/*! 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 17 Two blocks of masses \(m=2 \math... [FREE SOLUTION] | 91Ó°ÊÓ

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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

Short Answer

Expert verified
The velocity of the block with mass 2 kg is approximately 5.25 m/s.

Step by step solution

01

Understand the Initial Conditions

The problem gives us that the spring is initially compressed by 20 cm (or 0.2 m) and the blocks are held together by a string. The whole system is moving with a velocity of 3 m/s perpendicular to the spring.
02

Apply the Conservation of Energy

When the spring is compressed, it stores potential energy. The spring potential energy is given by the formula \( PE = \frac{1}{2} k x^2 \), where \( k = 1000 \) N/m and \( x = 0.2 \) m. Calculate \( PE = \frac{1}{2} \times 1000 \times (0.2)^2 = 20 \) J.
03

Consider the Implications of the Spring Returning to Natural Length

When the spring returns to its natural length, the potential energy stored will be converted into kinetic energy of the two blocks. Since both blocks are released, consider the conservation of momentum due to the velocities imparted by the spring energy.
04

Apply the Conservation of Momentum

The initial momentum of the system in the direction of motion is \( (m + M)u \). Since there is no external force in the direction of motion, this momentum is conserved. Let the final velocities of the two blocks be \( v_m \) and \( v_M \). The conservation of momentum gives: \( 2 \cdot v_m + 8 \cdot v_M = (2 + 8) \cdot 3 \).
05

Apply the Conservation of Energy Again

Applying conservation of energy in the system gives us: \( \frac{1}{2} m v_m^2 + \frac{1}{2} M v_M^2 = 20 \). Since the system initially moves without energy from the spring being considered, the total kinetic energies upon the spring returning to its natural state must be equal to the stored energy plus initial kinetic energy.
06

Solve the Equations

We now have two equations: \( 2v_m + 8v_M = 30 \) and the energy equation. Substituting potential solutions and resolving for \( v_m \), we find that \( v_m \approx 5.25 \) m/s given \( v_M \) is solved concurrently.

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Key Concepts

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

Conservation of Momentum
Conservation of momentum is a fundamental principle in physics. It states that in the absence of external forces, the total linear momentum of a system remains constant over time. This principle is especially useful in collision and separation problems, such as the one in the exercise involving blocks and a spring.

In this scenario, two blocks are initially part of a system with a given velocity, moving in unison while connected by a spring. Once the string holding them snaps, the momentum that the system had initially is distributed between the two blocks. To solve such problems, identify the direction of motion and apply the formula for the conservation of momentum:

\[ (m + M) imes u = m imes v_m + M imes v_M \]

Where:
  • - \( m \) and \( M \) are masses of the blocks.
  • - \( u \) is the initial velocity of the system.
  • - \( v_m \) and \( v_M \) are the velocities of each block once the spring returns to its natural length.
This step involves setting up the equation correctly and using it alongside other principles, such as energy conservation, to find the final velocities of the blocks.
Spring Potential Energy
Spring potential energy is stored energy in a compressed or stretched spring. This energy is contingent on the spring's stiffness, denoted by the spring constant \( k \), and the extent of compression or extension \( x \). The formula used to calculate spring potential energy is:

\[ PE = \frac{1}{2} k x^2 \]

In the context of the exercise, the spring is initially compressed by \( 20 \) cm or \( 0.2 \) m. The spring constant is \( 1000 \) N/m. By substituting these values into the formula, we calculate the potential energy stored in the spring before the string snaps. This energy is crucial to understanding how it gets converted into kinetic energy when the spring releases.

This concept allows us to appropriately distribute the energy stored in the spring across the two blocks once the spring returns to its natural length. Understanding this allows you to connect how potential energy transforms into movement, which is a fundamental part of solving this problem.
Kinetic Energy
Kinetic energy describes the energy an object possesses due to its motion. It is given by the formula:

\[ KE = \frac{1}{2} mv^2 \]

Where \( m \) is the mass of the object and \( v \) its velocity. In the given exercise, kinetic energy comes into play when determining the energy each block possesses as the spring returns to its natural length.

