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Chapter 9: Problem 41

(III) \(\mathrm{A} 3.0\) -kg block slides along a frictionless tabletop at 8.0 \(\mathrm{m} / \mathrm{s}\) toward a second block (at rest) of mass 4.5 \(\mathrm{kg}\) . A coil spring, which obeys Hooke's law and has spring constant \(k=850 \mathrm{N} / \mathrm{m},\) is attached to the second block in such a way that it will be compressed when struck by the moving block, Fig. 40 (a) What will be the maximum compression of the spring? (b) What will be the final velocities of the blocks after the collision? (c) Is the collision elastic? Ignore the mass of the spring.

Short Answer

Expert verified
Maximum compression: 0.63 m. Collision is not elastic as kinetic energy changes.

Step by step solution

01

Analyze the Collision

The moving 3.0 kg block will collide with the stationary 4.5 kg block, compressing the spring. We will first determine if the collision is perfectly elastic by checking the kinetic energy before and after the collision.
02

Apply Conservation of Momentum for Collision

Using the principle of conservation of momentum, consider the initial momentum of the system which is only due to the moving block. The total initial momentum is given by:\[ p_{i} = m_{1}v_{1i} + m_{2}v_{2i} = 3.0 imes 8.0 + 4.5 imes 0 = 24.0 \] This will be equal to the total momentum after the collision:\[ p_{f} = m_{1}v_{1f} + m_{2}v_{2f} \] Where \(v_{1f}\) and \(v_{2f}\) are the final velocities of blocks m1 and m2.
03

Apply Conservation of Mechanical Energy for Maximum Spring Compression

For maximum compression, all kinetic energy is temporarily converted into elastic potential energy of the spring:\[ \frac{1}{2} m_1 v_1^2 = \frac{1}{2} k x^2 \] Here, \(x\) is the maximum compression of the spring. Solve for \(x\):\[ x = \sqrt{\frac{m_1 v_1^2}{k}} = \sqrt{\frac{3.0 \times 8.0^2}{850}} \approx 0.63 \mathrm{m} \]
04

Check ELasticity of the Collision

To check if the collision is elastic, verify if the total kinetic energy before and after the collision is conserved. If not, the collision is inelastic.
05

Calculate Kinetic Energy before the Collision

The initial kinetic energy of the system is entirely due to the moving block:\[ KE_{initial} = \frac{1}{2} m_1 v_1^2 = \frac{1}{2} \times 3.0 \times 8.0^2 = 96.0 \, \mathrm{J} \]
06

Determine Kinetic Energy After the Collision

After the collision, part of this kinetic energy has been stored as potential energy in the compressed spring, making an exact solve for velocities complex. Typically, if the total energy initially stored as kinetic energy (\(96.0\) J) cannot entirely remain unchanged as kinetic energy, the collision is not perfectly elastic. Without exact velocity values post-collision, provide that in real scenarios typically not all kinetic energy is converted back. The difference between pre and post energy would certify if it鈥檚 elastic or not.

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Key Concepts

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

Conservation of Momentum
The principle of conservation of momentum is foundational in understanding collisions. It states that the total momentum of a closed system remains constant, provided no external forces act upon it.
In the given exercise, we have two blocks: one moving and one stationary. The initial momentum is entirely due to the moving 3.0 kg block, calculated as its mass times its velocity.
  • Before the collision: The momentum is 24 kg路m/s (since the second block is at rest).
  • After the collision: The system's total momentum must still sum to 24 kg路m/s.
This principle helps us set up equations to find the final velocities of the two blocks after the collision.
If the final velocities can be accurately determined, they will show that the interaction complies with the conservation of momentum. This principle is crucial in both physics problems and real-world scenarios, like car accidents, to predict post-collision movement.
Hooke's Law
Hooke's Law describes the behavior of springs when they are stretched or compressed. It states that the force exerted by a spring is proportional to the amount of stretch or compression. Mathematically, it is given by:
  • F = -kx
where:
  • F is the force exerted by the spring,
  • k is the spring constant,
  • x is the displacement of the spring from its equilibrium position.

In the exercise, we apply Hooke's Law to determine the maximum compression of the spring when the moving block strikes the stationary block attached to the spring.
The spring compresses, storing potential energy calculated with:
  • Potential Energy = \( \frac{1}{2} kx^2 \)
This law allows us to see how kinetic energy is temporarily stored as potential energy in the spring during maximum compression, giving us insights into elastic properties of materials.
Kinetic Energy
Kinetic energy is the energy possessed by an object due to its motion, calculated as \( KE = \frac{1}{2}mv^2 \).
Understanding kinetic energy is crucial for determining whether a collision is elastic.In elastic collisions, the total kinetic energy before collision equals the total kinetic energy after collision. This means energy efficiency is maximal, with no kinetic energy loss to other forms like sound or heat.
  • Initially, only the moving block contributes to the kinetic energy, calculated as 96 J.
  • After the collision, part of this energy may transform into potential energy as the spring compresses.
If post-collision calculations show a reduction in total kinetic energy, the collision is deemed inelastic. This concept helps differentiate between purely elastic collisions and those with energy dissipation, illustrating motion energy's transfer or conversion during interactions.

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Most popular questions from this chapter

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.

(II) An atomic nucleus of mass \(m\) traveling with speed \(v\) collides elastically with a target particle of mass \(2 m\) (initially at rest) and is scattered at \(90^{\circ} .(a)\) At what angle does the target particle move after the collision? \((b)\) What are the final speeds of the two particles? ( \(c\) ) What fraction of the initial kinetic energy is transferred to the target particle?

(II) The masses of the Earth and Moon are \(5.98 \times 10^{24} \mathrm{~kg}\) and \(7.35 \times 10^{22} \mathrm{~kg},\) respectively, and their centers are separated by \(3.84 \times 10^{8} \mathrm{~m} .\) (a) Where is the CM of this system located? (b) What can you say about the motion of the Earth-Moon system about the Sun, and of the Earth and Moon separately about the Sun?

(III) For an elastic collision between a projectile particle of mass \(m_{\mathrm{A}}\) and a target particle (at rest) of mass \(m_{\mathrm{B}}\), show that the scattering angle, \(\theta_{\mathrm{A}}^{\prime},\) of the projectile \((a)\) can take any value, 0 to \(180^{\circ},\) for \(m_{\mathrm{A}}m_{\mathrm{B}}\).

(II) Show that, in general, for any head-on one-dimensional elastic collision, the speeds after collision are $$ v_{\mathrm{B}}^{\prime}=v_{\mathrm{A}}\left(\frac{2 m_{\mathrm{A}}}{m_{\mathrm{A}}+m_{\mathrm{B}}}\right)+v_{\mathrm{B}}\left(\frac{m_{\mathrm{B}}-m_{\mathrm{A}}}{m_{\mathrm{A}}+m_{\mathrm{B}}}\right) $$ and $$ v_{\mathrm{A}}^{\prime}=v_{\mathrm{A}}\left(\frac{m_{\mathrm{A}}-m_{\mathrm{B}}}{m_{\mathrm{A}}+m_{\mathrm{B}}}\right)+v_{\mathrm{B}}\left(\frac{2 m_{\mathrm{B}}}{m_{\mathrm{A}}+m_{\mathrm{B}}}\right) $$ where \(v_{\mathrm{A}}\) and \(v_{\mathrm{B}}\) are the initial speeds of the two objects of mass \(m_{\mathrm{A}}\) and \(m_{\mathrm{B}}\)
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