/*! 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 66 Block 1, with mass \(m_{1}\) and... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 9: Problem 66

Block 1, with mass \(m_{1}\) and speed \(4.0 \mathrm{~m} / \mathrm{s}\), slides along an \(x\) axis on a frictionless floor and then undergoes a one-dimensional elastic collision with stationary block 2, with mass \(m_{2}=0.40 m_{1} .\) The two blocks then slide into a region where the coefficient of kinetic friction is \(0.50\); there they stop. How far into that region do (a) block 1 and \((b)\) block 2 slide?

Short Answer

Expert verified
Block 1 slides 0.298 m and block 2 slides 1.200 m.

Step by step solution

01

Conservation of Momentum

In an elastic collision, both momentum and kinetic energy are conserved. Apply the law of conservation of momentum to find the velocities post-collision. Initially, only block 1 moves, so total initial momentum is given by:\[ p_{initial} = m_1 \cdot v_1 = m_1 \cdot 4.0 \, \text{m/s} \]Post-collision, the equation for momentum conservation is:\[ m_1 \cdot v_{1}' + m_2 \cdot v_{2}' = m_1 \cdot 4.0 \, \text{m/s} \]where \(v_{1}'\) and \(v_{2}'\) are the velocities of block 1 and block 2 after the collision, respectively.
02

Conservation of Kinetic Energy

Since the collision is elastic, kinetic energy is also conserved:\[ \frac{1}{2}m_1(4)^2 = \frac{1}{2}m_1(v_{1}')^2 + \frac{1}{2}m_2(v_{2}')^2 \]Solving these equations simultaneously will give the velocities of both blocks after the collision.
03

Solve for Post-Collision Velocities

Using the two conservation equations, solve for \(v_1'\) and \(v_2'\):From momentum conservation:\[ m_1 \cdot v_1 + 0 = m_1 \cdot v_{1}' + 0.4m_1 \cdot v_{2}' \]From energy conservation:\[ 16m_1 = (v_{1}')^2 \cdot m_1 + 0.4(v_{2}')^2 \cdot m_1 \]Solutions:\[ v_1' = 1.71 \, \text{m/s}, \quad v_2' = 3.43 \, \text{m/s} \]
04

Calculate Deceleration Due to Friction

Apply the formula for friction force to get deceleration. The frictional force \(f_k\) is:\[ f_k = \mu_k \cdot m \cdot g \]Deceleration \(a\) due to friction is:\[ a = \frac{f_k}{m} = \mu_k \cdot g = 0.50 \cdot 9.8 \, \text{m/s}^2 = 4.9 \, \text{m/s}^2 \]
05

Calculate Distance Slid by Each Block

Use the kinematic equation to find the distance stopped due to the frictional force.For block 1:\[ v_f^2 = v_{1}'^2 - 2a d_1 \]Solving for \(d_1\):\[ 0 = (1.71)^2 - 2 \cdot 4.9 \cdot d_1 \Rightarrow d_1 = \frac{(1.71)^2}{9.8} \approx 0.298 \, \text{m} \]For block 2:\[ v_f^2 = v_{2}'^2 - 2a d_2 \]Solving for \(d_2\):\[ 0 = (3.43)^2 - 2 \cdot 4.9 \cdot d_2 \Rightarrow d_2 = \frac{(3.43)^2}{9.8} \approx 1.200 \, \text{m} \]

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.

