Problem 62 - College Physics

Question

A \(5.00 \mathrm{~g}\) bullet traveling horizontally at \(450 \mathrm{~m} / \mathrm{s}\) is shot through a \(1.00 \mathrm{~kg}\) wood block suspended on a string \(2.00 \mathrm{~m}\) long. If the center of mass of the block rises a distance of \(0.450 \mathrm{~cm},\) find the speed of the bullet as it emerges from the block.

Step-by-Step Solution

Answer

The bullet's speed as it emerges from the block is approximately 391 m/s.

Step 1: Determine Potential Energy Increase

First, calculate the potential energy increase of the block as it rises. Use the formula for gravitational potential energy: \( \Delta U = mgh \), where \( m \) is the mass of the block, \( g \) is the acceleration due to gravity (approximately \( 9.81 \mathrm{~m/s^2} \)), and \( h \) is the height it rises. Convert 0.450 cm to meters: \( h = 0.00450 \mathrm{~m} \). Substitute in the values: \( \Delta U = 1.00 \mathrm{~kg} \times 9.81 \mathrm{~m/s^2} \times 0.00450 \mathrm{~m} = 0.044145 \mathrm{~J} \).

Step 2: Use Conservation of Energy

By the conservation of energy, the initial kinetic energy transferred to the block by the bullet is equal to the potential energy at the highest point. The initial kinetic energy of the block \( K_i \) is \( \Delta U = 0.044145 \mathrm{~J} \). This is the kinetic energy imparted to the block by the bullet.

Step 3: Calculate Initial Velocity of Block

Use the kinetic energy formula \( K = \frac{1}{2} mv^2 \) to find the initial velocity \( v \) of the block. Set \( K = 0.044145 \mathrm{~J} \) and solve for \( v \): \( 0.044145 = \frac{1}{2} \times 1.00 \times v^2 \). Therefore, \( v = \sqrt{2 \times 0.044145} \approx 0.297 \mathrm{~m/s} \).

Step 4: Apply Conservation of Momentum

Apply the principle of conservation of momentum. Initial momentum of system = Final momentum of system. Initially, only the bullet contributes to momentum: \( p_i = 0.00500 \times 450 \). Final momentum = momentum of bullet after passing through + max momentum of block: \( p_f = 0.00500v_f + 1.00 \times 0.297 \). Equate and solve: \( 2.25 = 0.00500v_f + 0.297 \).

Step 5: Solve for Final Bullet Speed

Rearrange and solve for \( v_f \): \( 2.25 - 0.297 = 0.00500v_f \), thus \( v_f = \frac{1.953}{0.00500} = 390.6 \mathrm{~m/s} \).

Key Concepts

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion. It is directly proportional to the mass of the object and the square of its velocity. This is why faster-moving objects and those with more mass have significantly higher kinetic energy levels. In physics, we calculate kinetic energy using the formula:

• \( K = \frac{1}{2} mv^2 \)

where \( m \) is the mass and \( v \) is the velocity of the object.
In the exercise provided, when the bullet travels through the block, it initially transfers part of its kinetic energy to the block. This energy then gets transformed, leading to the block's movement. Therefore, analyzing changes in kinetic energy provides insights into how much energy is imparted or transformed in a dynamic system.

Gravitational Potential Energy

Gravitational potential energy refers to the energy an object has due to its position relative to the Earth or another gravitational field. It is determined by the height of the object and its mass, along with gravitational acceleration. The formula for gravitational potential energy is:

• \( U = mgh \)

where \( m \) is the mass, \( g \) is the gravitational acceleration (approximately \( 9.81 \, \text{m/s}^2 \) on Earth), and \( h \) is the height.When the wooden block in the exercise is pushed upward due to the bullet, it gains gravitational potential energy. This energy is a result of climbing against gravity, translated from the initial kinetic energy given to it by the bullet. This principle helps us understand positional energy and its dependence on height and mass.

Conservation of Energy

The law of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In closed systems, the total energy remains constant. This principle is pivotal in solving many physics problems, including the one involving the bullet and block. When the bullet transfers energy to the block, it is transformed from kinetic to potential energy as the block rises. The conservation of energy principle tells us that the kinetic energy given to the block's mass equates to the potential energy at the block's highest point. This forms the basis of the calculations in the exercise, ensuring the total energy before and after the event remains constant, confirming our solution's correctness.

Physics Problem Solving

Solving physics problems efficiently requires a clear strategic approach, often utilizing fundamental principles. Start by analyzing the problem, identifying known variables, and noting the unknowns. Use diagrams if necessary to better understand the problem setup.

• Apply relevant physics principles like conservation of momentum and energy.
• Translate physical scenarios into mathematical equations.
• Organize your solution into logical steps to trace how each part contributes to the overall solution.

In the exercise, these steps involved computing changes in potential energy and applying energy conservation principles. Then, using the conservation of momentum, we determined the bullet's final speed through mathematical equations. The careful balance of steps ensures that each principle is applied correctly, leading to accurate solutions.