/*! 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 31 Two blocks of masses \(5 \mathrm... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 1: Problem 31

Two blocks of masses \(5 \mathrm{~kg}\) and \(2 \mathrm{~kg}\) are placed on a frictionless surface and connected by a spring. An external kick gives a velocity of \(14 \mathrm{~m} / \mathrm{s}\) to the heavier block in the direction of lighter one. The magnitudes of velocities of two blocks in the centre of mass frame after the kick are, respectively, (1) \(4 \mathrm{~m} / \mathrm{s}, 4 \mathrm{~m} / \mathrm{s}\) (2) \(10 \mathrm{~m} / \mathrm{s}, 4 \mathrm{~m} / \mathrm{s}\) (3) \(4 \mathrm{~m} / \mathrm{s}, 10 \mathrm{~m} / \mathrm{s}\) (4) \(10 \mathrm{~m} / \mathrm{s}, 10 \mathrm{~m} / \mathrm{s}\)

Short Answer

Expert verified
The velocities in the center of mass frame are 4 m/s and 10 m/s.

Step by step solution

01

Determine the Initial Momentum

The initial momentum is given by the momentum of the heavier block, since only it receives a velocity. Assuming the lighter block starts at rest, the momentum of the system can be written as: \[ P_{ ext{initial}} = m_1 imes v_1 = 5 imes 14 = 70 \text{ kg m/s} \] where \( m_1 = 5 \text{ kg} \) and \( v_1 = 14 \text{ m/s} \) are the mass and velocity of the heavier block, respectively.
02

Calculate the Velocity of the Center of Mass

The velocity of the center of mass \( V_{cm} \) is found using the total mass of the system and its total momentum:\[ V_{cm} = \frac{P_{ ext{initial}}}{m_1 + m_2} = \frac{70}{5 + 2} = \frac{70}{7} = 10 \text{ m/s} \] This is the velocity with which the center of mass of the system is moving.
03

Calculate Velocities in the Center of Mass Frame

To find the velocities of each block in the center of mass frame, subtract the center of mass velocity from the individual velocities. For the heavier block, the velocity in the center of mass frame, \( v_{1_{cm}} \), is:\[ v_{1_{cm}} = v_1 - V_{cm} = 14 - 10 = 4 \text{ m/s} \] For the lighter block, since it was initially at rest, \( v_{2} = 0 \):\[ v_{2_{cm}} = v_2 - V_{cm} = 0 - 10 = -10 \text{ m/s} \] The negative sign indicates that it moves in the opposite direction to the center of mass.

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.

Momentum Conservation
Momentum conservation is a fundamental principle in physics, stating that in the absence of external forces, the total linear momentum of a closed system remains constant. In our scenario, where two blocks of mass are placed on a frictionless surface, this principle is crucial.
- Initially, only the heavier block receives a velocity due to an external kick, imparting momentum to the system. - The lighter block doesn't move at first, so it doesn't contribute to the initial momentum. Therefore, the initial momentum of the system is solely determined by the momentum of the heavier block. This can be expressed mathematically as the product of the heavier block's mass and its velocity.
By employing momentum conservation, once the external force (the kick) has acted, the total momentum of both blocks together still equals the initial momentum. This allows the distribution of momentum within the two-blocks system, giving us further clues to how each block will move.
Velocity Calculation
Calculating velocities is a key step in understanding how objects in motion interact, particularly in a system like ours consisting of two connected blocks. Here are some points essential to this process:
  • First, calculate the velocity of the system's center of mass, which includes both blocks. This velocity tells us how the entire system moves through space.
  • The formula to calculate the center of mass velocity, \(V_{cm}\), is derived from dividing the total momentum of the system by the combined mass of both blocks.
In our problem, with an initial momentum calculated as \(70 \text{ kg m/s}\) and masses of \(m_1 = 5 \text{ kg}\) and \(m_2 = 2 \text{ kg}\), the center of mass velocity turns out to be \(10 \text{ m/s}\).
Next, determine the individual velocities of the blocks relative to the center of mass frame. By subtracting the center of mass velocity from each block's velocity, we discover how each block behaves independently.
Inertial Frame of Reference
An inertial frame of reference is a perspective in which an observer sees objects moving at a constant velocity unless acted upon by a force. It is crucial for analyzing motion in terms of Newton's laws.
  • In our exercise, the center of mass frame acts as an inertial frame of reference. By observing the blocks from this frame, we understand their true relative motion ignoring the overall movement of the system.
  • The velocity calculations of each block in this frame help simplify the problem by focusing solely on internal motions, unaffected by external forces.
Using the inertial frame, we've calculated the following:
- For the heavier block, its velocity in the center of mass frame simplifies to \(4 \text{ m/s}\).- For the lighter block, after subtracting the center of mass velocity, we obtain \(-10 \text{ m/s}\) indicating movement opposite to the system's overall direction.
By applying an inertial frame approach, such simplified analysis lets us comprehend the dynamics between the blocks clearly.

