/*! 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 104 A massless spring with spring co... [FREE SOLUTION] | 91Ó°ÊÓ

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A massless spring with spring constant \(k\) is placed between a block of mass \(m\) and a block of mass \(3 m .\) Initially the blocks are at rest on a frictionless surface and they are held together so that the spring between them is compressed by an amount \(D\) from its equilibrium length. The blocks are then released and the spring pushes them off in opposite directions. Find the speeds of the two blocks when they detach from the spring.

Short Answer

Expert verified
The speed of the block with mass \(m\) is \(v_1 = \sqrt{\frac{kD^2}{4m}}\) and the speed of the block with mass \(3m\) is \(v_2 = \sqrt{\frac{kD^2}{12m}}\).

Step by step solution

01

Understand the Problem

We are given a system comprising two blocks and a spring between them. The blocks are initially at rest, and the spring is compressed by a displacement \(D\). When released, the spring will exert equal and opposite forces on the two blocks, pushing them apart along a frictionless surface. We need to determine the speeds of the two blocks after they detach from the spring.
02

Apply Conservation of Mechanical Energy

Initially, all mechanical energy is stored as potential energy in the compressed spring. The potential energy of a spring is given by \(PE = \frac{1}{2}kD^2\). When the blocks are released and detach from the spring, all this potential energy is converted into the kinetic energy of the two blocks together. Thus, the total energy conservation equation is: \(\frac{1}{2}kD^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}(3m)v_2^2\), where \(v_1\) and \(v_2\) are the velocities of block \(m\) and block \(3m\) respectively.
03

Apply Conservation of Momentum

Since there are no external forces, the conservation of momentum can be applied. Initially, the total momentum is zero because the blocks are at rest. Therefore, after release, the sum of the momenta of the blocks must also be zero: \(mv_1 = 3mv_2\). From this equation, we can express \(v_1\) in terms of \(v_2\): \(v_1 = 3v_2\).
04

Solve for \(v_2\)

Substitute \(v_1 = 3v_2\) into the energy equation: \(\frac{1}{2}kD^2 = \frac{1}{2}m(3v_2)^2 + \frac{1}{2}(3m)v_2^2\). Simplify this to get: \(\frac{1}{2}kD^2 = \frac{9}{2}mv_2^2 + \frac{3}{2}mv_2^2 = 6mv_2^2\). Solving for \(v_2\), we have: \(v_2 = \sqrt{\frac{kD^2}{12m}}\).
05

Solve for \(v_1\)

Now substitute \(v_2\) into the expression \(v_1 = 3v_2\). Thus, \(v_1 = 3\sqrt{\frac{kD^2}{12m}}\). Further simplifying, we get \(v_1 = \sqrt{\frac{kD^2}{4m}} = \sqrt{\frac{kD^2}{12m}} \times 3\). Therefore, \(v_1 = \sqrt{\frac{9kD^2}{4m}} = 3 \sqrt{\frac{kD^2}{12m}}\).
06

Conclusion

The speeds of the blocks when they detach are \(v_1 = \sqrt{\frac{kD^2}{4m}}\) for the block of mass \(m\) and \(v_2 = \sqrt{\frac{kD^2}{12m}}\) for the block of mass \(3m\). This ensures that the kinetic energy and momentum are conserved in the system as the spring pushes the blocks apart.

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Key Concepts

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

Conservation of Momentum
The conservation of momentum is a fundamental concept in physics that states that within a closed system, the total momentum before an event must equal the total momentum after the event. Momentum is defined as the mass of an object multiplied by its velocity. In formula terms, it is represented as \( p = mv \).

Understanding this concept is central to solving the problem of two blocks being pushed apart by a compressed spring. Initially, when the blocks are at rest, their total momentum is zero. This condition remains true throughout because there are no external forces acting on the system. Consequently, when the spring pushes the blocks apart, their momenta must be equal and opposite in order to maintain the total system momentum of zero.

