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Chapter 10: Problem 42

\(\mathrm{A} 1.00 \times 10^{-2}-\mathrm{kg}\) bullet is fired horizontally into a \(2.50-\mathrm{kg}\) wooden block attached to one end of a massless horizontal spring \((k=845 \mathrm{N} / \mathrm{m})\). The other end of the spring is fixed in place, and the spring is unstrained initially. The block rests on a horizontal, frictionless surface. The bullet strikes the block perpendicularly and quickly comes to a halt within it. As a result of this completely inelastic collision, the spring is compressed along its axis and causes the block/bullet to oscillate with an amplitude of \(0.200 \mathrm{m} .\) What is the speed of the bullet?

Short Answer

Expert verified
The bullet's speed was approximately 921 m/s before the collision.

Step by step solution

01

Understand the Problem

A bullet of mass \( m_1 = 1.00 \times 10^{-2} \ \text{kg} \) is fired into a block of mass \( m_2 = 2.50\ \text{kg} \), connected to a spring with spring constant \( k = 845 \ \text{N/m} \). The block and bullet move together after collision. The amplitude of oscillation is \( A = 0.200 \ \text{m} \). We need to find the initial speed of the bullet.
02

Apply Conservation of Momentum

Before the collision, only the bullet is moving with speed \( v \). After the collision, the bullet and block move together. From conservation of momentum: \[ m_1 v = \left( m_1 + m_2 \right) V \] where \( V \) is the velocity just after the collision.
03

Energy in the Spring System

The mechanical energy converts entirely into the potential energy of the spring at maximum compression due to the inelastic collision. The potential energy in spring is given by: \[ \frac{1}{2} k A^2 = \frac{1}{2} \left( m_1 + m_2 \right) V^2 \] Solve this for \( V \).
04

Solve for \( V \)

Rearrange the equation for potential energy, \[ V^2 = \frac{k A^2}{m_1 + m_2} \]Calculate \( V \):\[ \begin{align*} V^2 & = \frac{845 \times (0.200)^2}{2.51} \V^2 & = \frac{845 \times 0.04}{2.51} \V^2 & = \frac{33.8}{2.51} \V^2 & \approx 13.466 \ \ V & \approx 3.67\ \text{m/s} \end{align*} \]
05

Solve for the Speed of the Bullet

Using the conservation of momentum equation:\[ \begin{align*} m_1 v & = (m_1 + m_2) V \v & = \left( \frac{(m_1 + m_2) V}{m_1} \right) \v & = \left( \frac{(1.00 \times 10^{-2} + 2.50) \times 3.67}{1.00 \times 10^{-2}} \right) \v & = \left( \frac{2.51 \times 3.67}{1.00 \times 10^{-2}} \right) \v & = \frac{9.2117}{1.00 \times 10^{-2}} \v & \approx 921.17\ \text{m/s} \end{align*} \]
06

Conclusion

The speed of the bullet before the collision was approximately \( 921.17\ \text{m/s} \).

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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 principle in physics that states the total momentum of a closed system remains constant, provided that no external forces are acting on it. This is especially crucial in collision problems such as the one described in this exercise. Before the collision, only the bullet is moving, which means it has all the momentum. After the collision, the bullet and block move together as a single object. At this point, the system's total momentum must still match the initial momentum of the bullet.

