/*! 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 50 Two blocks \(A\) and \(B\) of ma... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 5: Problem 50

Two blocks \(A\) and \(B\) of mass \(m\) and \(2 \mathrm{~m}\) respectively are connected by a massless spring of force constant \(k\). They are placed on a smooth horizontal plane. They are stretched by an amount \(x\) and then released. The relative velocity of the blocks when the spring comes to its natural length is (A) \(x \sqrt{\frac{3 k}{2 m}}\) (B) \(x \sqrt{\frac{2 k}{3 m}}\) (C) \(x \sqrt{\frac{2 k}{m}}\) (D) \(x \sqrt{\frac{3 k}{m}}\)

Short Answer

Expert verified
The relative velocity of the blocks when the spring comes to its natural length is \(v_{rel} = x \sqrt{\frac{2 k}{3 m}}\).

Step by step solution

01

The potential energy stored in the spring when it is stretched by distance x is given by the formula: \[U = \frac{1}{2} k x^2\] #Step 2: Calculate the total initial energy#

At the initial position, both blocks are held at rest. So, the initial kinetic energy is zero. Thus, the total initial mechanical energy (E_initial) is equal to the potential energy stored in the spring. \[E_{initial} = U = \frac{1}{2} k x^2\] #Step 3: Calculate the final kinetic energies of both blocks when the spring comes to its natural length#
02

Let's denote the final velocities of block A and block B as \(v_A\) and \(v_B\). When the spring comes to its natural length, the potential energy stored in the spring becomes zero. Therefore, the total final mechanical energy (E_final) is equal to the sum of the kinetic energies of blocks A and B, given by the formula: \[E_{final} = \frac{1}{2} m v_A^2 + \frac{1}{2} (2m) v_B^2\] #Step 4: Apply the conservation of mechanical energy#

According to the conservation of mechanical energy: \[E_{initial} = E_{final}\] \[\frac{1}{2} k x^2 = \frac{1}{2} m v_A^2 + m v_B^2\] #Step 5: Find the relative velocity of the blocks#
03

The relative velocity of block A with respect to block B is denoted as \(v_{rel}\) and given by: \[v_{rel} = v_A - v_B\] We have to find the value of \(v_{rel}\) given the condition \(\frac{1}{2} k x^2 = \frac{1}{2} m v_A^2 + m v_B^2\). #Step 6: Use the conservation of momentum to relate the velocities#

Because there is no external force acting on the system, the total momentum should be conserved. The blocks initially are at rest, so their initial momentum is zero, and their final momentum should also be zero: \[m(v_A - v_B) = 0\] \[v_A = v_B\] #Step 7: Substitute the relation between velocities in the energy conservation equation#
04

Now we rewrite the energy conservation equation using the relation found in step 6: \[\frac{1}{2} k x^2 = \frac{1}{2} m v_A^2 + m (v_A)^2\] #Step 8: Solve the equation for the relative velocity#

Simplify the equation to find \(v_{rel}\): \[v_A = v_{rel} = \sqrt{\frac{k x^2}{3m}}\] And the relative velocity of the blocks when the spring comes to its natural length is \[v_{rel} = x \sqrt{\frac{2 k}{3 m}}\] The correct answer is option (B).

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!

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

In the Fig. \(5.35\) shown, a spring mass system is placed on a horizontal smooth surface in between two vertical rigid walls \(W_{1}\) and \(W_{2}\). One end of spring is fixed with wall \(W_{1}\) and other end is attached with mass \(m\) which is free to move. Initially, spring is tension free and having natural length \(l_{\mathrm{o}}\). Mass \(m\) is compressed through distance \(a\) and released. Taking the collision between wall \(W_{2}\) and mass \(m\) as elastic and \(K\) as spring constant, the average force exerted by mass \(m\) on wall \(W_{2}\) is (A) \(\frac{2 a K}{\pi}\) (B) \(\frac{a K}{\pi}\) (C) \(\frac{a K}{2 \pi}\) (D) \(\frac{2 a K}{3 \pi}\)

A moving body collides elastically with another body of equal mass at rest. Then (A) a part of the energy is dissipated as heat. (B) momentum is conserved but \(\mathrm{KE}\) is not conserved. (C) both masses start moving with the same velocity. (D) the moving mass transfers whole of its energy to the mass at rest.

Position of two particles are given by \(x_{1}=2 t\) and \(x_{2}=2+3 t\). The velocity of centre of mass at \(t=2 \mathrm{~s}\) is \(2 \mathrm{~m} / \mathrm{s}\). Velocity of centre of mass at \(t=4 \mathrm{~s}\) will be (A) \(2 \mathrm{~m} / \mathrm{s}\) (B) \(4 \mathrm{~m} / \mathrm{s}\) (C) \(1 \mathrm{~m} / \mathrm{s}\) (D) Zero

A particle moves in the \(x-y\) plane under the action of force \(\vec{F}\) such that the value of its linear momentum \(\vec{P}\) at time \(t\) is \(P_{x}=2 \cos t\) and \(P_{y}=2 \sin t .\) The angle \(\theta\) between \(\vec{F}\) and \(\vec{P}\) at time \(t\) will be (A) \(90^{\circ}\) (B) \(0^{\circ}\) (C) \(180^{\circ}\) (D) \(30^{\circ}\)

A body moving towards a finite body at rest collides with it. It is possible that (A) both the bodies come to rest. (B) both the bodies move after collision. (C) the moving body comes to rest and the stationary body starts moving. (D) the stationary body remains stationary; the moving body changes its velocity.
See all solutions

Recommended explanations on Physics Textbooks

Electricity and Magnetism

Read Explanation

Linear Momentum

Read Explanation

Magnetism

Read Explanation

Atoms and Radioactivity

Read Explanation

Force

Read Explanation

Electromagnetism

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.