/*! 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 107 You are standing on a concrete s... [FREE SOLUTION] | 91Ó°ÊÓ

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You are standing on a concrete slab that in turn is resting on a frozen lake. Assume there is no friction between the slab and the ice. The slab has a weight five times your weight. If you begin walking forward at 2.00 \(\mathrm{m} / \mathrm{s}\) relative to the ice, with what speed, relative to the ice, does the slab move?

Short Answer

Expert verified
The slab moves at 0.40 m/s in the opposite direction.

Step by step solution

01

Understand Conservation of Momentum

The first principle to understand is the conservation of momentum, which states that in the absence of external forces, the total momentum of a system remains constant. In this problem, the total momentum before and after you start walking must be the same because there is no friction (no external horizontal force).
02

Set Up the Initial Momentum Equation

Initially, both you and the slab are at rest, so the total initial momentum of the system is zero:\[ p_{initial} = m imes 0 + M imes 0 = 0 \]where \( m \) is your mass and \( M \) is the slab's mass.
03

Set Up the Final Momentum Equation

When you walk forward on the slab, you have a velocity \( v = 2.00 \ \mathrm{m/s} \) relative to the ice. According to conservation of momentum:\[ p_{final} = m imes v - M imes V = 0 \]Here, \( V \) is the velocity of the slab relative to the ice. The negative sign accounts for the slab moving in the opposite direction.
04

Relate Slab's Mass to Your Mass

The problem states that the slab has a weight five times your weight. This means:\[ M = 5m \]
05

Solve for the Slab's Velocity

Substitute \( M = 5m \) into the final momentum equation:\[ m imes v - 5m imes V = 0 \]Simplifying gives:\[ m imes 2.00 - 5m imes V = 0 \]\[ 2.00m = 5mV \]\[ V = \frac{2.00}{5} \]\[ V = 0.40 \ \mathrm{m/s} \]
06

Re-examine Units and Direction

The velocity \( V = 0.40 \ \mathrm{m/s} \) is the speed of the slab relative to the ice and in the opposite direction of your motion. The units are consistent with the velocities used in the calculation.

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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 fundamental principle in physics, especially when dealing with motion and collisions. Momentum, represented by the symbol \( p \), is defined as the product of an object's mass \( (m) \) and its velocity \( (v) \), or mathematically, \( p = m \times v \). In this exercise, both you and the concrete slab initially have zero momentum because you are at rest on the ice. When you begin to walk, you change the system's state. To maintain balance, the momentum equation ensures that the total momentum before and after your movement will be the same. This is summarized by the equation:
  • Initial momentum: \( 0 \) (since neither you nor the slab were moving initially)
  • Final momentum: \( m \times v - M \times V = 0 \) (you move one way, the slab moves the other way)
This tells us that if you move, the slab must also move, but in the opposite direction, to maintain overall momentum. The negative sign in the equation indicates this directionality.
Frictionless Surface
A frictionless surface, like the frozen lake in this problem, is an idealized concept used to simplify calculations. It implies that there are no resistive forces acting against movement on this surface. Such a condition allows us to analyze motion without needing to account for energy losses due to friction or other external forces. In practical terms, this means the only forces we need to consider are those within the system itself. The absence of friction makes it possible to use the conservation of momentum directly, without adjustments for losses or thermal effects.
  • There are no external horizontal forces. This ensures momentum conservation.
  • Your motion is unrestricted, and the slab reacts purely based on your movement.
  • This simplifies the calculations, as we do not need to compensate for drag or other forces.
Conceptually, a frictionless surface is best imagined as a perfectly smooth, polished plane with zero roughness. In physics problems, it helps learners focus on core dynamics without added complexity.
Velocity Calculation
Calculating velocity on a frictionless surface involves understanding how movement affects related objects in a system. Here, your walking speed and the slab's resulting speed need to be analyzed together.Initially, your velocity was zero, but upon beginning to walk with a speed of \( 2.00 \ \mathrm{m/s} \), there is a shift. The slab must move in the opposite direction to conserve momentum. Using the relationships given:
  • Your momentum: \( m \times 2.00 \)
  • Slab's momentum: \(- M \times V \)
From the equation \( m \times 2.00 = 5m \times V \), we can solve for the slab's speed \( V \). After simplifying, \( V = \frac{2.00}{5} = 0.40 \ \mathrm{m/s} \). The calculation shows how the speed of an object can be derived when another object's speed and their mass ratios are known. This result confirms the slab moves at \( 0.40 \ \mathrm{m/s} \), opposite to the direction you're walking.

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Most popular questions from this chapter

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