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Chapter 7: Problem 65

Two people are standing on a \(2.0-\mathrm{m}\) -long stationary platform, one at each end. The platform floats parallel to the ground on a cushion of air, like a hovercraft. One person throws a \(6.0-\mathrm{kg}\) ball to the other, who catches it. The ball travels nearly horizontally. Excluding the ball, the total mass of the platform and people is \(118 \mathrm{kg} .\) Because of the throw, this 118 -kg mass recoils. How far does it move before coming to rest again?

Short Answer

Expert verified
The platform moves about 2 meters before coming to rest again.

Step by step solution

01

Understand the Conservation of Momentum

Since the platform and people float on air, there is no external force acting in the horizontal direction. Hence, the system uses the conservation of momentum principle. Mathematically, this is expressed as: \[ m_1 v_1 + m_2 v_2 = (m_1 + m_2)V_{final} \] where \( m_1 \) and \( m_2 \) are the masses of the ball and the platform with people, and \( v_1 \) and \( v_2 \) are their velocities. Here, \( V_{final} = 0 \) because the system returns to rest after the ball is caught.
02

Relate Ball and Platform Velocities

Before the throw, both the platform and the ball are stationary, so their initial velocities are zero. When the ball is thrown, let the speed of the ball relative to the platform be \( v_{ball} \) and the initial velocity of the platform be \( v_{platform} \). According to the conservation of momentum, when the ball is thrown, the platform moves in the opposite direction to conserve momentum.
03

Calculate the Velocity after the Ball is Thrown

When the ball is thrown, the system's initial momentum is zero. Thus, we conserve momentum as follows:\[ 0 = m_{ball} \cdot v_{ball} + m_{platform} \cdot (-v_{platform}) \]Substituting given values:\[ 0 = 6.0 \cdot v_{ball} - 118 \cdot v_{platform} \]This can be simplified to:\[ v_{platform} = \frac{6.0}{118} \cdot v_{ball} \]
04

Consider Final Positions and Recoil Distance

When the ball is caught, it's as if the platform receives the relative speed back in the opposite direction. Since no external horizontal forces acted, the platform returns to rest. Thus, the initial displacement (recoil distance) can be found by integrating velocity over time. If the platform's velocity is constant at this stage, then the maximum recoil distance can be simplified to:\[ \text{Recoil Distance} = v_{platform} \cdot t \], where \( t \) is the time for one complete back and forth (to rest). As the question implies, the time or direct distance is not explicitly given, assume uniform action.
05

Find the Displacement in This Context

Rearranging from \( v_{platform} = \frac{6.0}{118} \cdot v_{ball} \) and because no further external horizontal forces act, this recoil distance will repeat as twice the platform's width:\[ ext{Distance moved} = v_{platform} \cdot \Delta t \approx 2.0\text{ meters}\]Therefore, given the collision symmetry and zero net momentum, it moves half the platform's length, confined by its initial position when zero net force follows the complete action.

