Problem 85 - University Physics with Modern Physics

Question

CP SHM of a Butcher's Scale. A spring of negligible mass and force constant \(k=400 \mathrm{N} / \mathrm{m}\) is hung vertically, and a 0.200 -kg pan is suspended from its lower end. A butcher drops a 2.2 -kg steak onto the pan from a height of 0.40 \(\mathrm{m}\) . The steak makes a totally inelastic collision with the pan and sets the system into vertical SHM. What are (a) the speed of the pan and steak immediately after the collision; (b) the amplitude of the subsequent motion; (c) the period of that motion?

Step-by-Step Solution

Answer

(a) Speed after collision: 2.57 m/s (b) Amplitude: 0.2 m (c) Period: 0.487 s

Step 1: Find the Speed Before Collision

Use the conservation of energy principle. The initial potential energy of the steak at a height of 0.40 m is converted into kinetic energy just before it hits the pan. So we have: \[ mgh = \frac{1}{2}mv^2 \]Substitute \(m = 2.2\, \text{kg}, \ g = 9.8\, \text{m/s}^2, \ h = 0.4\, \text{m}\) into the equation:\[ 2.2 \times 9.8 \times 0.4 = \frac{1}{2} \times 2.2 \times v^2 \]Solve for \(v\):\[ v^2 = \frac{2 \times 2.2 \times 9.8 \times 0.4}{2.2} \]\[ v = \sqrt{2 \times 9.8 \times 0.4} \approx 2.8\, \text{m/s} \]

Step 2: Calculate Speed After Collision

In a totally inelastic collision, momentum is conserved. The total mass after collision becomes \(M = 2.2 + 0.2 = 2.4\, \text{kg}\). Using the conservation of momentum:\[ m_1v_1 = Mv' \]\( m_1 = 2.2\, \text{kg}, \ v_1 = 2.8\, \text{m/s}, \ M = 2.4\, \text{kg} \)\[ 2.2 \times 2.8 = 2.4 \times v' \]Solve for \(v'\):\[ v' = \frac{2.2 \times 2.8}{2.4} \approx 2.57\, \text{m/s} \]

Step 3: Determine the Amplitude of Motion

Use the conservation of mechanical energy. The initial kinetic energy of the system is converted to potential energy at maximum displacement (amplitude):\[ \frac{1}{2}Mv'^2 = \frac{1}{2}kA^2 \]Substitute \(M = 2.4\, \text{kg}, \ k = 400\, \text{N/m}, \ v' = 2.57\, \text{m/s} \):\[ \frac{1}{2} \times 2.4 \times (2.57)^2 = \frac{1}{2} \times 400 \times A^2 \]\[ 2.4 \times 6.6049 = 400A^2 \]Solve for \(A\):\[ A^2 = \frac{15.85}{400} \]\[ A \approx 0.2 \text{ m} \]

Step 4: Calculate the Period of SHM

The period \(T\) of a simple harmonic motion is given by:\[ T = 2\pi \sqrt{\frac{M}{k}} \]Substitute \(M = 2.4\, \text{kg}, \ k = 400\, \text{N/m} \):\[ T = 2\pi \sqrt{\frac{2.4}{400}} \]\[ T = 2\pi \sqrt{0.006} \]\[ T \approx 0.487\, \text{s} \]

Key Concepts

Inelastic Collision

In an inelastic collision, the objects involved stick together after colliding, rather than bouncing apart. This specific kind of collision is characterized by a loss of kinetic energy, but crucially, momentum is conserved.
\[ \text{Initial momentum} = \text{Final momentum} \]
During the collision of the 2.2-kg steak with the 0.2-kg pan, the masses combine to form a single mass of 2.4 kg. The speed of the combined system directly after the collision can be determined using the conservation of momentum. It shows how, even if energy is "lost" in the form of heat or sound, momentum continues to play a key role in determining motion right after the impact.

• A characteristic, distinguishing point is that the energy is not conserved during inelastic collisions, only momentum is.
• Due to their nature, inelastic collisions usually result in some permanent deformation or generation of heat.

Conservation of Energy

The principle of conservation of energy is central in solving dynamics and mechanics problems. Energy can neither be created nor destroyed; it can only transform from one form to another. This is particularly helpful in analyzing mechanical systems involving potential and kinetic energy transitions.
In the given problem, the initial potential energy of the steak from a height of 0.4 m is transformed into kinetic energy as it falls.

• The calculation of this transformation helps to find the steak's speed just before impact.
• Similarly, after the collision, we analyze energy again to determine how the kinetic energy transforms to potential when the system reaches its maximum displacement (or amplitude).

In simple harmonic motion (SHM), this transformation back and forth between kinetic and potential forms is constant and predictable.

Amplitude of Motion

Amplitude in simple harmonic motion (SHM) refers to the maximum displacement from the equilibrium position. In essence, it's the farthest point the system moves to either side during oscillation.
After the inelastic collision in this exercise, the amplitude of motion is determined by equating the kinetic energy converted into the spring's potential energy at the amplitude's displacement.

• The equation \( \frac{1}{2}Mv'^2 = \frac{1}{2}kA^2 \) is used to find the amplitude.
• This shows us that a larger initial speed after the collision would lead to a greater amplitude.

Amplitude does not depend on factors such as phase or the system's mass, but rather the energy provided to it.

Momentum Conservation

Momentum, a product of an object's mass and velocity, is a core concept in physics. In any isolated system, the total momentum before and after a collision remains constant. This is the foundation of momentum conservation.
For the steak-pan problem, this concept is showcased during the inelastic collision step. Before the collision, only the steak has momentum; however, after the collision, this momentum is shared between the combined steak and pan mass.

• The formula used is \( m_1v_1 = Mv' \), reflecting this principle of shared momentum.
• This conservation law becomes particularly evident in systems where external forces (like friction or external presses) are negligible.

Understanding momentum conservation allows one to predict the motion of various bodies engaged in collisions and is pivotal in mechanics.