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 ball of mass \(m\) is projected with a speed \(v\) into the barrel of a spring gun of mass \(M\) initially at rest lying on a frictionless surface. The mass sticks in the barrel at the point of maximum compression in the spring. The fraction of kinetic energy of the ball stored in the spring is (1) \(\frac{m}{M}\) (2) \(\frac{M}{m+M}\) (3) \(\frac{m}{M+M}\) (4) none of these

A block ' \(A\) ' of mass \(m_{1}\) hits horizontally the rear side of a spring (ideal) attached to a block \(B\) of mass \(m_{2}\) resting on a smooth horizontal surface. After hitting, ' \(A\) ' gets attached to the spring. Some statements are given at any moment of time: i. If velocity of \(A\) is greater than \(B\), then kinetic energy of the system will be decreasing. ii. If velocity of \(A\) is greater than \(B\), then kinetic energy of the system will be increasing. iii. If velocity of \(A\) is greater than \(B\), then momentum of the system will be decreasing. iv. If velocity of \(A\) is greater than \(B\), then momentum of the system will be increasing. Now select correct alternative: (1) only iv (2) only i (3) ii and iv (4) \(\mathrm{i}\) and iii

A body of mass \(1 \mathrm{~kg}\) initially at rest, explodes and breaks into three fragments of masses in the ratio \(1: 1: 3\). The two pieces of equal mass fly off perpendicular to each other with a speed of \(15 \mathrm{~ms}^{-1}\) each. What is the velocity of the heavier fragment? (1) \(10 \sqrt{2} \mathrm{~m} \mathrm{~s}^{-1}\) (2) \(5 \sqrt{3} \mathrm{~ms}^{-1}\) (3) \(10 \sqrt{3} \mathrm{~ms}^{-1}\) (4) \(5 \sqrt{2} \mathrm{~ms}^{-1}\)

Two blocks of masses \(m=2 \mathrm{~kg}\) and \(M=8 \mathrm{~kg}\) are connectes to a spring of force constant \(k=1 \mathrm{kN} / \mathrm{m} .\) The spring is compressed by \(20 \mathrm{~cm}\) and the two blocks are held in thin position by a string. The system is placed on a horizonta smooth surface and given a velocity \(u=3 \mathrm{~m} / \mathrm{s}\) perpendicula to the spring. The string snaps while moving. The speed (i \(\mathrm{m} / \mathrm{s}\) ) of the block of mass \(m\) when the spring regains in natural length is

Two persons \(\mathrm{A}\) and \(\mathrm{B}\) of weight \(80 \mathrm{~kg}\) and \(50 \mathrm{~kg}\) respectivel are standing at opposite ends of a boat of mass \(70 \mathrm{~kg}\) a length \(10 \mathrm{~m}\) at rest. When they interchange their position then displacement of the boat will be (1) \(1.5 \mathrm{~m}\) towards right (2) \(1.0 \mathrm{~m}\) towards right (3) \(1.5 \mathrm{~m}\) towards left (4) stationary
See all solutions

Recommended explanations on Physics Textbooks

Particle Model of Matter

Read Explanation

Scientific Method Physics

Read Explanation

Engineering Physics

Read Explanation

Space Physics

Read Explanation

Further Mechanics and Thermal Physics

Read Explanation

Measurements

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.