Problem 42 - University Physics with Modern Physics

Question

A 5.00 -g bullet is fired horizontally into a 1.20 -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 initial speed of the bullet was approximately 228.09 m/s.

Step 1: Identify the Known Values and Formulas Needed

We are given the mass of the bullet, \(m_b = 5.00\, \text{g} = 0.005\, \text{kg}\), the mass of the block, \(m_B = 1.20\, \text{kg}\), the coefficient of kinetic friction, \(\mu_k = 0.20\), and the distance the block slides, \(d = 0.230\, \text{m}\). The initial velocity of the bullet is what we need to find.

Step 2: Calculate the Force of Friction

The force of kinetic friction \(f_k\) is given by the equation \(f_k = \mu_k \cdot m_B \cdot g\), where \(g = 9.81 \, \text{m/s}^2\) is the acceleration due to gravity. Substituting the values, we find \(f_k = 0.20 \times 1.20 \, \text{kg} \times 9.81 \, \text{m/s}^2 = 2.3544 \, \text{N}\).

Step 3: Calculate the Work Done by Friction

The work done by the friction force is \(W = f_k \cdot d\). Substituting the values from previous steps, \(W = 2.3544 \, \text{N} \times 0.230 \, \text{m} = 0.541512 \, \text{J}\).

Step 4: Use Conservation of Energy

The initial kinetic energy of the bullet and block system is equal to the work done by friction. The kinetic energy is \(\frac{1}{2} (m_b + m_B) v^2\), where \(v\) is the velocity of the bullet and block after the collision. \(\frac{1}{2} (0.005 + 1.20) v^2 = 0.541512\).

Step 5: Solve for the Initial Velocity

Solving for \(v\), we have \(v^2 = \frac{2 \times 0.541512}{0.005 + 1.20}\). Calculate \(v =\sqrt{\frac{2 \times 0.541512}{1.205}} = \sqrt{0.898363566}\approx 0.947\, \text{m/s}\). This is the speed of the block and bullet just after impact. The initial speed of the bullet before impact is found by considering momentum.

Step 6: Apply Conservation of Momentum

The initial momentum of the bullet \((m_b \cdot u)\) equals the final momentum of the combined block and bullet system \((m_b + m_B) \cdot v\). \(0.005\, \text{kg} \cdot u = (0.005 + 1.20) \times 0.947\). Solve for \(u\), the initial speed of the bullet: \(u = \frac{1.205 \times 0.947}{0.005} \approx 228.09\, \text{m/s}\).

Key Concepts

Kinetic Friction

When the bullet embeds itself into the wooden block, it causes the block to slide across a horizontal surface. As the block moves, it encounters a resistive force known as kinetic friction. This force acts opposite to the block's motion and eventually brings it to a stop.
The kinetic frictional force is dependent on two key factors: the coefficient of kinetic friction (\( \mu_k \)) and the normal force (usually the weight of the object sliding). The formula is:

• \( f_k = \mu_k \cdot m_B \cdot g \)

where \( g \) is the acceleration due to gravity. In this scenario, \( \mu_k = 0.20 \) and the block's mass, \( m_B \), is 1.20 kg. This results in a frictional force of 2.3544 N. Kinetic friction is crucial in calculating how far the block will travel after impact, as it determines the work done to stop the block.

Conservation of Momentum

The principle of conservation of momentum is fundamental in collisions like the one in this exercise. Momentum, a measure of motion, is given by the product of mass and velocity.

In this collision between the bullet and the block, momentum is conserved. This means the total momentum before the impact equals the total momentum after the impact:

• \( m_b \cdot u = (m_b + m_B) \cdot v \)

Before impact, the bullet has initial speed \( u \) and the block is at rest. After impact, both move together at speed \( v \).
Using the given masses and calculated speed \( v = 0.947 \) m/s, we found the bullet's initial speed \( u \) to be approximately 228.09 m/s. This conservation principle allows us to work backward and unveil the bullet's initial velocity.

Bullet-Block Collision

In this exercise, a bullet-block collision is an example of an inelastic collision. After the collision, the bullet and block move together as one mass. This type of collision is characterized by the objects sticking together.

Inelastic collisions conserve momentum but do not conserve kinetic energy. The sticking together of the objects and motion as a combined unit is why the velocity \( v \) of the system can be calculated using conservation of momentum. To fully understand inelastic collisions like this, one must apply both momentum and energy considerations, recognizing that some mechanical energy is transformed into other forms, like sound or heat.

Kinetic Energy

Kinetic energy is the energy an object has due to its motion. It is calculated as:

• \( KE = \frac{1}{2} mv^2 \)

In the bullet-block collision, the initial kinetic energy of the system is provided by the moving bullet. After the collision, the block and bullet together have their kinetic energy due to movement.
The kinetic energy plays a critical role in calculating how much work the friction force does to stop the system. Here, the initial kinetic energy converts completely to the work needed to stop the block after sliding 0.230 m. This relationship allowed us to set up equations necessary to find the post-collision speed \( v \).

Conservation of Energy

While momentum is conserved in this collision, kinetic energy is not entirely conserved. However, conservation of energy as a principle is still vital in understanding the outcome of the collision.

The overall idea is that the total amount of energy - kinetic and other forms - in an isolated system remains constant. In our case, the initial kinetic energy ultimately transforms into other forms such as heat due to friction and sound.

• Total initial energy = Total energy after accounting for work done (mainly due to kinetic friction)

This transformation principle underpins the entire event from when the bullet strikes until the block stops sliding. Thus, while kinetic energy in the conventional form is not conserved, energy conservation as a whole governs the changes occurring throughout the collision and subsequent sliding.