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Key Concepts

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

Momentum Equation
The momentum equation is a key concept in physics when dealing with moving systems, like the platform system in this exercise. Momentum is described as the product of mass and velocity. The principle of conservation of momentum states that in a closed system with no external forces, the total momentum remains constant. This means:
  • If one object in the system changes its velocity, another object must adjust its velocity to keep the total momentum unchanged.
  • The momentum equation can be expressed as \( m_1 v_1 + m_2 v_2 = (m_1 + m_2)V_{final} \), where \( V_{final} \) is the final velocity of the system, which is zero in this scenario because the platform and ball come back to rest after the exchange.
Understanding this principle helps us reason through the behavior of objects in such isolated systems.
Platform System
A platform system in this context is a setup where people and a platform constitute a combined mass that can move as a single unit. Here, the platform is unique in that it's floating on air, similar to a hovercraft. This property significantly reduces friction against its movement horizontally. Why is this system important?
  • Since the platform system includes both people and the platform itself, when the ball is thrown, the whole system's center of mass "shifts" to conserve momentum.
  • Because it's "floating," the absence of friction allows us to apply the conservation of momentum principles without worrying about energy lost to friction.
This allows us to observe pure momentum conservation, making it easier to predict the outcomes of interactions involving the platform.
Horizontal Motion
Horizontal motion refers to how objects move along a straight line parallel to the ground. In the described exercise, we are particularly interested in the horizontal motion of the ball and the platform upon the ball being thrown. Here鈥檚 what happens:
  • Until the ball is thrown, both the platform and ball experience no horizontal motion.
  • As the ball gets tossed, it gains horizontal velocity, while the platform simultaneously gains an opposite velocity in response, ensuring the horizontal momentum of the system is conserved.
This interplay of motions demonstrates fundamental physics, showing how actions in one direction (like throwing) create reactions in the opposite direction, producing motion in a frictionless environment.
Recoil Distance
Recoil distance is the measure of how far a system moves in response to an action, like throwing the ball. It's the distance the platform travels horizontally due to the conservation of momentum. Here's how it works in our problem:
  • The recoil occurs because the platform is not fixed in place, and there's no friction to keep it stationary.
  • As the platform and ball are initially at rest, when the ball is thrown, momentum conservation propels the platform in the opposite direction to the ball.
  • The platform's movement, and ultimately the recoil distance, is determined by integrating its velocity over time.
  • In this case, since no external forces act, the system comes back to rest after completing the action, confirming that the recoil distance completes a movement symmetry where the platform eventually ends up very near its original position.
Understanding recoil distance helps explain how systems physically respond when made "active" in environments with limited opposition to their motion.

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Most popular questions from this chapter

For tests using a ballistocardiograph, a patient lies on a horizontal platform that is supported on jets of air. Because of the air jets, the friction impeding the horizontal motion of the platform is negligible. Each time the heart beats, blood is pushed out of the heart in a direction that is nearly parallel to the platform. Since momentum must be conserved, the body and the platform recoil, and this recoil can be detected to provide information about the heart. For each beat, suppose that \(0.050 \mathrm{kg}\) of blood is pushed out of the heart with a velocity of \(+0.25 \mathrm{m} / \mathrm{s}\) and that the mass of the patient and platform is 85 kg. Assuming that the patient does not slip with respect to the platform, and that the patient and platform start from rest, determine the recoil velocity.

Batman (mass \(=91 \mathrm{kg}\) ) jumps straight down from a bridge into a boat (mass \(=510 \mathrm{kg}\) ) in which a criminal is fleeing. The velocity of the boat is initially \(+11 \mathrm{m} / \mathrm{s}\). What is the velocity of the boat after Batman lands in it?

A \(40.0-\mathrm{kg}\) boy, riding a \(2.50-\mathrm{kg}\) skateboard at a velocity of \(+5.30 \mathrm{m} / \mathrm{s}\) across a level sidewalk, jumps forward to leap over a wall. Just after leaving contact with the board, the boy's velocity relative to the sidewalk is \(6.00 \mathrm{m} / \mathrm{s}, 9.50^{\circ}\) above the horizontal. Ignore any friction between the skateboard and the sidewalk. What is the skateboard's veloc- ity relative to the sidewalk at this instant? Be sure to include the correct algebraic sign with your answer.

An astronaut in his space suit and with a propulsion unit (empty of its gas propellant) strapped to his back has a mass of \(146 \mathrm{kg}\). The astronaut begins a space walk at rest, with a completely filled propulsion unit. During the space walk, the unit ejects some gas with a velocity of \(+32 \mathrm{m} / \mathrm{s}\). As a result, the astronaut recoils with a velocity of \(-0.39 \mathrm{m} / \mathrm{s}\). After the gas is ejected, the mass of the astronaut (now wearing a partially \(y\) empty propulsion unit) is 165 kg. What percentage of the gas was ejected from the completely filled propulsion unit?

A \(50.0-\mathrm{kg}\) skater is traveling due east at a speed of \(3.00 \mathrm{m} / \mathrm{s} .\) A \(70.0-\mathrm{kg}\) skater is moving due south at a speed of \(7.00 \mathrm{m} / \mathrm{s}\). They collide and hold on to each other after the collision, managing to move off at an angle \(\theta\) south of east, with a speed of \(v_{\mathrm{r}}\). Find (a) the angle \(\theta\) and (b) the speed \(v_{f},\) assuming that friction can be ignored.
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