From the conservation of momentum, we set up the equation \( mv_1 = 3mv_2 \). This relationship ensures that when one block moves to the right with a momentum of \( mv_1 \), the other block moves to the left with a momentum of \( 3mv_2 \), thus balancing the total momentum at zero after the blocks separate.
Spring Potential Energy
Spring potential energy is the energy stored in a spring when it is compressed or stretched from its natural length. This energy is determined by the spring constant \( k \) and the displacement from the equilibrium position \( D \). The formula used to calculate the spring's potential energy is \( PE = \frac{1}{2} k D^2 \).

In this particular exercise, the spring is initially compressed by an amount \( D \), storing potential energy that is later converted into kinetic energy as the spring expands and releases the blocks. This transformation is an essential aspect of the conservation of energy principle, as energy is conserved in the form of mechanical energy within a closed system.

The calculation of spring potential energy helps us understand the initial energy available to be transformed into kinetic energy for the two blocks. By knowing \( PE = \frac{1}{2} k D^2 \), we can determine how much total kinetic energy the spring transfers to the blocks once they're released.
Kinetic Energy
Kinetic energy relates to the energy of an object in motion, which is crucial once the blocks start moving away from the spring. It is given by the formula \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass of the object and \( v \) is its velocity.

In this exercise, the total initial mechanical energy is stored as spring potential energy in the compressed spring. Upon release, this energy transforms into kinetic energy for the two blocks, adhering to the conservation of energy principle. To find the velocities of the blocks, we equate the total initial potential energy of the spring to the sum of the kinetic energies of both blocks.

The energy conservation equation is set as \( \frac{1}{2}kD^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}(3m)v_2^2 \). By solving this equation alongside the momentum equation, we deduce that the kinetic energies of each block correspond to their respective speeds, \( v_1 \) and \( v_2 \), providing insights into how energy is evenly distributed via the spring's release.

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Most popular questions from this chapter

(II) A 145-g baseball, moving along the \(x\) axis with speed \(30.0 \mathrm{~m} / \mathrm{s}\), strikes a fence at a \(45^{\circ}\) angle and rebounds along the \(y\) axis with unchanged speed. Give its change in momentum using unit vector notation.

(II) The jet engine of an airplane takes in \(120 \mathrm{~kg}\) of air per second, which is burned with \(4.2 \mathrm{~kg}\) of fuel per second. The burned gases leave the plane at a speed of \(550 \mathrm{~m} / \mathrm{s}\) (relative to the plane). If the plane is traveling \(270 \mathrm{~m} / \mathrm{s}(600 \mathrm{mi} / \mathrm{h}),\) determine: \((a)\) the thrust due to ejected fuel; \((b)\) the thrust due to accelerated air passing through the engine; and ( \(c\) ) the power (hp) delivered.

(II) A 22-g bullet traveling \(210 \mathrm{~m} / \mathrm{s}\) penetrates a \(2.0-\mathrm{kg}\) block of wood and emerges going \(150 \mathrm{~m} / \mathrm{s}\). If the block is stationary on a frictionless surface when hit, how fast does it move after the bullet emerges?

(II) Car A hits car B (initially at rest and of equal mass) from behind while going \(35 \mathrm{~m} / \mathrm{s}\). Immediately after the collision, car B moves forward at \(25 \mathrm{~m} / \mathrm{s}\) and car \(\mathrm{A}\) is at rest. What fraction of the initial kinetic energy is lost in the collision?

A hockey puck of mass \(4 m\) has been rigged to explode, as part of a practical joke. Initially the puck is at rest on a frictionless ice rink. Then it bursts into three pieces. One chunk, of mass \(m,\) slides across the ice at velocity \(v \hat{\mathbf{i}}\). Another chunk, of mass \(2 m\), slides across the ice at velocity \(2 v \hat{\mathbf{j}} .\) Determine the velocity of the third chunk.

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