Problem 12 - Applied Physics

Question

A \(16.0\) -g bullet is shot into a wooden block at rest with mass \(4550 \mathrm{~g}\) on a frictionless surface. The block moves \(1.20 \mathrm{~m} / \mathrm{s}\) after the bullet strikes and becomes lodged in the block. Find the speed of the bullet before striking the block.

Step-by-Step Solution

Answer

The bullet's speed before impact is approximately 342 m/s.

Step 1: Identify Given Values and Convert Units

We are given the masses of the bullet and the block as well as the final velocity of the block-bullet system. Convert the masses from grams to kilograms for consistency in units: 16.0 g = 0.016 kg and 4550 g = 4.550 kg. The final velocity is already given as 1.20 m/s.

Step 2: Understand the Conservation of Momentum

In an isolated system, the total momentum before the interaction is equal to the total momentum after the interaction. This is expressed as: \( m_{bullet} \times v_{bullet,initial} = (m_{bullet} + m_{block}) \times v_{final} \), where \( v_{bullet,initial} \) is the speed of the bullet before striking the block.

Step 3: Set Up the Equation Using Momentum Conservation

Based on the conservation of momentum, substitute the known values into the equation: \( (0.016) \times v_{bullet,initial} = (0.016 + 4.550) \times 1.20 \). Simplify this to find \( v_{bullet,initial} \).

Step 4: Solve for Initial Bullet Speed

Calculate the right side of the equation to find the momentum after the collision: \((0.016 + 4.550) \times 1.20 = 5.4664 \). To find \( v_{bullet,initial} \), solve the equation: \( v_{bullet,initial} = \frac{5.4664}{0.016} \).

Step 5: Compute Final Value

Perform the division to find the initial speed of the bullet. \( v_{bullet,initial} = 341.65 \). Conclude that the speed of the bullet before striking the block is approximately 342 m/s.

Key Concepts

Bullet and Block Collision

When a bullet collides with a block, it creates a captivating real-world example of momentum in action. Imagine a small-sized bullet traveling at high speed suddenly striking a much heavier wooden block. What happens next? The bullet lodges itself into the block, and together, they move in the same direction over a frictionless surface.

A crucial aspect of this setup is the interaction between the two objects during the collision process. Since the bullet embeds into the block, they form a single system post-collision. Hence, despite a significant difference in masses, both move as one unit after the collision. This transition from two separate bodies into a single entity affects how their velocities are evaluated.

• The bullet, being of only 16 grams, embeds into a 4550-gram block.
• The post-collision velocity of the combined mass is given as 1.20 m/s.

These points highlight that the velocity of the resulting system relies heavily on the initial conditions — which leads to the principle that helps analyze this motion.

Initial Speed Calculation

Understanding how to find an object's speed before a collision requires us to delve into the calculations themselves. We start with known variables; in this exercise, we're given the masses of the bullet and block and the final system velocity. First, ensure all units are consistent, typically converting grams to kilograms for mass.

To find the initial speed of the bullet, we work with the momentum conservation equation given by \[m_{bullet} \times v_{bullet,initial} = (m_{bullet} + m_{block}) \times v_{final}\].Substituting the known values is the next step:

• Bores down to: \((0.016) \times v_{bullet,initial} = (0.016 + 4.550) \times 1.20\)
• Calculating the right side: \((4.566) \times 1.20 = 5.4664\)
• Finally, solve for \(v_{bullet,initial}\): \[v_{bullet,initial} = \frac{5.4664}{0.016} \approx 341.65 \, \text{m/s}\]

And there we have it, illustrating the full power of algebraic manipulation married to the principles of physics!

Momentum Conservation Principle

The Momentum Conservation Principle is a cornerstone in understanding how objects interact during collisions. It tells us that in an isolated system, like our bullet and block example, the total momentum before the collision is exactly equal to the total momentum after.

Let's break down the components:

• **Momentum before collision**: Only the bullet is moving, so \[Momentum_{before} = m_{bullet} \times v_{bullet,initial}\]
• **Momentum after collision**: The bullet and block move together, hence \[Momentum_{after} = (m_{bullet} + m_{block}) \times v_{final}\]

By applying the momentum conservation principle, we establish a direct relation that lets us compute what happened before the collision from what we observe afterward. In many physics problems, this principle acts as an unbreakable rule, simplifying otherwise complex problems by maintaining equality of momentum. It effectively serves as a lens through which initial conditions can be predicted or backtracked, allowing us to solve for unknowns like the bullet's initial speed in frictionless environments.