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Chapter 9: Problem 28

In tae-kwon-do, a hand is slammed down onto a target at a speed of \(13 \mathrm{~m} / \mathrm{s}\) and comes to a stop during the \(5.0 \mathrm{~ms}\) collision. Assume that during the impact the hand is independent of the arm and has a mass of \(0.70 \mathrm{~kg} .\) What are the magnitudes of the (a) impulse and (b) average force on the hand from the target?

Short Answer

Expert verified
Impulse: 9.1 Ns; Average force: 1820 N.

Step by step solution

01

Identify the Known Variables

We have the initial speed of the hand as \(v_i = 13 \text{ m/s}\), the final speed as \(v_f = 0 \text{ m/s}\) since it comes to a stop, the time duration of the collision as \(t = 5.0 \text{ ms} = 0.005 \text{ s}\), and the mass of the hand as \(m = 0.70 \text{ kg}\).
02

Calculate the Impulse

Impulse is the change in momentum, which can be calculated using the formula: \[\text{Impulse} = m(v_f - v_i)\]Substituting the values we have:\[\text{Impulse} = 0.70 \times (0 - 13) = -9.1 \text{ Ns}\]The impulse is \(-9.1 \text{ Ns}\), but the magnitude is \(9.1 \text{ Ns}\).
03

Use Impulse to Find Average Force

The impulse is also equal to the average force multiplied by the time interval during which the force acts. This is expressed as:\[\text{Impulse} = F_{\text{avg}} \times t\]Solving for the average force \(F_{\text{avg}}\):\[F_{\text{avg}} = \frac{9.1}{0.005} = 1820 \text{ N}\]The magnitude of the average force is \(1820 \text{ N}\).

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Key Concepts

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

Collision Dynamics
Collision dynamics involve the interaction of two objects colliding with each other, where their motion and forces play a crucial role. When objects collide, forces are exchanged rapidly between them over a brief time. This is especially true in sports like tae-kwon-do, where quick impacts are common.
  • In a collision, the duration is typically short, such as milliseconds, making the forces involved very large.
  • During a collision, objects exert equal and opposite forces on one another due to Newton's Third Law of Motion.
Understanding collision dynamics requires evaluating different physical parameters, like velocity, mass, and the duration of impact. These factors together highlight how objects exchange forces in the blink of an eye, which can be analyzed by calculating momentum and impulse values.
Momentum
Momentum is a key concept in mechanics that describes the quantity of motion an object has. It is calculated as the product of an object's mass and its velocity and has both magnitude and direction, making it a vector quantity.
  • The formula for momentum is: \ [ p = m imes v ] \ where \(m\) is the mass and \(v\) is the velocity of the object.
  • In a collision, momentum often changes rapidly due to external forces acting over a short duration.
  • The change in momentum during a collision is referred to as impulse. Impulse quantifies the effect of a force acting over time.
For the tae-kwon-do problem, the hand moves initially with a velocity, creating initial momentum, then comes to rest, implying a significant change in momentum measured as impulse.
Studying momentum in collisions helps us predict and analyze the outcome of such events accurately.
Average Force Calculation
The calculation of average force is an essential aspect of understanding collision dynamics, especially in short-duration impacts like those in martial arts. The average force is the total impulse (or change in momentum) divided by the time period over which it acts.
  • The formula to find average force is given by: \ [ F_{\text{avg}} = \frac{\text{Impulse}}{t} ] \ where "Impulse" is the change in momentum and \(t\) is the time duration of the force application.
  • This provides a way to measure how powerful an interaction was in terms of force, considering how quickly the interaction was completed.
In the context of the exercise, such an analysis shows the relation between speed, stopping time, and exerted force, reflecting real-world forces experienced in martial arts training and competition.

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Most popular questions from this chapter

A rocket that is in deep space and initially at rest relative to an inertial reference frame has a mass of \(2.55 \times 10^{5} \mathrm{~kg}\), of which \(1.81 \times 10^{5} \mathrm{~kg}\) is fuel. The rocket engine is then fired for \(250 \mathrm{~s}\) while fuel is consumed at the rate of \(480 \mathrm{~kg} / \mathrm{s}\). The speed of the exhaust products relative to the rocket is \(3.27 \mathrm{~km} / \mathrm{s}\). (a) What is the rocket's thrust? After the \(250 \mathrm{~s}\) firing, what are (b) the mass and (c) the speed of the rocket?

A \(0.25 \mathrm{~kg}\) puck is initially stationary on an ice surface with negligible friction. At time \(t=0\), a horizontal force begins to move the puck. The force is given by \(\vec{F}=\left(12.0-3.00 t^{2}\right) \hat{1}\), with \(\vec{F}\) in newtons and \(t\) in seconds, and it acts until its magnitude is zero. (a) What is the magnitude of the impulse on the puck from the force between \(t=0.500 \mathrm{~s}\) and \(t=1.25 \mathrm{~s} ?\) (b) What is the change in momentum of the puck between \(t=0\) and the instant at which \(F=0 ?\)

In a game of pool, the cue ball strikes another ball of the same mass and initially at rest. After the collision, the cue ball moves at \(3.50 \mathrm{~m} / \mathrm{s}\) along a line making an angle of \(22.0^{\circ}\) with the cue ball's original direction of motion, and the second ball has a speed of \(2.00 \mathrm{~m} / \mathrm{s}\). Find (a) the angle between the direction of motion of the second ball and the original direction of motion of the cue ball and (b) the original speed of the cue ball. (c) Is kinetic energy (of the centers of mass, don't consider the rotation) conserved?

In February 1955 , a paratrooper fell \(370 \mathrm{~m}\) from an airplane without being able to open his chute but happened to land in snow, suffering only minor injuries. Assume that his speed at impact was \(56 \mathrm{~m} / \mathrm{s}\) (terminal speed), that his mass (including gear) was \(85 \mathrm{~kg}\), and that the magnitude of the force on him from the snow was at the survivable limit of \(1.2 \times 10^{5} \mathrm{~N}\). What are (a) the minimum depth of snow that would have stopped him safely and (b) the magnitude of the impulse on him from the snow?

A \(75 \mathrm{~kg}\) man rides on a \(39 \mathrm{~kg}\) cart moving at a velocity of \(2.3 \mathrm{~m} / \mathrm{s}\) He jumps off with zero horizontal velocity relative to the ground. What is the resulting change in the cart's velocity, including sign?
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