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Chapter 1: Problem 48

Block \(A\) of mass \(M=2 \mathrm{~kg}\) is connected to another block \(B\) of mass \(m=1 \mathrm{~kg}\) with a string and a spring of force constant \(k=600 \mathrm{~N} / \mathrm{m}\) as shown in figure. Initially, spring is compressed to \(10 \mathrm{~cm}\) and whole system is moving on a smooth surface with a velocity \(v=1 \mathrm{~m} / \mathrm{s}\). At any time, thread is burnt, the velocity of block \(A\), when \(B\) is having maximum velocity w.r.t. ground, is (1) zero (2) \(1 \mathrm{~m} / \mathrm{s}\) (3) \(3 \mathrm{~m} / \mathrm{s}\) (4) none of these

Short Answer

Expert verified
The velocity of block A is approximately 0.27 m/s, which is not listed in the options.

Step by step solution

01

Understand the System

Initially, both blocks are moving together with a velocity of \( v = 1 \ \mathrm{m/s} \). The spring is initially compressed by \( x = 0.1 \ \mathrm{m} \). The problem asks for the velocity of block \( A \) when block \( B \) reaches its maximum velocity with respect to the ground after the thread is burnt.
02

Analyze Energy Conservation

After the thread is burnt, the spring forces separate the blocks. When block \( B \) is at maximum velocity, all the potential energy stored in the spring is converted into kinetic energy of block \( B \). The potential energy of the spring is \( U = \frac{1}{2}kx^2 \), where \( k = 600 \ \mathrm{N/m} \) and \( x = 0.1 \ \mathrm{m} \). Calculate \( U \).
03

Calculate Initial Spring Potential Energy

Calculate the initial potential energy stored in the spring using the formula for potential energy: \[ U = \frac{1}{2}kx^2 = \frac{1}{2} \times 600 \times (0.1)^2 = 3 \ \mathrm{J} \].
04

Apply Momentum Conservation

Initially, the momentum of the entire system is given by:\[ \text{Initial momentum} = (M + m)v = 3 \ \mathrm{kg} \cdot 1 \ \mathrm{m/s} = 3 \ \mathrm{kg \cdot m/s} \].After burning the thread, the final momentum will still equal the initial momentum, i.e.\[ Mv_A + mv_B = 3 \ \mathrm{kg \cdot m/s} \].
05

Calculate Maximum Velocity of Block B

The maximum kinetic energy of block \( B \) is equal to the potential energy stored in the spring:\[ \frac{1}{2}mv_B^2 = 3 \ \mathrm{J} \].Solving for \( v_B \):\[ v_B = \sqrt{\frac{2 \times 3}{1}} = \sqrt{6} \ \mathrm{m/s} \].
06

Find the Velocity of Block A

Using the conservation of momentum calculated previously \[ Mv_A + mv_B = 3 \].substitute \( v_B = \sqrt{6} \):\[ 2v_A + 1 \times \sqrt{6} = 3 \].Solve for \( v_A \):\[ v_A = \frac{3 - \sqrt{6}}{2} \approx 0.27 \ \mathrm{m/s} \].
07

Conclude Solution

From the above calculation, the velocity of block \( A \) when block \( B \) is at maximum velocity is approximately \( 0.27 \ \mathrm{m/s} \). This value is not listed among the multiple-choice options provided.

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Key Concepts

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

Hooke's Law
Hooke's Law is a fundamental principle in physics that relates the force needed to compress or stretch a spring to the distance it is compressed or stretched. The law states that the force exerted by the spring is directly proportional to the displacement, which in math terms is expressed as: \( F = -kx \).
Here:
  • \( F \) is the force exerted by the spring.
  • \( k \) is the spring constant, which measures the stiffness of the spring.
  • \( x \) is the displacement of the spring from its equilibrium position.
The negative sign indicates that the force exerted by the spring is in the opposite direction of the displacement. This means, if you compress the spring, the spring pushes back, and if you stretch it, the spring pulls back towards its equilibrium.
In the example problem, Hooke's Law helps in determining the force when the spring is compressed by 0.1 meters. The spring constant given is 600 N/m, implying the spring is fairly stiff.
Kinetic Energy
Kinetic energy is the energy that an object possesses due to its motion. It is one of the major types of energy forms and is crucial when analyzing systems in motion. The formula for calculating kinetic energy is:\( KE = \frac{1}{2} mv^2 \).
Here:
  • \( KE \) stands for kinetic energy.
  • \( m \) is the mass of the object.
  • \( v \) is the velocity of the object.
In the given exercise, when block \( B \) reaches its maximum velocity, all the stored potential energy from the spring is converted into kinetic energy for block \( B \).
The calculated kinetic energy (3 Joules in this case) originates from when the spring was initially compressed. Understanding this transformation is crucial for solving problems involving energy, as it links the potential energy stored in springs with the kinetic energy observed when released.
Spring Force
The spring force is the restoring force offered by a spring when it is compressed or stretched. It is what makes springs useful for storing and releasing energy. Mathematically, it aligns with Hooke鈥檚 Law, expressed as \( F = -kx \).
  • Springs are conservative forces, meaning the work done by them depends only on the initial and final positions, not on the path taken.
  • The work done by or against a spring is stored as potential energy, which can later be converted into other types of energy, such as kinetic.
In our dynamic system example, as soon as the thread breaks, the stored energy in the compressed spring acts as the force pushing block \( B \) away, allowing it to gain maximum velocity. This process showcases how spring force plays a significant role in converting stored potential energy into kinetic energy, driving systems from a state of rest or lower energy to higher kinetic states.

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Most popular questions from this chapter

A particle of mass \(4 m\) is projected from the ground at some angle with horizontal. Its horizontal range is \(R\). At the highest point of its path it breaks into two pieces of masses \(m\) and \(3 m\), respectively, such that the smaller mass comes to rest. The larger mass finally falls at a distance \(x\) from the point of projection, where \(x\) is equal to (1) \(\frac{2 R}{3}\) (2) \(\frac{7 R}{6}\) (3) \(\frac{5 R}{4}\) (4) none of these

A ring of mass \(M=90\) gram and radius \(R=3\) meter is kept on a frictionless horizontal surface such that its plane is parallel to horizontal plane. A particle of mass \(m=10\) gram is placed in contact with the inner surface of ring as shown figure. An initial velocity \(v=\sqrt{2} \mathrm{~m} / \mathrm{s}\) is given to the particle along the tangent of the ring. Find the magnitude of the force of interaction (in milli-newton) between the particle and ring.

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) \(5 \sqrt{2} \mathrm{~ms}^{-1}\)

Figure shows a hollow cube of side ' \(a\) ' and volume ' \(V\). There is a small chamber of volume \(V / 4\) in the cube as shown. The chamber is completely filled by \(m \mathrm{~kg}\) of water. Water leaks through a hole \(H\) and spreads in the whole cube. Then the work done by gravity in this process assuming that the complete water finally lies at the bottom of the cube is (1) \(\frac{1}{2} m g a\) (2) \(\frac{3}{8} m g a\) (3) \(\frac{5}{8} m g a\) (4) \(\frac{1}{8} m g a\)

A man is standing on a plank which is placed on smooth horizontal surface. There is sufficient friction between feet of man and plank. Now man starts running over plank, correct statement is/are (1) Work done by friction on man with respect to ground is negative. (2) Work done by friction on man with respect to ground is positive (3) Work done by friction on plank with respect to ground is positive. (4) Work done by friction on man with respect to plank is zero.
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