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Chapter 8: Problem 33

On a very muddy football field, a \(110-\mathrm{kg}\) linebacker tackles an \(85-\mathrm{kg}\) 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?

Short Answer

Expert verified
The players move together at approximately 5.89 m/s in a northeast direction, 32.6掳 east of north.

Step by step solution

01

Understand the Problem

We have two players colliding: a 110 kg linebacker moving north at 8.8 m/s, and an 85 kg halfback moving east at 7.2 m/s. After tackling, they move together as one body. We need to find their combined velocity and direction.
02

Conservation of Momentum

The law of conservation of momentum states that the total momentum before a collision is equal to the total momentum after the collision. We'll calculate initial momentum components for each player and find the final velocity components for the combined mass.
03

Calculate Northward Momentum

The momentum of the linebacker in the northward direction is given by \( p_{N} = m_{N} \cdot v_{N} \). So, \( p_{N} = 110 \, \mathrm{kg} \times 8.8 \, \mathrm{m/s} = 968 \, \mathrm{kg \cdot m/s} \). The halfback has no northward momentum, so we only use the linebacker's value here.
04

Calculate Eastward Momentum

The momentum of the halfback in the eastward direction is \( p_{E} = m_{E} \cdot v_{E} \). Hence, \( p_{E} = 85 \, \mathrm{kg} \times 7.2 \, \mathrm{m/s} = 612 \, \mathrm{kg \cdot m/s} \). The linebacker has no eastward momentum.
05

Calculate Combined Mass

After the collision, both players move together. So, the combined mass \( M \) is \( M = m_{N} + m_{E} = 110 \, \mathrm{kg} + 85 \, \mathrm{kg} = 195 \, \mathrm{kg} \).
06

Calculate Combined Velocity Components

The final northward velocity component \( v_{N_f} \) is \( \frac{p_{N}}{M} = \frac{968}{195} \approx 4.96 \, \mathrm{m/s} \). The final eastward velocity component \( v_{E_f} \) is \( \frac{p_{E}}{M} = \frac{612}{195} \approx 3.14 \, \mathrm{m/s} \).
07

Resultant Velocity Magnitude

Use the Pythagorean theorem to calculate the magnitude of the velocity: \( v = \sqrt{v_{N_f}^2 + v_{E_f}^2} = \sqrt{(4.96)^2 + (3.14)^2} \approx 5.89 \, \mathrm{m/s} \).
08

Determine Direction Angle

The direction angle \( \theta \) relative to north can be found using \( \tan(\theta) = \frac{v_{E_f}}{v_{N_f}} \). Thus, \( \theta = \tan^{-1}\left(\frac{3.14}{4.96}\right) \approx 32.6^\circ \). The movement is in the northeast direction.

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Key Concepts

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

Physics Problem Solving
Physics problems often require us to find solutions by breaking down the scenario into manageable steps. In this problem, two football players collide on a muddy field, reminding us how real-world physics can be both challenging and relatable. The essence of solving any physics problem like this involves:
  • Understanding the given data and the question.
  • Identifying physics laws or equations that apply.
  • Breaking down the problem using these laws into step-by-step resolutions.
The key here is to not only interpret the problem correctly but to also apply the conservation of momentum as the dominant principle governing the interaction between the players. By calculating separate momentum components for the collision, we can find both the magnitude and the direction of their combined velocity after they move together.
Collision Mechanics
In the realm of collision mechanics, understanding how objects interact during a collision is crucial. Here, a collision between a linebacker and a halfback on the field is a prime example of an inelastic collision.

During an inelastic collision:
  • The two objects stick together, moving as a single entity after the collision.
  • While the momentum is conserved, kinetic energy is generally not, leading to a shared direction and velocity.
  • The problem becomes about understanding how the individual velocities combine into a new, single velocity.
The challenge for students lies in visualizing and computing these changes. By understanding how initial momenta from separate directions blend into one resultant direction, one can grasp how combined movements in physics operate, with both directional and speed changes.
Momentum Components
Momentum is a vector quantity, meaning it has both magnitude and direction. Consequently, when dealing with objects moving in different directions, it is vital to consider the momentum components separately.

In this problem:
  • The northward and eastward movements of the two players create separate momentum components.
  • The northward momentum is calculated by multiplying the linebacker鈥檚 mass by its velocity, while the eastward momentum results from the halfback鈥檚 velocity and mass.
  • Once these components are calculated separately, they can be combined using vector addition to find the total momentum vector.
  • The result uses the Pythagorean theorem to determine the magnitude of their combined velocity and trigonometry to find the direction angle.
This approach illustrates how complex movement can be simplified by breaking it into understandable parts, showing how vector mathematics can be employed to solve practical physics problems.

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Most popular questions from this chapter

(a) What is the magnitude of the momentum of a \(10,000-k g\) truck whose speed is 12.0 \(\mathrm{m} / \mathrm{s} ?\) (b) What speed would a \(2,000-\mathrm{kg}\) SUV have to attain in order to have (i) the same momentum? (ii) the same kinetic energy?

(a) Show that the kinetic energy \(K\) and the momentum magnitude \(p\) of a particle with mass \(m\) are related by \(K=p^{2} / 2 m .\) (b) A \(0.040-\mathrm{kg}\) cardinal (Richmondena cardinalis) and a \(0.145-\mathrm{kg}\) baseball have the same kinetic energy. Which has the greater magnitude of momentum? What is the ratio of the cardinal's magnitude of momentum to the bascball's? A \(700-N\) man and a \(450-N\) woman have the same momentum. Who has the greater kinetic energy? What is the ratio of the man's kinetic energy to that of the woman?

A 20.0 -kg projectile is fired at an angle of \(60.0^{\circ}\) above the horizontal with a speed of 80.0 \(\mathrm{m} / \mathrm{s}\) . At the highest point of its trajectory, the projectile explodes into two fragments with equal mass, one of which falls vertically with zero initial speed. You can ignore air resistance. (a) How far from the point of firing does the other fragment strike if the terrain is level? (b) How much energy is released during the explosion?

Three coupled railroad cars roll along and couple with a fourth car, which is initially at rest. These four cars roll along and couple with a fifth car initially at rest. This process continues until the speed of the final collection of railroad cars is one-fifth the speed of the initial three railroad cars. All the cars are identical. Ignoring friction, how many cars are in the final collection?

An atomic nucleus suddenly bursts apart (fissions) into two pieces. Piece \(A,\) of mass \(m_{A},\) travels off to the left with speed \(v_{A}\) . Piece \(B\) , of mass \(m_{B}\) , travels off to the right with speed \(v_{B}\) . (a) Use conservation of momenturn to solve for \(v_{B}\) in terms of \(m_{A}, m_{B},\) and \(v_{A}-(b)\) Use the results of part (a) to show that \(K_{A} / K_{B}=m_{B} / m_{A}\) where \(K_{\mathrm{A}}\) and \(K_{B}\) are the kinetic energies of the two pieces.
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