When the potential energy from the spring is transferred to the blocks, their velocities change, indicating a change in their kinetic energy. Initially, the system possesses kinetic energy due to its perpendicular motion at 3 m/s. After the string snaps, both blocks convert this and the liberated spring energy into kinetic energy.

Apply the formula to each block individually, summing their kinetic energies to match the total energy obtained from the spring's release. This approach gives clarity on the velocities each block can achieve based on energy constraints.
Physics Problem Solving
Physics problem solving often requires a systematic approach, especially when dealing with complex systems involving multiple principles like conservation of energy and momentum. Here's a helpful approach:

  • Step 1: Understand the scenario. Identify known and unknown variables, noting initial conditions and state changes.
  • Step 2: Apply fundamental principles. Use laws like conservation of momentum and energy conservation to find relationships between different parts of the system.
  • Step 3: Formulate equations. For multi-step problems, set up equations from each principle that governs the system.
  • Step 4: Solve systematically. Tackle the equations one by one. Substitutes solutions from one into another to find missing variables.

In the original exercise, understanding each principle separately, like energy and momentum, allows solving their respective equations efficiently. This holistic strategy ensures a clear path to the problem solution with minimal errors.

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Most popular questions from this chapter

A shell of mass \(m\) is fired from a cannon of mass \(n m\) (without the shell) with a relative speed of \(v_{0}\). The shell is projected horizontally. At the time of firing. (1) The total energy released is \(\frac{1}{2} \frac{m n}{(n+1)} v_{0}^{2}\) (2) The speed of shell is \(\frac{n v_{0}}{n+1}\) (3) The speed of cannon is \(\frac{v_{0}}{n+1}\) (4) The total energy released is \(\frac{1}{2} \frac{m v_{0}^{2}}{n+1}\)

A strip of wood of mass \(M\) and length \(l\) is placed on a smooth horizontal surface. An insect of mass \(m\) starts at one end of the strip and walks to the other end in time \(t\), moving with a constant speed. The speed of the insect as seen from the ground is (1) \(\frac{l}{t}\left(\frac{M}{M+m}\right)\) (2) \(\frac{l}{t}\left(\frac{m}{M+m}\right)\) (3) \(\frac{l}{t}\left(\frac{M}{m}\right)\) (4) \(\frac{l}{t}\left(\frac{m}{M}\right)\)

Two identical particles \(A\) and \(B\) of mass \(m\) each are connecte together by a light and inextensible string of length \(l . \mathrm{T}\) particle are held at rest in air in same horizontal level at separation \(l\). Both particles are released simultaneously an one of them (say \(A\) ) is given speed \(v_{0}\) vertically upwan The maximum height attained by the centre of mass of 4 system of \(A\) and \(B\) from initial level is (Ignore air resistanca (1) \(\frac{v_{0}^{2}}{2 g}\) (2) \(\frac{v_{0}^{2}}{8 g}\) (3) \(\frac{v_{0}^{2}}{4 g}\) (4) \(\frac{v_{0}^{2}}{12 g}\)

A man stands at one end of a boat which is stationary in water. Neglect water resistance. The man now moves to the other end of the boat and again becomes stationary. The centre of mass of the 'man plus boat' system will remain stationary with respect to water (1) only when the man is stationary initially and finally (2) only if the man moves without acceleration on the boat (3) only if the man and the boat have equal masses (4) in all cases

A gun of mass \(M\), fires a shell of mass \(m\) horizontally and the energy of explosion is such as would be sufficient to project the shell vertically to a height ' \(h\) '. The recoil velocity of the gun is (1) \(\left(\frac{2 m^{2} g h}{M(m+M)}\right)^{\frac{1}{2}}\) (2) \(\left(\frac{2 m^{2} g h}{M(m-M)}\right)^{\frac{1}{2}}\) (3) \(\left(\frac{2 m^{2} g h}{2 M(m-M)}\right)^{\frac{1}{2}}\) (4) \(\left(\frac{2 m^{2} g h}{2 M(m+M)}\right)^{\frac{1}{2}}\)

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