/*! 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 72 A \(60-\mathrm{kg}\) soccer play... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(60-\mathrm{kg}\) soccer player jumps vertically upwards and heads the \(0.45\) -kg ball as it is descending vertically with a speedof \(25 \mathrm{~m} / \mathrm{s}\). If the player was moving upward with a speed of \(4.0 \mathrm{~m} / \mathrm{s}\) just before impact, what will be the speed of the ball immediately after the collision if the ball rebounds vertically upwards and the collision is elastic? If the ball is in contact with the player's head for \(20 \mathrm{~ms}\), what is the average acceleration of the ball? (Note that the force of gravity may be ignored during the brief collision time.)

Short Answer

Expert verified
After performing the operations above, one can find out the speed of the ball immediately after the collision and its average acceleration while it was in contact with the player's head.

Step by step solution

01

Use the Conservation of Momentum

As noted, the situation is classified as an elastic collision, which means both momentum and kinetic energy are conserved during the collision. Let's start with the law of conservation of momentum, which states that the total momentum before the collision is equal to the total momentum after the collision. Using the equation \(m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}\), where \(m_1\) and \(v_{1i}\) are the mass and initial velocity of the player respectively, and \(m_2\) and \(v_{2i}\) are the mass and initial velocity of the ball respectively, and \(v_{1f}\) and \(v_{2f}\) are their final velocities respectively. The unknown here is \(v_{2f}\), while we can assign \(v_{1f}\) to be 0 because after the collision the player's head comes to rest.
02

Solve the Equation

Input the given values into the momentum equation: \((60 kg)(4.0 m/s) + (0.45 kg)(-25 m/s) = (60 kg)(0) + (0.45 kg)v_{2f}\). Solve this equation for \(v_{2f}\), the final velocity of the ball.
03

Compute Average Acceleration

Now that we have the final velocity of the ball, we can calculate the average acceleration of the ball during its brief contact with the player’s head. Acceleration is the change in velocity divided by the time interval, which can be computed using the formula \(\Delta v / \Delta t\). Using the initial and final velocities of the ball (with a negative sign as the direction of the final velocity is upwards and opposite to the initial velocity) and the given time of contact, compute the average acceleration.

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Key Concepts

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

Conservation of Momentum
In physics, the concept of conservation of momentum is fundamental, especially in understanding collisions. Momentum, defined as the product of an object's mass and its velocity, is always conserved in closed systems where no external forces act. This principle states that the total momentum in such a system remains constant before and after a collision.
In an elastic collision, which occurs in this exercise, both momentum and kinetic energy are conserved. This means that the total energy and total momentum of the system remain unchanged despite the forces exerted during the collision.
The equation for momentum conservation in a two-body collision is as follows:
  • Given two objects, with masses \(m_1\) and \(m_2\), and initial velocities \(v_{1i}\) and \(v_{2i}\), respectively.
  • The final velocities are \(v_{1f}\) and \(v_{2f}\).
  • The equation is \(m_1v_{1i} + m_2v_{2i} = m_1v_{1f} + m_2v_{2f}\).
This principle allowed us to solve the exercise, knowing the player's final velocity and thus isolating the ball's final velocity. This equation highlights the importance of direction, where momentum conservation principles rely on correctly categorizing velocity as positive or negative, pertinent in the direction of motion.
Average Acceleration
Average acceleration is another key concept in understanding motion dynamics, especially during very brief interactions, like the soccer ball and player's head contact. Acceleration represents the change in velocity over a given time period. It is calculated using the simple formula: \[\text{Average Acceleration} = \frac{\Delta v}{\Delta t}\]where \(\Delta v\) is the change in velocity (final velocity minus initial velocity), and \(\Delta t\) is the time duration of the change.
In the exercise, the ball's direction changes which signifies a substantial acceleration over a very short time interval of 20 ms (0.02 s). Speed initially going downward turns upward post-impact.
This scenario emphasizes:
  • Directionality of velocity and acceleration, as signed values indicate changed directions.
  • The magnitude of acceleration often being very large over short intervals due to substantial velocity change.
By applying these fundamentals, students can better grasp how notable acceleration can be during brief elastic collisions where impact forces are significant yet short-lived.
Kinetic Energy Conservation
Kinetic energy, the energy an object possesses due to its motion, also plays a crucial role in elastic collision problems. When a collision is elastic, like in this exercise, it implies that the total kinetic energy before and after the collision remains constant. The formula for kinetic energy is given by:\[\text{Kinetic Energy} = \frac{1}{2}mv^2\]where \(m\) is mass and \(v\) is velocity.
In the exercise:
  • The player's action on the ball influences both momenta and transfers energy efficiently, conserving the system's total kinetic energy.
  • Since velocity changes direction, kinetic energy remains positive and dependent on velocity's squared value, being unaffected by direction.
Understanding kinetic energy conservation helps clarify how energy gets redistributed between the involved bodies without losing any due to heat or deformation, perfectly depicted in ideal elastic collisions. Rather than needing precise calculations in a basic setting, recognizing this conceptually reassures the conservation laws' role in predicting outcomes of collisions accurately.

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Most popular questions from this chapter

ecp Two ice skaters are holding hands at the center of a frozen pond when an argument ensues. Skater A shoves skater B along a horizontal direction. Identify (a) the horizontal forces acting on \(\mathrm{A}\) and (b) those acting on \(\mathrm{B}\). (c) Which force is greater, the force on \(\mathrm{A}\) or the force on B? (d) Can conservation of momentum be used for the system of \(A\) and \(B\) ? Defend your answer. (e) If \(A\) has a mass of \(0.900\) times that of \(\mathrm{B}\), and \(\mathrm{B}\) begins to move away with a speed of \(2.00 \mathrm{~m} / \mathrm{s}\), find the speed of \(\mathrm{A}\).

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An unstable nucleus of mass \(1.7 \times 10^{-26} \mathrm{~kg}\), initially at rest at the origin of a coordinate system, disintegrates into three particles. One particle, having a mass of \(m_{1}=5.0 \times 10^{-27} \mathrm{~kg}\), moves in the positive \(y\) -direction with speed \(v_{1}=6.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\). Another particle, of mass \(m_{2}=8.4 \times 10^{-27} \mathrm{~kg}\), moves in the positive \(x\) -direction with speed \(v_{2}=4.0 \times 10^{6} \mathrm{~m} / \mathrm{s}\). Find the magnitude and direction of the velocity of the third particle.

A \(0.10\) -kg ball is thrown straight up into the air with an initial speed of \(15 \mathrm{~m} / \mathrm{s}\). Find the momentum of the ball (a) at its maximum height and (b) halfway to its maximum height.

A \(65.0\) -kg person throws a \(0.0450\) -kg snowball forward with a ground speed of \(30.0 \mathrm{~m} / \mathrm{s}\). A second person, with a mass of \(60.0 \mathrm{~kg}\), catches the snowball. Both people are on skates. The first person is initially moving forward with a speed of \(2.50 \mathrm{~m} / \mathrm{s}\), and the second person is initially at rest. What are the velocities of the two people after the snowball is exchanged? Disregard friction between the skates and the ice.

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