Problem 9 - College Physics

Question

\(A\) 5.0-g bullet with a speed of \(200 \mathrm{~m} / \mathrm{s}\) is fired horizontally into a \(0.75-\mathrm{kg}\) wooden block at rest on a table. If the block containing the bullet slides a distance of \(0.20 \mathrm{~m}\) before coming to rest, (a) what is the coefficient of kinetic friction between the block and the table? (b) What fraction of the bullet's energy is dissipated in the collision?

Step-by-Step Solution

Answer

(a) Coefficient of kinetic friction is 0.089. (b) 99.34% of bullet's energy is dissipated.

Step 1: Calculate the momentum before and after the collision

Initially, the momentum of the bullet is given by the formula: \[ p_{initial} = m_{bullet} \cdot v_{bullet} = 0.005 \, \text{kg} \times 200 \, \text{m/s} = 1 \, \text{kg} \cdot \text{m/s}. \]The wooden block is initially at rest, so its initial momentum is 0.After the collision, the bullet and block move together with a combined mass of \( m_{total} = 0.005 \, \text{kg} + 0.75 \, \text{kg} = 0.755 \, \text{kg} \). Let the velocity of the block and bullet system after collision be \( V \).According to the conservation of momentum:\[ p_{initial} = p_{final} \] \[ 1 \, \text{kg} \cdot \text{m/s} = 0.755 \, \text{kg} \cdot V \]\[ V = \frac{1}{0.755} \approx 1.324 \, \text{m/s}. \]

Step 2: Calculate the coefficient of kinetic friction

After the collision, the block with the bullet embedded comes to rest due to the force of kinetic friction. We use the work-energy principle, where the work done by friction equals the kinetic energy lost:\[ F_{k} \cdot d = \frac{1}{2} m_{total} V^2 \]Let \( F_{k} = \mu_k \cdot n \), where \( n = m_{total} \cdot g = 0.755 \, \text{kg} \times 9.8 \, \text{m/s}^2 \).The work done by friction is:\[ W = \mu_k \cdot m_{total} \cdot g \cdot d \]Equating to kinetic energy:\[ \mu_k \cdot m_{total} \cdot g \cdot d = \frac{1}{2} m_{total} \cdot V^2 \]\[ \mu_k \times 9.8 \times 0.755 \times 0.2 = 0.5 \times 0.755 \times (1.324)^2 \]Solving for \( \mu_k \):\[ \mu_k = \frac{0.5 \times 0.755 \times 1.324^2}{9.8 \times 0.755 \times 0.2} \approx 0.089 \]

Step 3: Calculate the initial kinetic energy of the bullet

The initial kinetic energy of the bullet is:\[ KE_{initial} = \frac{1}{2} m_{bullet} v_{bullet}^2 = \frac{1}{2} \times 0.005 \, \text{kg} \times (200 \, \text{m/s})^2 \]\[ KE_{initial} = 0.5 \times 0.005 \times 40000 = 100 \, \text{J} \]

Step 4: Calculate the kinetic energy after collision

The kinetic energy of the block and bullet system right after the collision is:\[ KE_{final} = \frac{1}{2} m_{total} V^2 \]\[ KE_{final} = 0.5 \times 0.755 \times (1.324)^2 \approx 0.66 \, \text{J} \]

Step 5: Calculate the fraction of the bullet's energy dissipated

The energy dissipated during the collision is the difference between initial and final kinetic energy:\[ E_{dissipated} = KE_{initial} - KE_{final} \]\[ E_{dissipated} = 100 \, \text{J} - 0.66 \, \text{J} = 99.34 \, \text{J} \]The fraction of energy dissipated is:\[ \text{Fraction} = \frac{E_{dissipated}}{KE_{initial}} = \frac{99.34}{100} \approx 0.9934 \]

Step 6: Finalize the Answers

(a) The coefficient of kinetic friction between the block and the table is approximately 0.089. (b) Around 99.34% of the bullet's initial kinetic energy is dissipated during the collision.

Key Concepts

Conservation of Momentum

When it comes to collision problems like this one, understanding how momentum works is key. Momentum, which is the product of mass and velocity, must be conserved in a closed system.
This means that the total momentum before the collision is equal to the total momentum after the collision. Before our bullet hits the block, only the bullet carries momentum, calculated as 1 kg·m/s.
After the collision, the bullet embeds into the block, forming a single system that keeps the initial momentum but now involves a single mass moving together. Thus, they share the velocity calculated using conservation laws.
Important takeaway here? Momentum conservation helps us find velocities post-collision.

• Initial momentum is calculated with the bullet's mass and speed.
• Total momentum remains the same after collision.
• Momentum is shared by new combined mass after collision.

Kinetic Energy

Kinetic energy is all about motion. It quantifies the energy that an object has due to its movement.
In our scenario, the bullet initially has kinetic energy because it is moving. This is calculated as 100 Joules (J).
After the collision, kinetic energy still exists but is reduced because it's now distributed between greater combined mass. The idea here is not to confuse kinetic energy with momentum; while both involve mass and motion, they tell different tales about the system.
These calculations help us understand how much motion energy the bullet had initially and how it changes due to the collision. After collision, the energy is smaller at 0.66 J, showing a loss in energy due to internal force during collision.

• Kinetic energy is calculated with an object's speed and mass before collision.
• Some energy is lost as heat or sound, which is why post-collision energy is less.
• Understanding this change is essential for further analysis on other topics like energy dissipation.

Kinetic Friction

Friction is what eventually brings the sliding block and bullet to a stop. Kinetic friction acts on moving objects and opposes the motion.
The coefficient of kinetic friction, in this context, represents the force ratio needed to slide two surfaces across each other. By using the work-energy principle, which ties the change in kinetic energy to the work done by friction, we determine how much force is involved.
Essentially, kinetic friction converts kinetic energy into thermal energy or heat during sliding, and in our solution, it's calculated with a coefficient of 0.089.

• Kinetic friction is a force that resists the sliding of two surfaces.
• The coefficient of kinetic friction explains how "sticky" or resistant a surface pair is.
• Finds application in calculating how long or far an object will slide before stopping.

Energy Dissipation

Whenever energy gets "lost" in a process, we call it dissipation. It doesn't mean energy disappears entirely but typically changes form; such as, transforming into heat or sound.
In our example, we track dissipation through the difference in kinetic energy before and after the collision.
This bullet-barrel system lost most of its moving energy, precisely 99.34 J, through dissipation. Learning about energy dissipation is crucial for understanding inefficiencies in physical systems, helping one troubleshoot why less energy seemed active by end.

• Energy dissipation usually results in heating up the environment, not destroying energy.
• It helps explain why post-collision kinetic energy is much lower.
• Understanding dissipation is critical for analyzing real-world applications where energy "loss" is involved.

Work-Energy Principle

The work-energy principle connects the idea of work (force applied over distance) with changes in kinetic energy.
In practical terms, work done on an object is translated into energy changes within the system. In this problem, work done by friction literally means how much energy the block needs to lose to stop.
By applying this principle, it allows us to calculate how forces like friction can halt kinetic energy flow, by measuring how much work is equivalent to the change observed.

• Work-energy principle helps link energy loss or gain to how a force acts over a distance.
• For our setup, it shows us how friction’s work turns kinetic energy into thermal energy.
• A powerful tool to demonstrate how external forces influence a system's energy capacity.