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On a very muddy football field, a \(110 \mathrm{~kg}\) linebucker tackles an \(85 \mathrm{~kg}\) halthnck. Immediately hefore the collision, the linchacker is slipping with a velocity of \(8.8 \mathrm{~m} / \mathrm{s}\) north and the halthack 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 cullision?

Short Answer

Expert verified
The two football players move together at a speed of 5.87 m/s at an angle of 32.39 degrees east of north.

Step by step solution

01

Determine Initial Momentum

Calculate the momentum separately for each player before the collision using the formula momentum = mass × velocity. These will be vector quantities. For the linebacker: \(P_{L} = m_{L} * v_{L} = 110 kg * 8.8 m/s = 968 kg*m/s\) North. For the halfback: \(P_{H} = m_{H} * v_{H} = 85 kg * 7.2 m/s = 612 kg*m/s\) East.
02

Calculate Total Initial Momentum

The total initial momentum is the vector sum of the individual momenta. In our case, the momentum vectors are perpendicular to each other (since one is moving North and the other East), so we can use the Pythagorean theorem to calculate their resultant. So, the total momentum \(P_{total}\) = \(\sqrt{(P_{L}^2 + P_{H}^2)} = \sqrt{(968^2 + 612^2)} = 1145 kg*m/s\).
03

Calculate Total Mass

The total mass of the system after collision is the sum of the individual masses, i.e., \( m_{total} = m_{L} + m_{H} = 110 kg + 85 kg = 195 kg \)
04

Calculate Final Velocity

The final velocity (magnitude) can be calculated using the conservation of momentum (since no external force was applied, the momentum before and after the collision is the same) and is given by \( v_{final} = P_{total} / m_{total} = 1145 kg*m/s / 195 kg = 5.87 m/s \).
05

Calculate Final Direction

The direction of the final velocity (θ) can be calculated by taking the inverse tangent of the ratio of the halfback's momentum and the linebacker's momentum. i.e., \(θ = tan^{-1} (P_{H} / P_{L}) = tan^{-1} (612 / 968 ) = 32.39 degrees \). This angle is measured relative to the northward direction.

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Key Concepts

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

Collision Physics
In the world of physics, collisions are events where two or more bodies exert forces on each other for a relatively short time. In this exercise, we're dealing with a perfectly inelastic collision, where a linebacker tackles a halfback on a football field.

In such cases, the two players stick together after colliding. The main principle ruling this event is the conservation of momentum. Momentum is a vector quantity, meaning it has both magnitude and direction. In collision physics, we often assume no external forces are acting during the short time span of the collision. Therefore, the total momentum of the system before the collision is equal to the total momentum after the collision.

Specifically, the key steps involve calculating the momentum of each player before they collide and combining these to find the resulting motion of the system. This example highlights the importance of understanding vector nature of momentum in collision scenarios.
Vector Addition
Adding vectors is a crucial step in solving the problem of two football players colliding. Each player has an initial momentum vector, which reflects both their speed and direction.

In our exercise, the linebacker moves north, and the halfback moves east. Thus, their momentum vectors are perpendicular. When vectors are perpendicular, we use the Pythagorean theorem to find their resultant. This gives us a new vector that represents the total momentum.
  • The magnitude of the total momentum is calculated as \( \sqrt{(P_{L}^2 + P_{H}^2)} \).
  • This resultant vector shows the direction and magnitude of the combined movement after collision.

Vector addition helps us merge these two separate momenta into a single vector, crucial for further calculations concerning velocity and direction after a collision.
Momentum Calculation
Calculating momentum involves using the fundamental formula: \( \text{momentum} = \text{mass} \times \text{velocity} \). Each player in the exercise has their momentum calculated prior to the collision.

For the linebacker, the momentum is determined by multiplying his mass (110 kg) by his velocity (8.8 m/s), resulting in a northward momentum vector. Similarly, the halfback's momentum is found by multiplying his mass (85 kg) with his velocity (7.2 m/s), resulting in an eastward vector.
  • Linebacker: \( P_{L} = 968 \, \text{kg}\cdot\text{m/s (north)} \)
  • Halfback: \( P_{H} = 612 \, \text{kg}\cdot\text{m/s (east)} \)

These vector quantities are essential to understanding the complete picture of the collision. They represent how the different parts of the system are moving and allow us to apply conservation laws to find post-collision behaviors.
Angle Determination
Determining the angle of the resultant velocity vector helps us understand the direction the players move together after the collision.

Using the tangent function, we can find the angle \( \theta \) between the resultant vector and the northward axis (the line of the linebacker's initial movement). This involves calculating the arctangent of the ratio of the eastward and northward momentum components. Specifically:
  • The formula used is \( \theta = \tan^{-1} \left( \frac{P_{H}}{P_{L}} \right) \).
  • In this exercise, it's calculated as \( \theta = \tan^{-1}(\frac{612}{968}) \).
  • The resulting angle provides a measure of how much the combined motion veers from the north direction into the east.

This angle determination is critical in giving us a complete understanding not just of the speed, but also of the new direction the players collectively follow after the collision has occurred.

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Most popular questions from this chapter

A small wooden block with mass \(0.800 \mathrm{~kg}\) is suspended from the lower end of a light cord that is \(1.60 \mathrm{~m}\) long. The block is initially at rest. A bullet with mass \(12.0 \mathrm{~g}\) is fired at the block with a horizontal velocity \(c_{0}\) - The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on the end of the cord. When the block has risen a vertical height of \(0.800 \mathrm{~m}\), the tension in the cord is \(4.80 \mathrm{~N}\). What was the initial speed \(t_{0}\) of the bullct?

Starting at \(t=0\). a horizontal net externul force \(F=(0.280 \mathrm{~N} / \mathrm{s}) t \hat{\imath}+\left(-0.450 \mathrm{~N} / \mathrm{s}^{2}\right) r^{2} \hat{\jmath}\) is applied to a box that has an initial momentum \(\vec{p}=(-3.00 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}) \hat{\imath}+(4.00 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}) \hat{\jmath} .\) What is the momentum of the box at \(t=2.00 \mathrm{~s} ?\)

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 \(4.00 \mathrm{~m} / \mathrm{s}\), 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?

A baseball has mass \(0.145 \mathrm{~kg}\). (a) If the velocity of a pitched ball has a magnitude of \(45.0 \mathrm{~m} / \mathrm{s}\) and the batted ball's velocity is \(55.0 \mathrm{~m} / \mathrm{s}\) in the opposite direction, find the magnitude of the change in momentum of the ball and of the impulse applied to it by the bat. (b) If the ball remains in contact with the bat for \(2.00 \mathrm{~ms}\), find the magnitude of the average force applied by the bat.

Jonathan and Jane are sitting in a sleigh that is at rest on frictionless ice. Jonathan's wcight 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 of \(5.00 \mathrm{~m} / \mathrm{s}\) at \(30.0^{\circ}\) above the horizontal (relative to the ice), 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.

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