Conservation of Momentum
Momentum is the product of an object's mass and its velocity. The principle of conservation of momentum states that the total momentum of a closed system remains constant if it is not affected by external forces. In the context of our elastic collision problem,
  • Initially, only Block 1 is moving, so its momentum is calculated as the product of its mass and velocity, i.e., \( m_1 \times 4.0 \, \text{m/s} \).
  • During the collision, the total momentum before and after must be equal, allowing us to write the conservation equation: \[ m_1 \cdot v_{1}' + m_2 \cdot v_{2}' = m_1 \times 4.0 \, \text{m/s} \]
This equation is crucial in determining the velocities of both blocks after the collision. Since no external force acts in the horizontal direction, momentum is conserved.
Conservation of Kinetic Energy
Kinetic energy is the energy of motion, given by the formula \( \frac{1}{2} m v^2 \). In elastic collisions, not only is momentum conserved, but kinetic energy is also conserved. This means the total kinetic energy before and after the collision remains the same.
  • The initial kinetic energy of Block 1 is \( \frac{1}{2}m_1(4)^2 \).
  • After the collision, the kinetic energy is distributed between the two blocks, expressed as: \[ \frac{1}{2}m_1(v_{1}')^2 + \frac{1}{2}m_2(v_{2}')^2 \]
The conservation of both momentum and kinetic energy provides two equations that can be solved simultaneously to find the velocities \(v_1'\) and \(v_2'\) after collision.
Frictional Force
Friction is the resistive force acting between surfaces in contact, opposing the relative motion or impending motion between them. Here, when the blocks enter a region with friction, they experience a frictional force that brings them to a stop. The kinetic frictional force \(f_k\) is computed using the formula:
  • \( f_k = \mu_k \cdot m \cdot g \)
  • Where \( \mu_k = 0.50 \) and \( g \) is the acceleration due to gravity, \( 9.8 \, \text{m/s}^2 \).
This frictional force causes a uniform deceleration, calculated using:\[ a = \frac{f_k}{m} = \mu_k \cdot g \]resulting in \( a = 4.9 \, \text{m/s}^2 \). This deceleration is then used to find out how far each block slides into the frictional region.
Kinematic Equations
Kinematics deals with the motion of objects without considering the forces that cause the motion. To find out how far the blocks slide before stopping, we use kinematic equations, specifically one that relates velocity, acceleration, and distance:
  • \(v_f^2 = v_i^2 - 2ad \) where \( v_f \) (final velocity) is zero because the blocks stop, \( v_i \) is the initial velocity (either \(v_1'\) or \(v_2'\)), \( a \) is the deceleration due to friction, and \( d \) is the distance slid.
This equation simplifies to solve for \( d \), yielding separate distances for block 1 as \( d_1 \approx 0.298 \, \text{m} \) and block 2 as \( d_2 \approx 1.200 \, \text{m} \). This showcases how both the initial velocities and the frictional force determine the stopping distances.

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 vessel at rest at the origin of an \(x y\) coordinate system explodes into three pieces. Just after the explosion, one piece, of mass \(m\), moves with velocity \((-30 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}\) and a second piece, also of mass \(m\), moves with velocity \((-30 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}} .\) The third piece has mass \(3 \mathrm{~m} .\) Just after the explosion, what are the (a) magnitude and (b) direction of the velocity of the third piece?

The last stage of a rocket, which is traveling at a speed of \(7600 \mathrm{~m} / \mathrm{s}\), consists of two parts that are clamped together: a rocket case with a mass of \(290.0 \mathrm{~kg}\) and a payload capsule with a mass of \(150.0 \mathrm{~kg}\). When the clamp is released, a compressed spring causes the two parts to separate with a relative speed of \(910.0 \mathrm{~m} / \mathrm{s}\). What are the speeds of (a) the rocket case and (b) the payload after they have separated? Assume that all velocities are along the same line. Find the total kinetic energy of the two parts (c) before and (d) after they separate. (e) Account for the difference.

Two blocks of masses \(1.0 \mathrm{~kg}\) and \(3.0 \mathrm{~kg}\) are connected by a spring and rest on a frictionless surface. They are given velocities toward each other such that the \(1.0 \mathrm{~kg}\) block travels initially at \(1.7 \mathrm{~m} / \mathrm{s}\) toward the center of mass, which remains at rest. What is the initial speed of the other block?

At time \(t=0\), force \(\vec{F}_{1}=(-4.00 \hat{\mathrm{i}}+5.00 \hat{\mathrm{j}}) \mathrm{N}\) acts on an initially stationary particle of mass \(2.00 \times 10^{-3} \mathrm{~kg}\) and force \(\vec{F}_{2}=(2.00 \hat{i}-4.00 \mathrm{j}) \mathrm{N}\) acts on an initially stationary particle of mass \(4.00 \times 10^{-3} \mathrm{~kg}\). From time \(t=0\) to \(t=2.00 \mathrm{~ms}\), what are the (a) magnitude and (b) angle (relative to the positive direction of the \(x\) axis) of the displacement of the center of mass of the twoparticle system? (c) What is the kinetic energy of the center of mass at \(t=2.00 \mathrm{~ms}\) ?

A big olive \((m=0.50 \mathrm{~kg})\) lies at the origin of an \(x y\) coordinate system, and a big Brazil nut \((M=1.5 \mathrm{~kg})\) lies at the point \((1.0,2.0) \mathrm{m} .\) At \(t=0\), a force \(\vec{F}_{o}=(2.0 \hat{\mathrm{i}}+3.0 \mathrm{j}) \mathrm{N}\) begins to act on the olive, and a force \(\vec{F}_{n}=(-3.0 \hat{\mathrm{i}}-2.0 \hat{\mathrm{j}}) \mathrm{N}\) begins to act on the nut. In unit-vector notation, what is the displacement of the center of mass of the olive-nut system at \(t=4.0 \mathrm{~s}\), with respect to its position at \(t=0 ?\)
See all solutions

Recommended explanations on Physics Textbooks

Classical Mechanics

Read Explanation

Physical Quantities And Units

Read Explanation

Fundamentals of Physics

Read Explanation

Conservation of Energy and Momentum

Read Explanation

Engineering Physics

Read Explanation

Electricity

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.