Problem 25 - College Physics

Question

. A \(\mathrm{A} 5.00\) g bullet is fired horizontally into a 1.20 \(\mathrm{kg}\) wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is \(0.20 .\) The bullet remains embedded in the block, which is observed to slide 0.230 \(\mathrm{m}\) along the surface before stopping. What was the initial speed of the bullet?

Step-by-Step Solution

Answer

The bullet's initial speed was approximately 229 m/s.

Step 1: Understand the Problem

A bullet is embedded into a block, leading to motion. We need to find the initial speed of the bullet before the collision, considering the friction between the block and surface.

Step 2: Determine the System's Final Velocity

Since the bullet becomes embedded in the block, we treat the bullet-block system as one complete mass. When the block slides to a stop, its final velocity is 0.

Step 3: Use Work-Energy Principle for Sliding

The kinetic friction force stops the block. Calculate work done by this force to find initial kinetic energy of the bullet-block system:\[Work = Friction \times Distance = \mu kg(m_{block} + m_{bullet})g \times \Delta x\]where \(\mu = 0.2\), \(\Delta x = 0.230\) m, and \(g = 9.8\) m/s².

Step 4: Calculate Frictional Work

\[Work = 0.20 \times (1.20 + 0.005) \text{ kg} \times 9.8 \text{ m/s}^2 \times 0.230 \text{ m} = 0.55092 \, ext{J}\]

Step 5: Set Initial Kinetic Energy Equal to Work Done by Friction

The work done by friction is equal to the initial kinetic energy of the bullet-block system:\[\frac{1}{2}(m_{block} + m_{bullet})v^2 = 0.55092\]

Step 6: Calculate Initial Velocity of Bullet-Block System

Solve for the initial velocity \(v\) of the bullet-block system after the collision: \[\frac{1}{2}(1.20 + 0.005) v^2 = 0.55092\] This gives:\[ v = \sqrt{\frac{2 \times 0.55092}{1.205}} = 0.949 \, ext{m/s} \]

Step 7: Use Conservation of Momentum for Bullet and Block System

The momentum before collision (bullet alone) equals momentum after collision (bullet+block):\[ m_{bullet}v_{bullet} = (m_{bullet} + m_{block})v \]Given \(m_{bullet} = 0.005\) kg, solve for bullet's initial speed \(v_{bullet}\) :\[ 0.005 v_{bullet} = 1.205 \times 0.949\]\[ v_{bullet} = \frac{1.205 \times 0.949}{0.005} \approx 229 \, ext{m/s}\]

Key Concepts

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion. Any object in motion—whether it's a car speeding down the highway or a tiny bullet flying through the air—has kinetic energy. The amount of kinetic energy (\[KE\]) an object has depends on two factors:

• Its mass (\[m\]) - the more massive the object, the more kinetic energy it possesses.
• Its velocity (\[v\]) - the faster the object moves, the more kinetic energy it has.

The formula for kinetic energy is given by:\[KE = \frac{1}{2}mv^2\]In the context of our exercise, the bullet and block system has kinetic energy just after the collision. This kinetic energy is entirely due to the initial speed of the bullet before it embeds itself in the block. Once the block begins to slide, this energy is gradually converted to other forms, primarily due to friction. Understanding kinetic energy helps us calculate how fast the bullet-block system was moving immediately after the bullet hit the block.

Work-Energy Principle

The work-energy principle is a powerful concept in physics that connects the work done on an object to its change in kinetic energy. In simple terms, it states that the work done on an object by external forces changes its kinetic energy. When work is done on an object, energy is transferred to or from that object.In the bullet and block problem:

• The bullet embeds itself into the block, giving the system a certain amount of kinetic energy as it starts sliding across the surface.
• As the bullet-block system slides, the kinetic friction force does work on it, gradually reducing its kinetic energy until it comes to a rest.

We calculate the work done by friction using the formula:\[Work = Friction \times Distance = \mu(m_{block} + m_{bullet})g \times \Delta x\]where \(\mu\) is the coefficient of friction, \(g\) is the acceleration due to gravity, and \(\Delta x\) is the distance slid. The work done by the friction is equal to the initial kinetic energy, allowing us to find the initial velocity of the bullet-block system.

Friction

Friction is the force that opposes motion between two surfaces that are in contact. In physics, it comes into play in various scenarios, influencing how objects move or stop. Friction can be divided into two main types:

• Static friction, which prevents objects from starting to move.
• Kinetic friction, which acts on objects that are already moving.

In this exercise, kinetic friction is what slows down the bullet-block system. When the bullet embeds in the block and they slide across the surface, the kinetic friction force works against their movement, reducing their speed until they stop.The force of friction depends on two primary factors:

• The coefficient of friction \((\mu)\), which is a measure of how "grippy" the surfaces are with respect to each other.
• The normal force, which is the perpendicular contact force exerted by a surface, often slightly affected by the mass of the object.

In this situation, the frictional force was given by:\[Friction = \mu kg (m_{block} + m_{bullet})\]Despite seeming like a nuisance, friction is essential as it is responsible for many real-world applications, such as writing with a pencil or coming to a stop in a car. Understanding how it impacts motion and energy is crucial for solving mechanical problems.