When dealing with collisions, especially inelastic ones where the colliding objects stick together, the principle of momentum conservation can be mathematically represented by:
  • Initial momentum: \( m_1 v \)
  • Final momentum: \( (m_1 + m_2) V \)
  • Conservation equation: \( m_1 v = (m_1 + m_2) V \)
In this specific problem, solving this equation gives us the ability to relate the initial speed of the bullet to the velocity of the combined bullet and block just after the collision.
Spring Mechanics
Understanding spring mechanics is critical in this exercise, as it relates to how the potential energy is stored and released by the spring. A spring has a constant, \( k \), which quantifies its stiffness, and is measured in newtons per meter (N/m). The compression or extension of the spring changes its potential energy. This is defined by Hooke's Law, which states that the force exerted by the spring is proportional to its displacement:
  • Hooke's Law: \( F = -kx \)
  • Potential energy in a spring: \( PE = \frac{1}{2} k x^2 \)
In this collision problem, the spring is unstrained at the start, and becomes compressed as a result of the collision. At its maximum compression, it stores the entire kinetic energy of the block and bullet, now converted into potential energy. This conversion is key to determining how the energies balance after the inelastic collision.
Potential Energy
Potential energy is the energy stored in an object due to its position or configuration. In this problem, we specifically talk about elastic potential energy stored in the compressed spring. When the block and bullet come to a temporary rest at maximum compression, all kinetic energy has been converted into potential energy:
  • Potential Energy in Spring: \( PE = \frac{1}{2} k A^2 \)
This potential energy dictates the movement of the bullet and block system, showing peak energy when the spring's compression reaches its maximum (which is the amplitude \( A \) in the problem). Afterward, this energy will reconvert back to kinetic energy, causing the system to oscillate. This interplay between kinetic and potential energy is crucial for understanding mechanics, as it governs a myriad of physical phenomena, from simple pendulums to complex molecular bonds.

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Most popular questions from this chapter

A 68.0 -kg bungee jumper is standing on a tall platform \(\left(h_{0}=\right.\) \(46.0 \mathrm{m}),\) as indicated in the figure. The bungee cord has a natural length of \(L_{0}=9.00 \mathrm{m}\) and, when stretched, behaves like an ideal spring with a spring constant of \(k=66.0 \mathrm{N} / \mathrm{m}\). The jumper falls from rest, and it is assumed that the only forces acting on him are his weight and, for the latter part of the descent, the elastic force of the bungee cord. Concepts: (i) Can we use the conservation of mechanical energy to find his speed at any point along the descent? Explain your answer. (ii) What type of energy does he have when he is standing on the platform? (iii) What types of energy does he have at point A? (iv) What types of energy does he have at point \(\mathrm{B} ?\) Calculations: What is his speed when he is at the following heights above the water: (a) \(h_{\mathrm{A}}=37.0 \mathrm{m},\) and (b) \(h_{\mathrm{B}}=15.0 \mathrm{m} ?\)

A 6.8 -kg bowling ball is attached to the end of a nylon cord with a cross- sectional area of \(3.4 \times 10^{-5} \mathrm{m}^{2} .\) The other end of the cord is fixed to the ceiling. When the bowling ball is pulled to one side and released from rest, it swings downward in a circular arc. At the instant it reaches its lowest point, the bowling ball is \(1.4 \mathrm{m}\) lower than the point from which it was released, and the cord is stretched \(2.7 \times 10^{-3} \mathrm{m}\) from its unstrained length. What is the unstrained length of the cord? (Hint: When calculating any quantity other than the strain, ignore the increase in the length of the cord.)

A horizontal spring is lying on a frictionless surface. One end of the spring is attached to a wall, and the other end is connected to a movable object. The spring and object are compressed by \(0.065 \mathrm{m}\), released from rest, and subsequently oscillate back and forth with an angular frequency of 11.3 rad/s. What is the speed of the object at the instant when the spring is stretched hy \(0.048 \mathrm{m}\) relative to its \(\mathrm{un}\) strained length?

A spring is resting vertically on a table. A small box is dropped onto the top of the spring and compresses it. Suppose the spring has a spring constant of 450 N/m and the box has a mass of 1.5 kg. The speed of the box just before it makes contact with the spring is 0.49 m/s. (a) Determine the magnitude of the spring鈥檚 displacement at an instant when the acceleration of the box is zero. (b) What is the magnitude of the spring鈥檚 displacement when the spring is fully compressed?

Multiple-Concept Example 6 presents a model for solving this problem. As far as vertical oscillations are concerned, a certain automobile can be considered to be mounted on four identical springs, each having a spring constant of \(1.30 \times 10^{5} \mathrm{N} / \mathrm{m} .\) Four identical passengers sit down inside the car, and it is set into a vertical oscillation that has a period of 0.370 s. If the mass of the empty car is 1560 kg, determine the mass of each passenger. Assume that the mass of the car and its passengers is distributed evenly over the springs.
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