/*! 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 3 Two skaters have weight in the r... [FREE SOLUTION] | 91Ó°ÊÓ

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Two skaters have weight in the ratio \(4: 5\) and are \(9 \mathrm{~m}\) apart. on a smooth frictionless surface. They pull on a rope stretched between them. The ratio of the distance covered by them when they meet each other will be (1) \(5: 4\) (2) \(4: 5\) (3) \(25: 16\) (4) \(16: 25\)

Short Answer

Expert verified
The ratio of distances covered by the skaters is \(5:4\).

Step by step solution

01

Understanding the Problem

We're given two skaters with weights in the ratio \(4:5\) and they're \(9\) meters apart, pulling a rope on a frictionless surface. We need to determine the ratio of their distances covered when they meet based on their relative weights.
02

Applying the Concept of Center of Mass

On a frictionless surface, the center of mass of the system (two skaters) doesn't move. This means the distance each skater covers is inversely proportional to their weight.
03

Setting Up the Proportional Relation

Since the weight ratio is given as \(4:5\), the distance each skater covers from their initial positions to the meeting point will be inversely proportional, i.e., \(d_1:d_2 = 5:4\), where \(d_1\) is the distance covered by the first skater and \(d_2\) by the second.
04

Verifying the Total Distance

The total distance between the skaters is 9 meters. Hence, we need to confirm that the ratio \(5:4\) fits this scenario by recognizing the inverse of mass proportionality correctly answers the initial setup.

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Key Concepts

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

Frictionless Surface
Imagine a world where there is no friction—this means no resistance against motion. A frictionless surface is a theoretical concept used in physics to simplify problems, often used to analyze motion without the complexities introduced by friction forces.
Friction is usually responsible for opposing motion and slowing things down. So, in this special environment, moving objects won't slow down unless acted upon by other forces. For our skaters pulling on a rope, the ideal frictionless condition ensures that their interactions depend solely on their action of pulling, not any opposing floor force.
  • Objects can glide effortlessly.
  • The energy converts without loss to heat due to friction—perfect for energy conservation discussions.
  • Useful in understanding idealized motion and scenarios like ice skating on extremely smooth surfaces.
The absence of friction allows the description of motion to significantly focus on external forces and interactions, such as the tension in the rope between the skaters.
Relative Motion
Relative motion is the concept of observing motion concerning a certain frame of reference. In our example, when two skaters pull on a rope between them, their movements are determined relative to each other and can be analyzed from different viewpoints.
When we say one skater moves a certain distance, it's often described in relation to the other skater's position or movement.
  • This concept allows us to predict where and when they will meet based on how they move towards each other.
  • It provides a dynamic view where motion depends on the observer's point of view, aiding the understanding of interactions without fixed positions.
  • Knowing each skater’s relative motion helps in calculating the meeting point considering their different weights and initial positions.
Thus, by focusing on relative rather than absolute motion, we smartly determine the distance ratios covered by each skater as they pull themselves towards one another.
Proportionality in Physics
Proportionality in physics helps us understand how different quantities relate to each other. It can determine how one variable affects another. In the skater problem, it shows how the distances each skater covers relate inversely to their weights.
When given a situation like this, understanding that weights and distances have a certain proportional relationship allows for complex problem-solving through simplification.
  • A direct proportion means increasing one quantity increases another at a constant rate.
  • An inverse proportion, as seen here, means increasing one quantity decreases another. As the skater problem shows, a skater with more weight will cover less distance and vice versa.
  • Using proportionality clarifies how initial conditions, like mass ratio, impact final outcomes, such as meeting points.
Through these ratios, we precisely determine the individual distances covered, with the general rule: more weight less distance in inverse flight.

