Question

A \(4.00-\mathrm{kg}\) block is suspended from a spring with a force constant of \(5.00 \mathrm{~N} / \mathrm{cm}\). A \(50.0-\mathrm{g}\) bullet is fired into the block from below with a speed of \(150 \mathrm{~m} / \mathrm{s}\) and comes to rest in the block. (a) Find the amplitude of the resulting simple harmonic motion. (b) What fraction of the original kinetic energy of the bullet appears as mechanical energy in the oscillator?

Short answer

The amplitude of the resulting simple harmonic motion is \(x\) and the fraction of the original kinetic energy of the bullet that appears as mechanical energy in the oscillator is \(frac{ME_o}{KE_b}\).

Step-by-step solution

Conservation of linear momentum

The momentum of the block and bullet before the bullet hits the block is equal to the momentum of the block and bullet after the bullet hits the block. This gives the equation \(m_b*v_b = (m_b + m)*v\) where \(m_b\) is the mass of the bullet, \(m\) is the mass of the block, \(v_b\) is the initial velocity of the bullet and \(v\) is the velocity of the block and bullet system after the bullet hits the block. We can solve this equation for \(v\).

Calculate the maximum height

The kinetic energy of the block-bullet system is converted into potential energy when the system reaches its maximum height. This gives the equation \(0.5*(m_b + m)*v^2 = (m_b + m)*g*h + 0.5*k*x^2\) where \(g\) is the acceleration due to gravity, \(h\) is the initial height, \(k\) is the spring constant and \(x\) is the maximum displacement of the system which equals the amplitude. Because the spring was initially at rest, \(h\) is zero and we can solve this equation for \(x\).

Calculate the fraction of original kinetic energy

The initial kinetic energy of the bullet is given by \(KE_b=0.5*m_b*v_b^2\) and the mechanical energy in the oscillator is \(ME_o=0.5*(m_b + m)*v^2\). The fraction of the original kinetic energy appearing as mechanical energy in the oscillator is given by \(frac{ME_o}{KE_b}\). This fraction can then be calculated.

Key concepts

Conservation of Momentum

When a collision occurs, like the one between the bullet and the block, an important principle called the conservation of momentum comes into play. This principle states that the total momentum of an isolated system remains constant if no external forces act on it.
To apply this concept, consider the momentum before and after the bullet hits the block. The initial momentum of the bullet is simply its mass multiplied by its velocity, or \( m_b \times v_b \). After the collision, both the bullet and the block move together with a new velocity \( v \). The combined mass of the system is \( m_b + m \). Therefore, the momentum after the collision is \((m_b + m) \times v\).
Using the conservation of momentum, we equate these two expressions:

• Initial momentum: \( m_b \times v_b \)
• Final momentum: \((m_b + m) \times v\)

Solving this equation helps us find the new velocity of the block and bullet system. This value is crucial for further calculations, such as determining the amplitude of the resulting motion.

Kinetic Energy

Kinetic energy, which is the energy of motion, plays a vital role here. Initially, the bullet possesses kinetic energy owing to its high speed. This energy is given by the equation for kinetic energy: \( KE_b = 0.5 \times m_b \times v_b^2 \).
When the bullet becomes embedded in the block, the kinetic energy of the bullet isn't entirely lost. Instead, some of it gets transferred to the block-bullet system, causing it to move. This new kinetic energy of the system can be expressed as \( 0.5 \times (m_b + m) \times v^2 \).
However, not all the initial kinetic energy is retained as kinetic energy post-collision. We can calculate how much of this energy is converted into mechanical energy in the system by comparing the initial kinetic energy of the bullet to the kinetic energy of the combined mass post-collision. This comparison can help us find the fraction of energy conserved as mechanical energy.

Spring-Mass System

A spring-mass system is a classic model used to understand simple harmonic motion (SHM), where an object is attached to a spring and oscillates back and forth. In our scenario, the block (with the bullet inside) connected to the spring behaves as such a system.
The key component of this system is the spring constant \( k \), which measures the stiffness of the spring. The spring's potential energy at maximum displacement equals the combined kinetic energy of the block and bullet. In mathematical terms, this relationship is given by \( 0.5 \times k \times x^2 \), where \( x \) is the maximum displacement, also known as the amplitude of oscillation.
As the bullet-block system moves upward, kinetic energy converts into spring potential energy, reaching its maximum at the peak of oscillation. Knowing the spring constant \( k \) and using energy conservation, we can solve for \( x \), determining the swing's amplitude, providing insights into the system's behavior.

Oscillator Energy

Once the block and bullet constitute a single unit moving together, they form an oscillator. Oscillator energy comprises two main types: kinetic energy and potential energy.
In an ideal SHM system, energy smoothly transitions between kinetic and potential forms. As the block-bullet system compresses the spring, the kinetic energy reduces while the potential energy stored in the spring increases. Conversely, as the system moves towards its equilibrium position, potential energy transforms back into kinetic energy, propelling the system and maintaining oscillation.
We calculate the energy involved by evaluating mechanical energy post-collision as \( ME_o = 0.5 \times (m_b + m) \times v^2 \). By comparing this with the initial kinetic energy, we discern how much energy sustains the motion, thus providing essential insights into how energy dynamics govern simple harmonic motion.