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Chapter 7: Problem 11

A student \((m=63 \mathrm{kg})\) falls freely from rest and strikes the ground. During the collision with the ground, he comes to rest in a time of \(0.040 \mathrm{s} .\) The average force exerted on him by the ground is \(+18000 \mathrm{N},\) where the upward direction is taken to be the positive direction. From what height did the student fall? Assume that the only force acting on him during the collision is that due to the ground.

Short Answer

Expert verified
The student fell from a height of approximately 6.66 meters.

Step by step solution

01

Understand the Problem

A student falls from rest, and we're given his mass and the time during which he stops upon hitting the ground. The goal is to find the height from which he fell by using the average force exerted during the collision.
02

Calculate the Impulse Exerted on the Student

Impulse is given by the formula \( J = F_{avg} \times \Delta t \). With the average force \( F_{avg} = 18000 \mathrm{N} \) and the collision time \( \Delta t = 0.040 \mathrm{s} \), we calculate the impulse:\[J = 18000 \times 0.040 = 720 \mathrm{Ns}.\]
03

Use Impulse to Find Change in Momentum

The impulse is equal to the change in momentum \( \Delta p \). The student comes to rest, so the final momentum is zero. Thus, \( \Delta p = -m \times v_0 \), where \( m = 63 \mathrm{kg} \) and \( v_0 \) is the initial velocity before impact. We have:\[J = -m \times v_0 \Rightarrow 720 = -63 \times v_0.\]Solve for \( v_0 \):\[v_0 = -\frac{720}{63} \approx -11.43 \mathrm{m/s}.\]
04

Relate Velocity to Height

The velocity of the student before impact is due to falling from a height. The relationship between velocity and height when an object falls freely is given by the equation \( v_0 = \sqrt{2gh} \), where \( g = 9.81 \mathrm{m/s^2} \) is the acceleration due to gravity.\[v_0^2 = 2gh \Rightarrow (11.43)^2 = 2 \times 9.81 \times h.\]Solve for \( h \):\[130.68 = 19.62 \times h \Rightarrow h = \frac{130.68}{19.62} \approx 6.66 \mathrm{m}.\]

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Key Concepts

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

Free Fall
Free fall describes the motion of any object solely under the influence of gravity. In this scenario, the student is in free fall as they descend towards the ground. This is a special case of motion where air resistance is neglected, and the only force acting on the falling object is gravity itself.
- All objects in free fall on Earth experience an acceleration downwards of approximately 9.81 m/s虏.
- The initial velocity in this scenario is zero because the student falls from rest.
- Free fall conditions help us easily link velocity, acceleration, and height, which are crucial to solving this problem.
By understanding free fall, we can relate the descent height to the velocity just before the student hits the ground.
Average Force
Average force is the overall force exerted during an event, given in this problem by the ground during the collision. While colliding, the student exerts a force on the ground, and the ground exerts an equal and opposite force back, as described by Newton's third law.
- The average force here is 18,000 N, specified upwards, indicating it acts opposite to the free fall direction.
- The formula for impulse is used to find change in momentum, which involves multiplying this average force by the time interval over which the collision occurs, 0.04 seconds.
Calculating the impulse helps us determine how rapidly the velocity changes upon impact, which connects directly to understanding how forces are distributed over the short collision time.
Collision Impact
During the collision impact between the student and the ground, momentum plays a crucial role. The collision happens in a very short time span of 0.04 seconds, during which the student comes to a complete stop.
- Momentum is the product of mass and velocity. Before impact, the student has an initial downward velocity, which changes to zero during impact.
- Impulse, the product of average force and time, equals the change in momentum, allowing us to find the pre-impact velocity.
This conversion of velocity to a stopping condition amplifies how short collision times require larger forces for stopping, which we observe through the calculated large average force value. This understanding helps us find the initial velocity, key to determining the fall height.
Velocity-Height Relation
The relationship between velocity and height in free fall is governed by gravitational acceleration. This principle helps connect the speed just before impact to the height from which the object fell.
- The formula used is derived from energy conservation: \[ v_0 = \sqrt{2gh}\]- Using the previously found initial velocity of -11.43 m/s, we relate it to height using gravity's constant force (9.81 m/s虏).
- By squaring the velocity, dividing by twice the gravitational acceleration, we solve for the height, finding it to be around 6.66 meters.
Understanding this relation allows us to see how energy conversions during free fall impact speed and height calculations, a fundamental concept in physics.

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Most popular questions from this chapter

A \(2.3-\mathrm{kg}\) cart is rolling across a frictionless, horizontal track toward a \(1.5-\mathrm{kg}\) cart that is held initially at rest. The carts are loaded with strong magnets that cause them to attract one another. Thus, the speed of each cart increases. At a certain instant before the carts collide, the first cart's velocity is \(+4.5 \mathrm{m} / \mathrm{s},\) and the second cart's velocity is \(-1.9 \mathrm{m} / \mathrm{s}\) (a) What is the total momentum of the system of the two carts at this instant? (b) What was the velocity of the first cart when the second cart was still at rest?

Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic collision. Momentum is conserved. Object A has a mass of \(m_{A}=17.0 \mathrm{kg}\) and an initial velocity of \(\overrightarrow{\mathbf{v}}_{a_{A}}=8.00 \mathrm{m} / \mathrm{s},\) due east. Object \(\mathrm{B},\) however, has a mass of \(m_{\mathrm{B}}=29.0 \mathrm{kg}\) and an initial velocity of \(\overrightarrow{\mathbf{v}}_{\mathrm{eg}}=5.00 \mathrm{m} / \mathrm{s},\) due north. Find the magnitude and direction of the total momentum of the two-object system after the collision.

Adolf and Ed are wearing harnesses and are hanging at rest from the ceiling by means of ropes attached to them. Face to face, they push off against one another. Adolf has a mass of \(120 \mathrm{kg},\) and \(\mathrm{Ed}\) has a mass of \(78 \mathrm{kg} .\) Following the push, Adolf swings upward to a height of \(0.65 \mathrm{m}\) above his starting point. To what height above his own starting point does Ed rise?

A ball is attached to one end of a wire, the other end beingfastened to the ceiling. The wire is held horizontal, and the ball is released from rest (see the drawing). It swings downward and strikes a block initially at rest on a horizontal frictionless surface. Air resistance is negligible, and the collision is elastic. The masses of the ball and block are, respectively, \(1.60 \mathrm{kg}\) and \(2.40 \mathrm{kg},\) and the length of the wire is \(1.20 \mathrm{m} .\) Find the velocity (magnitude and direction) of the ball (a) just before the collision, and (b) just after the collision.

John's mass is \(86 \mathrm{kg}\), and Barbara's is \(55 \mathrm{kg}\). He is standing on the \(x\) axis at \(x_{1}=+9.0 \mathrm{m},\) while she is standing on the \(x\) axis at \(x_{\mathrm{B}}=12.0 \mathrm{m}\) They switch positions. How far and in which direction does their center of mass move as a result of the switch?
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