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In a common but dangcrous prank, a chair is pulled away as a person is moving downward to sit on it, causing the victim to land hard on the floor. Suppose the victim falls by \(0.50 \mathrm{~m}\), the mass that moves downward is \(70 \mathrm{~kg}\). and the collision on the floor lasts \(0.082 \mathrm{~s}\). What are the magnitudes of the (a) impulse and (b) average force acting on the victim from the floor during the collision?

Short Answer

Expert verified
Impulse is 219.1 Ns; average force is 2672.0 N.

Step by step solution

01

Calculate the Final Velocity

To solve this problem, we first need to find the final velocity of the victim just before hitting the floor. We can use the equation for motion under gravity: \[ v^2 = u^2 + 2gh \]where initial velocity \( u = 0 \) (since they start from rest), \( g = 9.8\, \text{m/s}^2 \), and \( h = 0.50\, \text{m} \).\[ v^2 = 0 + 2 \times 9.8 \times 0.50 \ = 9.8 \]\[ v = \sqrt{9.8} \ ≈ 3.13\, \text{m/s} \]
02

Calculate the Impulse

Impulse \( J \) is the change in momentum, which can be calculated using the formula:\[ J = \Delta p = m \Delta v = m(v - u) \]where \( m = 70\, \text{kg} \), \( v = 3.13\, \text{m/s} \), and \( u = 0 \).\[ J = 70 \times (3.13 - 0) = 219.1\, \text{Ns} \]
03

Calculate the Average Force

The average force \( F_{avg} \) can be found using the impulse and the collision time \( \Delta t = 0.082 \text{s} \):\[ F_{avg} = \frac{J}{\Delta t} = \frac{219.1}{0.082} \]\[ F_{avg} ≈ 2672.0\, \text{N} \]

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Key Concepts

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

Impulse
Impulse is a fundamental concept in physics that describes the change in momentum of an object when a force is applied over a period of time. This concept can be a bit tricky to wrap your head around, but let's break it down to make it simple.
  • Think of momentum as the "oomph" of a moving object, which is determined by the object's mass and velocity.
  • Impulse is what causes a change in this "oomph" when an external force is applied.
In our example, the person falling to the ground experiences impulse during the collision with the floor. Impulse (\( J \)) is calculated using the formula:
\[J = m \Delta v\]where:
  • \( m \) is the mass of the object (70 kg in this example),
  • \( \Delta v \) is the change in velocity.
In this situation, the velocity changes from zero to the velocity just before impact (3.13 m/s). By applying the impulse formula, we find that the impulse experienced by the victim is 219.1 Ns.Impulse not only tells us how strong a force acts but also how it acts over time.
Average Force
Average force plays a crucial role when dealing with impacts as it tells us how much force was, on average, exerted during the course of a collision or contact. The average force can be determined using the impulse found earlier and the time interval over which the force acts, as seen in our exercise:
\[F_{avg} = \frac{J}{\Delta t}\]where:
  • \( J \) is the impulse (219.1 Ns in this example),
  • \( \Delta t \) is the time duration of the impact (0.082 s).
By dividing the impulse by the time, we find that the average force acting on the victim from the floor is approximately 2672.0 N. This average force indicates how hard the floor pushes against the person to stop their fall in such a short time. It explains why a sudden impact can feel so jarring on our bodies.
Motion under Gravity
The principle of motion under gravity is critical in understanding how objects behave when they are freely falling towards the Earth. In this concept, gravity is the only force acting on the object, causing it to accelerate downwards.
  • For a freely falling body under gravity, the acceleration \( g \) is approximately 9.8 m/s².
  • This constant acceleration is what makes objects increase their speed as they fall.
In the example exercise, the victim falls a distance of 0.50 m. We use the equation:
\[v^2 = u^2 + 2gh\]where:
  • \( u \) is the initial velocity (0 m/s here, since starting at rest),
  • \( v \) is the final velocity just before contacting the ground,
  • \( h \) is the height fallen (0.50 m).
This equation helps us find that the final velocity just before impact is approximately 3.13 m/s. Understanding motion under gravity allows us to predict the behavior of falling objects and calculate their parameters, such as velocity and time taken to fall a certain distance, which is crucial for solving physics problems involving free falling objects.

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Most popular questions from this chapter

A \(5.20 \mathrm{~g}\) bullet moving at \(672 \mathrm{~m} / \mathrm{s}\) strikes a \(700 \mathrm{~g}\) wooden block at rest on a frictionless surface. The bullet emerges, traveling in the same direction with its speed reduced to \(428 \mathrm{~m} / \mathrm{s}\). (a) What is the resulting speed of the block? (b) What is the speed of the bullet-block center of mass?

block \(A\) (mass \(1.6 \mathrm{~kg}\) ) slides into block \(B\) (mass \(2.4 \mathrm{~kg}\) ), along a frictionless surface. The directions of three velocities before \((i)\) and after \((f)\) the collision are indicated; the corresponding speeds are \(v_{A i}=5.5 \mathrm{~m} / \mathrm{s}\) \(v_{B i}=2.5 \mathrm{~m} / \mathrm{s},\) and \(v_{B f}=4.9 \mathrm{~m} / \mathrm{s} .\) What are the (a) speed and (b) direction (left or right) of velocity \(\vec{v}_{A f} ?\) (c) Is the collision elastic?

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?

Suppose a gangster sprays Superman's chest with 3 g bullets at the rate of 100 bullets/min, and the speed of cach bullet is \(500 \mathrm{~m} / \mathrm{s}\). Suppose too that the bullets rebound straight back with no change in speed. What is the magnitude of the average force on Superman's chest?

A force in the negative direction of an \(x\) axis is applied for \(27 \mathrm{~ms}\) to a \(0.40 \mathrm{~kg}\) ball initially moving at \(14 \mathrm{~m} / \mathrm{s}\) in the positive direction of the axis. The force varies in magnitude, and the impulse has magnitude \(32.4 \mathrm{~N} \cdot \mathrm{s}\). What are the ball's (a) speed and (b) dircction of travel just after the force is applied? What are (c) the average magnitude of the force and (d) the direction of the impulse on the ball?

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