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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 kg m/s; Average force: 1820 N.

Step by step solution

01

Identify Given Data

We need to list all the given data for this problem: The initial speed of the hand, \( v_i = 13 \ \text{m/s} \), the final speed \( v_f = 0 \ \text{m/s} \) since it comes to a stop, the collision time \( \Delta t = 5.0 \ \text{ms} = 5.0 \times 10^{-3} \ \text{s} \), and the mass of the hand \( m = 0.70 \ \text{kg} \).
02

Calculate Impulse

Impulse can be calculated using the formula \( J = \Delta p \) where \( \Delta p \) is the change in momentum. Since momentum \( p = mv \), \( J = m(v_f - v_i) = 0.70 \ (0 - 13) = -9.1 \ \text{kg m/s} \). The impulse's magnitude is \( 9.1 \ \text{kg m/s} \).
03

Calculate Average Force

Average force can be found using the relation between impulse and force: \( J = F_{avg} \times \Delta t \). Solving for \( F_{avg} \), \( F_{avg} = \frac{J}{\Delta t} = \frac{-9.1}{5.0 \times 10^{-3}} = -1820 \ \text{N} \). 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.

Momentum
Momentum is a key concept in understanding how objects move and interact. It's essentially the "oomph" that an object carries due to its mass and velocity. Imagine a bowling ball rolling down the lane; it has momentum because of its mass and the speed at which it's moving. Mathematically, momentum (\( p \)) is calculated as the product of mass (\( m \)) and velocity (\( v \)). So, \[ p = mv \]
  • Mass: This is how much "stuff" an object has; think of it as the object's weight, but without gravity.
  • Velocity: This is how fast something is moving in a specific direction.
In the tae-kwon-do example, when the hand hits the target, its momentum changes drastically. It goes from moving rapidly to stopping in a very short period. This quick change in momentum is crucial to consider when thinking about the force involved in such impacts.
Average Force
Average force is different from the instantaneous force, which might vary during an impact. Here, average force (\( F_{avg} \)) represents the "steady push" or "steady pull" that, over a defined time period, could achieve the same change in momentum as the actual varying forces that occur. This is invaluable for understanding the overall effect of a collision.The formula connecting impulse (\( J \)) and average force is:\[ F_{avg} = \frac{J}{\Delta t} \]Where:
  • Impulse (\( J \)): Represents the total change in momentum.
  • Collision Time (\( \Delta t \)): The time interval over which the collision or impact occurs.
During the tae-kwon-do collision, average force becomes a simpler way to express the force that causes the hand's momentum to change from high to zero in an efficiently expressed number, allowing us to understand and communicate the impact's magnitude more clearly.
Collision Time
Collision time is the duration during which two objects are in contact during a collision. It's essentially the blink-of-an-eye moment when forceful interactions happen, like the hand striking the target in tae-kwon-do. This brief window of time is remarkably significant because it helps us determine the average force exerted during the impact.The collision time (\( \Delta t \)) is crucial because:
  • It influences the magnitude of the average force. A shorter \( \Delta t \) often results in a larger force.
  • Understanding \( \Delta t \) helps in analyzing safety equipment and materials that are designed to extend the impact time, thus reducing force (e.g., airbags).
In the exercise, the collision time of \(5.0 \ ext{ms}\) is what lets us use the impulse equation effectively to calculate the average force. It's a key detail that tells us how swiftly the hand's velocity drops to zero.

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Most popular questions from this chapter

A railroad car moves under a grain elevator at a constant speed of \(3.20 \mathrm{~m} / \mathrm{s}\). Grain drops into the car at the rate of 540 \(\mathrm{kg} / \mathrm{min}\). What is the magnitude of the force needed to keep the car moving at constant speed if friction is negligible?

A pellet gun fires ten \(2.0 \mathrm{~g}\) pellets per second with a speed of \(500 \mathrm{~m} / \mathrm{s}\). The pellets are stopped by a rigid wall. What are (a) the magnitude of the momentum of each pellet, (b) the kinetic energy of each pellet, and (c) the magnitude of the average force on the wall from the stream of pellets? (d) If each pellet is in contact with the wall for \(0.60 \mathrm{~ms}\), what is the magnitude of the average force on the wall from each pellet during contact? (e) Why is this average force so different from the average force calculated in (c)?

Basilisk lizards can run across the top of a water surface (Fig. 9-52). With each step, a lizard first slaps its foot against the water and then pushes it down into the water rapidly enough to form an air cavity around the top of the foot. To avoid having to pull the foot back up against water drag in order to complete the step, the lizard withdraws the foot before water can flow into the air cavity. If the lizard is not to sink, the average upward impulse on the lizard during this full action of slap, downward push, and withdrawal must match the downward impulse due to the gravitational force. Suppose the mass of a basilisk lizard is \(90.0 \mathrm{~g}\), the mass of each foot is \(3.00 \mathrm{~g}\), the speed of a foot as it slaps the water is \(1.50\) \(\mathrm{m} / \mathrm{s}\), and the time for a single step is \(0.600 \mathrm{~s}\). (a) What is the magnitude of the impulse on the lizard during the slap? (Assume this impulse is directly upward.) (b) During the \(0.600 \mathrm{~s}\) duration of a step, what is the downward impulse on the lizard due to the gravitational force? (c) Which action, the slap or the push, provides the primary support for the lizard, or are they approximately equal in their support?

A collision occurs between a \(2.00 \mathrm{~kg}\) particle traveling with velocity \(\vec{v}_{1}=(-4.00 \mathrm{~m} / \mathrm{s}) \mathrm{i}+(-5.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) and a \(4.00 \mathrm{~kg}\) particle traveling with velocity \(\vec{v}_{2}=(6.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(-2.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\). The collision connects the two particles. What then is their velocity in (a) unit- vector notation and as a (b) magnitude and (c) angle?

Block 1, with mass \(m_{1}\) and speed \(4.0 \mathrm{~m} / \mathrm{s}\), slides along an \(x\) axis on a frictionless floor and then undergoes a one-dimensional elastic collision with stationary block 2, with mass \(m_{2}=0.40 m_{1}\). The two blocks then slide into a region where the coefficient of kinetic friction is \(0.50 ;\) there they stop. How far into that region do (a) block 1 and \((b)\) block 2 slide?
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