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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?

Short Answer

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

Step by step solution

01

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.
02

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, \( p = 0.012 \, \text{kg} \times 130 \, \text{m/s} = 1.56 \, \text{kg}\cdot\text{m/s}. \)
03

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, \( I = 1.56 \, \text{kg}\cdot\text{m/s} \).
04

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 \( a = -211250 \, \text{m/s}^2 \).
05

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 \( t \approx 0.0006154 \, \text{seconds} \).
06

Calculate Average Force on the Target (Part b)

The average force \( F \) is given by impulse divided by time: \( F = \frac{I}{t} = \frac{1.56 \, \text{kg}\cdot\text{m/s}}{0.0006154 \, \text{s}} \approx 2535.12 \, \text{N}. \)
07

Conclusion

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.

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Most popular questions from this chapter

When bunting, a baseball player uses the bat to change both the speed and direction of the baseball. (a) Will the magnitude of the change in momentum of the baseball before and after the bunt be (1) greater than the magnitude of the momentum of the baseball either before or after the bunt, (2) equal to the difference between the magnitudes of momenta of the baseball before and after the bunt, or (3) equal to the sum of the magnitudes of momenta of the baseball before and after the bunt? Why? (b) The baseball has a mass of \(0.16 \mathrm{~kg}\); its speeds before and after the bunt are \(15 \mathrm{~m} / \mathrm{s}\) and \(10 \mathrm{~m} / \mathrm{s}\) respectively; the bunt lasts \(0.025 \mathrm{~s}\). What is the change in momentum of the baseball? (c) What is the average force on the ball by the bat?

A \(4.0-\mathrm{kg}\) ball with a velocity of \(4.0 \mathrm{~m} / \mathrm{s}\) in the \(+x\) -direction collides head-on elastically with a stationary \(2.0-\mathrm{kg}\) ball. What are the velocities of the balls after the collision?

A volleyball is traveling toward you. (a) Which action will require a greater force on the volleyball, your catching the ball or your hitting the ball back? Why? (b) A 0.45 -kg volleyball travels with a horizontal velocity of \(4.0 \mathrm{~m} / \mathrm{s}\) over the net. You jump up and hit the ball back with a horizontal velocity of \(7.0 \mathrm{~m} / \mathrm{s}\). If the contact time is \(0.040 \mathrm{~s}\), what was the average force on the ball?

Two runners of mass \(70 \mathrm{~kg}\) and \(60 \mathrm{~kg}\), respectively, have a total linear momentum of \(350 \mathrm{~kg} \cdot \mathrm{m} / \mathrm{s}\). The heavier runner is running at \(2.0 \mathrm{~m} / \mathrm{s}\). Determine the possible velocities of the lighter runner.

A ballistic pendulum is a device used to measure the velocity of a projectile- for example, the muzzle velocity of a rifle bullet. The projectile is shot horizontally into, and becomes embedded in, the bob of a pendulum, as illustrated in \(>\) Fig. \(6.35 .\) The pendulum swings upward to some height \(h,\) which is measured. The masses of the block and the bullet are known. Using the laws of momentum and energy, show that the initial velocity of the projectile is given by \(v_{\mathrm{o}}=[(m+M) / m] 22 g h\).

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