally by a \(5.00-\mathrm{kg}\) steel dart that embeds itself in the lead sphere. What must be the minimum initial speed of the dart so that the combination makes a complete circular loop after the collision?

On a frictionless, horizontal air table, puck \(A\) (with mass 0.250 \(\mathrm{kg}\) is moving toward puck \(B\) (with mass 0.350 \(\mathrm{kg}\) ), which is initially at rest. After the collision, puck \(A\) has a velocity of 0.120 \(\mathrm{m} / \mathrm{s}\) to the left, and puck \(B\) has a velocity of 0.650 \(\mathrm{m} / \mathrm{s}\) to the right. (a) What was the speed of puck \(A\) before the collision? (b) Calculate the change in the total kinetic energy of the system that occurs during the collision.

Two skaters collide and grab on to each other on frictionless ice. One of them, of mass \(70.0 \mathrm{kg},\) is moving to the right at 2.00 \(\mathrm{m} / \mathrm{s}\) , while the other. of mass 65.0 \(\mathrm{kg}\) , is moving to the left at 2.50 \(\mathrm{m} / \mathrm{s}\) . What are the magnitude and direction of the velocity of these skaters just after they collide?

\(0.100-\mathrm{kg}\) stone rests on a friction-less, horizontal surface. A bullet of mass 6.00 \(\mathrm{g}\) , traveling horizontally at 350 \(\mathrm{m} / \mathrm{s}\) , strikes the stone and rebounds horizontally at right angles to its original direction with a speed of 250 \(\mathrm{m} / \mathrm{s}\) . (a) Compute the magnitude and direction of the velocity of the stone after it is struck. (b) Is the collision perfectly elastic?

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