Problem 32 - College Physics

Question

At a shooting competition, a contestant fires and a 12.0-g bullet leaves the rifle with a muzzle speed of \(130 \mathrm{~m} / \mathrm{s}\). The bullet hits the thick target backing and stops after traveling \(4.00 \mathrm{~cm}\). Assuming a uniform acceleration, (a) what is the impulse on the target? (b) What is the average force on the target?

Step-by-Step Solution

Answer

Impulse is 1.56 Ns; average force is about 2535.12 N.

Step 1: Define Impulse and Force

Impulse on an object is given by the change in momentum of the object. For a bullet, the initial momentum is the mass times its initial velocity, and the final momentum is zero (since it stops). Average force is determined using the relationship between impulse and force over time.

Step 2: Calculate Bullet's Initial Momentum

First, let's calculate the initial momentum of the bullet using the formula: \( p = m imes v \), where \( m = 12.0 \, \text{g} = 0.012 \, \text{kg} \) and \( v = 130 \, \text{m/s} \). Thus, \( p = 0.012 \, \text{kg} \times 130 \, \text{m/s} = 1.56 \, \text{kg}\cdot\text{m/s}. \)

Step 3: Calculate Impulse on the Target (Part a)

The impulse \( I \) can be calculated as the change in momentum. Since the bullet stops, the impulse is equal to the initial momentum. Thus, \( I = 1.56 \, \text{kg}\cdot\text{m/s} \).

Step 4: Determine Deceleration

We can use the kinematic equation to find the deceleration: \( v^2 = u^2 + 2as \), where \( v = 0 \), \( u = 130 \, \text{m/s} \), and \( s = 0.04 \, \text{m} \). Substitute to find \( a \): \( 0 = (130)^2 + 2a(0.04) \) solving gives \( a = -211250 \, \text{m/s}^2 \).

Step 5: Calculate Time of Contact

Using the relation \( v = u + at \), where \( v = 0 \), \( u = 130 \, \text{m/s} \), and \( a = -211250 \, \text{m/s}^2 \), solve for \( t \): \( 0 = 130 + (-211250)t \), gives \( t \approx 0.0006154 \, \text{seconds} \).

Step 6: Calculate Average Force on the Target (Part b)

The average force \( F \) is given by impulse divided by time: \( F = \frac{I}{t} = \frac{1.56 \, \text{kg}\cdot\text{m/s}}{0.0006154 \, \text{s}} \approx 2535.12 \, \text{N}. \)

Step 7: Conclusion

The impulse on the target is \( 1.56 \, \text{Ns} \), and the average force exerted on the target is approximately \( 2535.12 \, \text{N}. \)

Key Concepts

Uniform Acceleration

In physics, uniform acceleration refers to a constant rate of change in velocity for an object. This means that an object undergoing uniform acceleration will speed up or slow down at a fixed amount over each unit of time. Think of it like an elevator that either moves up or down at a steady speed once it starts moving, regardless of the direction.

When dealing with problems such as this one involving a bullet hitting a target, uniform acceleration can help us model the change in the bullet's movement as it goes from speeding along a path to a complete stop.

In this particular exercise, the bullet, which is initially moving very fast, slows down uniformly until it halts, which is an example of negative acceleration, also known as deceleration. Understanding this concept is crucial because it allows us to calculate important details like the time it takes for the bullet to stop and the force applied on the target.

Kinematic Equations

Kinematic equations are formulas that describe the basic motion of objects. They help us calculate things like velocity, acceleration, and displacement without having to account for the forces behind the motion.

In this exercise, we use the equation \( v^2 = u^2 + 2as \) to find the acceleration (or deceleration in this case) acting on the bullet as it transitions from an initial speed of 130 m/s to a stop over a distance of 4 cm.

• \( v \) is the final velocity, which is 0 because the bullet stops.
• \( u \) is the initial velocity, 130 m/s for the bullet.
• \( a \) is the acceleration, which we are trying to find.
• \( s \) is the displacement, 0.04 m.

By substituting these values into the formula, we solve for \( a \), finding it to be a large negative number indicating rapid deceleration.

Kinematic equations are powerful tools in physics problems, making them fundamental for solving scenarios involving constant acceleration.

Average Force

To find the average force exerted by the bullet on the target, we use the relationship between impulse and force. Impulse is essentially a measure of change in momentum and can be determined by calculating the product of mass and velocity change, which in this case is equal to the initial momentum of the bullet since it stops.

Given that impulse \( I \) is also equal to the product of force \( F \) and the time \( t \) it takes for the change in velocity, you can rearrange the formula to find force.

• Formula: \( F = \frac{I}{t} \)
• Impulse \( I = 1.56 \text{ kg m/s} \)
• Time \( t = 0.0006154 \text{ s} \)
• This results in an average force of about 2535.12 N.

The average force is a fundamental concept when determining how objects interact over time, like the bullet pressing against the target.

Physics Problem Solving

Solving physics problems can seem daunting at first, but it becomes manageable once you break it down into smaller steps. In the context of this exercise, understanding and applying core concepts such as momentum, impulse, and uniform acceleration are key.

Successful problem solving in physics involves a few essential strategies: - Clearly define the problem and identify what is being asked. - Draw on known values and conceptual formulas (like kinematic equations) that apply to your situation. - Sequentially solve for the unknown variables. - Double-check calculations and ensure they align with physical reality.

Moreover, practicing with problems such as this one enhances your critical thinking and equips you with the capacity to tackle increasingly complex scenarios you might encounter not only in academic settings but real-world situations too. Understanding the process and developing a methodology is more beneficial than just reaching the correct answer.