/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 A ball kept in a closed box move... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 33

A ball kept in a closed box moves in the box making collisions with the walls. The box is kept on a smooth surface. The velocity of the centre of mass (1) of the box remains constant (2) of the (box + ball) system remains constant (3) of the ball remains constant (4) of the ball relative to the box remains constant

Short Answer

Expert verified
The velocity of the center of mass of the (box + ball) system remains constant.

Step by step solution

01

Understanding the Problem

We have a ball inside a box, both moving on a smooth surface. The ball can collide with the box walls, which affects its motion. We need to determine the motion properties of the center of mass for certain components of this system.
02

Analyzing the System

Since the box is on a smooth surface, there are no external horizontal forces acting on the system of the box and the ball. By the conservation of linear momentum, the velocity of the center of mass of the whole system (box + ball) should remain constant.
03

Considering Individual Components

For the center of mass of just the box, without external horizontal forces, its velocity individually can change due to internal forces (collisions with the ball), unlike the box + ball system. Thus, the center of mass of only the box does not necessarily remain constant.
04

Analyzing the Ball's Motion

The ball undergoes collisions with the box walls, which can change its velocity. Therefore, the velocity of the center of mass of the ball will not remain constant.
05

Relative Motion Analysis

The velocity of the ball relative to the box changes due to collisions, as these collisions impart momentum transfer between the ball and the box. Thus, the ball's velocity relative to the box does not remain constant.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

Key Concepts

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

Center of Mass
The center of mass of a system is the point where the total mass of the system can be thought to be concentrated. When analyzing the box and ball system on a smooth surface, understanding the center of mass helps in applying the conservation laws effectively. The velocity of the center of mass of a system remains constant if no external forces are acting on the system. Since the box is on a smooth surface, which minimizes friction and external forces, the velocity of the center of mass for the combined box and ball system is constant. This principle stems from the conservation of linear momentum. The position of the center of mass depends on the distribution of mass within the system. If the box and the ball have different masses, changes in their individual positions affect the center of mass, but without external forces, the velocity remains unaffected.
Collision Mechanics
Collision mechanics is the study of how objects interact and exchange momentum and energy when they collide. In the context of the ball in the box, each collision between the ball and the box walls temporarily alters the motion of the ball as well as the box. While these collisions can change the velocity of individual parts of the system (like just the box or just the ball), they don't influence the whole system (box + ball) because of conserved momentum. The internal forces involved during collisions don't alter the overall motion of the system. This internal exchange of momentum is critical in understanding why the velocity of the box + ball system's center of mass remains unchanged, even though separate components like only the box or only the ball experience changes in motion.
Relative Velocity
Relative velocity refers to how fast an object is moving in relation to another. In the scenario with a ball inside a moving box, both are situated on a smooth surface. The relative velocity between the box and the ball changes as the ball bounces off the walls of the box.
  • As the ball hits the walls of the box, it transfers momentum, which causes changes in its speed and direction in comparison to the box.
  • Since collisions affect speed and direction, the relative velocity between the ball and box is not constant.
Understanding relative velocity is important for grasping the internal dynamics within a system, specifically how individual components seem to move when observed from each other's frame of reference.
Smooth Surface Dynamics
Smooth surface dynamics involves analyzing motion scenarios where friction and external resistance are minimal. This is crucial when observing the ball and box system, as the system rests on a smooth surface that offers negligible resistance. Such conditions ensure there are no external horizontal forces acting upon the system.
  • Because friction and other resistances are absent, external forces do not interfere with the motion of the system.
  • This makes it a perfect environment to study the conservation of momentum without distractions from frictional forces.
In this frictionless environment, the velocity of the system's components might change, but the velocity of the system's center of mass persists unaltered as long as all forces are internal, highlighting the elegance of conservation laws in ideal conditions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

A frog sits on the end of a long board of length \(L=5 \mathrm{~m}\). The board rests on a frictionless horizontal table. The frog wants to jump to the opposite end of the board. What is the minimum take-off speed (in \(\mathrm{m} / \mathrm{s}\) ), i.e., relative to ground that allows the frog to do the trick? The board and the frog have equal masses.

A strip of wood of mass \(M\) and length \(l\) is placed on a smooth horizontal surface. An insect of mass \(m\) starts at one end of the strip and walks to the other end in time \(t\), moving with a constant speed. The speed of the insect as seen from the ground is (1) \(\frac{l}{t}\left(\frac{M}{M+m}\right)\) (2) \(\frac{l}{t}\left(\frac{m}{M+m}\right)\) (3) \(\frac{l}{t}\left(\frac{M}{m}\right)\) (4) \(\frac{l}{t}\left(\frac{m}{M}\right)\)

A man of mass \(m\) walks from end \(A\) to the other end \(B\) of plank of mass \(M\) and length \(l\), placed on a smooth horizonta surface. The coefficient of friction between man and plank is \(\mu\) as no external force in horizontal direction. (1) If man runs at his maximumacceleration, the acceleration of plank is \(\frac{m}{M} \mu g\) (2) Time taken by man to reach other end of the plank is \(\sqrt{\frac{2 M l}{(M+m) \mu g}}\) (3) Magnitude of displacement of centre of mass of plank is \(\frac{M l}{m+M}\) (4) Velocity of centre of mass of (man + plank) is zero.

Two identical particles \(A\) and \(B\) of mass \(m\) each are connecte together by a light and inextensible string of length \(l . \mathrm{T}\) particle are held at rest in air in same horizontal level at separation \(l\). Both particles are released simultaneously an one of them (say \(A\) ) is given speed \(v_{0}\) vertically upwan The maximum height attained by the centre of mass of 4 system of \(A\) and \(B\) from initial level is (Ignore air resistanca (1) \(\frac{v_{0}^{2}}{2 g}\) (2) \(\frac{v_{0}^{2}}{8 g}\) (3) \(\frac{v_{0}^{2}}{4 g}\) (4) \(\frac{v_{0}^{2}}{12 g}\)

A man standing on the edge of the terrace of a high rise building throws a stone, vertically up with a speed of \(20 \mathrm{~m} / \mathrm{s}\). Two seconds later, an identical stone is thrown vertically downwards with the same speed of \(20 \mathrm{~m} / \mathrm{s}\). Then (1) the relative velocity between the two stones remains constant till one hits the ground (2) both will have the same kinetic energy, when they hit the ground (3) the time interval between their hitting the ground is \(2 \mathrm{~s}\) (4) if the collision on the ground is perfectly elastic, both will rise to the same height above the ground
See all solutions

Recommended explanations on Physics Textbooks

Radiation

Read Explanation

Circular Motion and Gravitation

Read Explanation

Quantum Physics

Read Explanation

Electrostatics

Read Explanation

Electricity

Read Explanation

Magnetism

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.