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Most popular questions from this chapter

A horizontal force \(F\) pulls a ring of mass \(m_{1}\) such that \(\theta\) remains constant with time. The ring is constrained to move along a smooth rigid horizontal wire. A bob of mass \(m_{2}\) hangs from \(m_{1}\) by an inextensible light string. Then match the entries of Column I with that of Column II. $$ \begin{array}{|c|c|c|} {\text { Column I }} & {\text { Column II }} \\ \hline \text { i. } \begin{array}{l} \text { \(F\)} \\ \end{array} & \begin{array}{c} \text { a. } \\ \text { \(\left(m_{1}+m_{2}\right) g\) } \end{array} \\ \hline \text { ii. } \begin{array}{l} \text { Force acting on \(m_{2}\) is } \end{array} & \text { b. } \begin{array}{l} \text {\(m_{2} g \sec \theta\) } \\ \end{array} \\ \hline \text { iii. } \begin{array}{l} \text {Tension in the string is} \\ \end{array} & \text { c. } \begin{array}{l} \text {\(m_{2} \frac{F}{m_{1}+m_{2}}\) } \\ \end{array} \\ \hline \text { iv. } \begin{array}{l} \text { Force acting on \(m_{1}\) by the wire is} \\ \end{array} & \text { d. } \begin{array}{l} \text { \(\left(m_{1}+m_{2}\right) g \tan \theta\)} \end{array} \\ \hline \end{array} $$

A block of weight \(9.8 \mathrm{~N}\) is placed on a table. The table surface exerts an upward force of \(10 \mathrm{~N}\) on the block. Assume \(g=9.8 \mathrm{~m} / \mathrm{s}^{2} .\) Select the correct statement(s). (1) The block exerts a force of \(10 \mathrm{~N}\) on the table (2) The block exerts a force of \(19.8 \mathrm{~N}\) on the table (3) The block exerts a force of \(9.8 \mathrm{~N}\) on the table (4) The block has an upward acceleration

Inside a horizontally moving box, an experimenter finds that when an object is placed on a smooth horizontal table and is released, it moves with an acceleration of \(10 \mathrm{~ms}^{-2}\). In this box, if \(1-\mathrm{kg}\) body is suspended with a light string, the tension in the string in equilibrium position. (w.r.t. experimenter) will be (take \(g=10 \mathrm{~m} \mathrm{~s}^{-2}\) ) (1) \(10 \mathrm{~N}\) (2) \(10 \sqrt{2} \mathrm{~N}\) (3) \(20 \mathrm{~N}\) (4) Zero

Column I describes the motion of the object and one or more of the entries of Column II may be the cause of motions described in Column I. Match the entries of Column I with the entries of Column II. $$ \begin{array}{|c|c|c|} {\text { Column I }} & {\text { Column II }} \\ \hline \text { i. } \begin{array}{l} \text { An object is moving } \\ \text { towards east. } \end{array} & \begin{array}{c} \text { a. } \\ \text { The net force acting } \\ \text { on the object must } \\ \text { be towards east. } \end{array} \\ \hline \text { ii. } \begin{array}{l} \text { An object is moving } \\ \text { towards east with } \\ \text { constant acceleration. } \end{array} & \text { b. } \begin{array}{l} \text { At least one force } \\ \text { must act towards } \\ \text { east. } \end{array} \\ \hline \text { iii. } \begin{array}{l} \text { An object is moving } \\ \text { towards east with } \\ \text { varying acceleration. } \end{array} & \text { c. } \begin{array}{l} \text { No force may act } \\ \text { towards east. } \end{array} \\ \hline \text { iv. } \begin{array}{l} \text { An object is moving } \\ \text { towards east with } \\ \text { constant velocity. } \end{array} & \text { d. } \begin{array}{l} \text { No force may act } \\ \text { on the object. } \end{array} \\ \hline \end{array} $$

The upper surface of block \(C\) is horizontal and its right part is inclined to the horizontal at angle \(37^{\circ} .\) The mass of block \(A\) and \(B\) are \(m_{1}=1.4 \mathrm{~kg}\) and \(m_{2}=5.5 \mathrm{~kg}\), respectively. Neglet friction and mass of the pulley. Calculate acceleration with which block \(C\) should be moved to the right so that and \(B\) can remain stationary relative to it.

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