/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 16 A 4.25 g bullet traveling horizo... [FREE SOLUTION] | 91Ó°ÊÓ

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A 4.25 g bullet traveling horizontally with a velocity of magnitude \(375 \mathrm{~m} / \mathrm{s}\) is fired into a wooden block with mass \(1.12 \mathrm{~kg},\) initially at rest on a level frictionless surface. The bullet passes through the block and emerges with its speed reduced to \(122 \mathrm{~m} / \mathrm{s} .\) How fast is the block moving just after the bullet emerges from it?

Short Answer

Expert verified
The block moves at 0.959 m/s after the bullet emerges.

Step by step solution

01

Identify the known values

We start by listing all the given information. The mass of the bullet is \( m_b = 4.25 \text{ g} = 0.00425 \text{ kg} \), the initial velocity of the bullet is \( v_{b_i} = 375 \text{ m/s} \), the final velocity of the bullet is \( v_{b_f} = 122 \text{ m/s} \), and the mass of the wooden block is \( m_B = 1.12 \text{ kg} \). The initial velocity of the block is \( v_{B_i} = 0 \text{ m/s} \) as it is initially at rest.
02

Apply the principle of conservation of momentum

Since the system is isolated and the block's surface is frictionless, the total momentum before and after the collision must be the same. The momentum before the collision is given by \( p_{initial} = m_b \times v_{b_i} + m_B \times v_{B_i} \). The momentum after the collision is \( p_{final} = m_b \times v_{b_f} + m_B \times v_{B_f} \), where \( v_{B_f} \) is the velocity of the block after the bullet emerges.
03

Set up the conservation of momentum equation

The equation becomes: \[ m_b \times v_{b_i} = m_b \times v_{b_f} + m_B \times v_{B_f} \]Substitute the values into the equation:\[(0.00425 \times 375) = (0.00425 \times 122) + (1.12 \times v_{B_f})\]
04

Solve for the block's final velocity

First calculate the momentum of the bullet before and after the collision: - Before = \( 0.00425 \times 375 = 1.59375 \text{ kg m/s} \)- After = \( 0.00425 \times 122 = 0.5185 \text{ kg m/s} \)Substitute these values into the conservation equation and solve for \( v_{B_f} \):\[ 1.59375 = 0.5185 + 1.12 \times v_{B_f} \]\[ 1.59375 - 0.5185 = 1.12 \times v_{B_f} \]\[ 1.07525 = 1.12 \times v_{B_f} \]Finally, solve for \( v_{B_f} \):\[ v_{B_f} = \frac{1.07525}{1.12} = 0.959 \text{ m/s} \]
05

Conclusion

The speed of the block just after the bullet passes through it is \( v_{B_f} = 0.959 \text{ m/s} \).

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Key Concepts

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

Bullet and Block Collision
In the given physics problem, we have a fascinating scenario of a bullet emerging from a block after a collision. This situation reflects an important application of the conservation of momentum principle. A 4.25 g bullet with a high velocity penetrates a stationary block on a frictionless surface. Such experiments help us understand how objects interact dynamically.
This bullet-block system offers an excellent example of how momentum transfers occur when two objects collide. The bullet's velocity shifts from 375 m/s before the impact to a reduced 122 m/s after passing through the block. Meanwhile, we need to determine the velocity of the block, which was initially at rest. Both the bullet and block experience changes in motion and energy. Through this understanding, students learn that the forces involved can considerably change the way objects move and behave after a collision.
Momentum Principles
The principle central to this exercise is the conservation of momentum. Momentum, which is the product of mass and velocity, remains unchanged in an isolated system. During the bullet and block collision, no external forces act on the system, thus conserving momentum.
  • The initial total momentum includes the bullet's momentum since the block is initially stationary.
  • After the collision, the total momentum comprises both the bullet's and the block's contributions.

Mathematically, this conservation is expressed as:\[ m_b v_{b_i} = m_b v_{b_f} + m_B v_{B_f} \]Here, the bullet and block exchange momentum, leading to changes in their respective velocities.
Through analyzing these quantities, students can calculate the block's velocity after the collision. Understanding this principle enables students to solve complex real-world problems, predicting how systems react following interactions and energy exchanges.
Physics Problem Solving
Solving physics problems like this bullet and block collision involves systematic approaches. First, students should clearly identify the known and unknown variables, like bullet mass, initial and final velocities, and the block's mass. Then, engaging in problem-solving with the correct application of physics principles ensures successful results.
The process starts with identifying the type of collision and calculating initial and final momentums. Students should substitute known values into the conservation of momentum equation. After simplifications, solving for the unknown variable, like the block's final velocity, becomes straightforward.
  • Identify Known Values: This defines the parameters of the problem.
  • Define Equations: Use physics laws, like momentum conservation, fitting the situation.
  • Compute: Use algebra to isolate and find the unknowns.
  • Conclusion: Review the calculated results within the physical context.
This step-by-step method enables students to approach not just this problem, but any similar physical interactions with confidence, ultimately fostering analytical skills crucial for further scientific pursuits.

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Most popular questions from this chapter

On a very muddy football field, a \(110 \mathrm{~kg}\) linebacker tackles an \(85 \mathrm{~kg}\) halfback. Immediately before the collision, the linebacker is slipping with a velocity of \(8.8 \mathrm{~m} / \mathrm{s}\) north and the halfback is sliding with a velocity of \(7.2 \mathrm{~m} / \mathrm{s}\) east. What is the velocity (magnitude and direction) at which the two players move together immediately after the collision?

A rocket is fired in deep space, where gravity is negligible. If the rocket has an initial mass of \(6000 \mathrm{~kg}\) and ejects gas at a relative velocity of magnitude \(2000 \mathrm{~m} / \mathrm{s}\), how much gas must it eject in the first second to have an initial acceleration of \(25.0 \mathrm{~m} / \mathrm{s}^{2} ?\)

2On a highly polished, essentially frictionless lunch counter, a \(0.500 \mathrm{~kg}\) submarine sandwich moving \(3.00 \mathrm{~m} / \mathrm{s}\) to the left collides with a \(0.250 \mathrm{~kg}\) grilled cheese sandwich moving \(1.20 \mathrm{~m} / \mathrm{s}\) to the right. (a) If the two sandwiches stick together, what is their final velocity? (b) How much mechanical energy dissipates in the collision? Where did this energy go?

A \(2 \mathrm{~kg}\) block is moving at \(5 \mathrm{~m} / \mathrm{s}\) along a frictionless table and collides with a second \(2 \mathrm{~kg}\) block that is initially at rest. After the collision, the two blocks stick together and then slide up a \(45^{\circ}\) frictionless inclined plane, as shown in Figure \(8.41 .\) Calculate the maximum distance \(L\) that the two blocks travel up the incline.

A block of ice with a mass of \(2.50 \mathrm{~kg}\) is moving on a frictionless, horizontal surface. At \(t=0,\) the block is moving to the right with a velocity of magnitude \(8.00 \mathrm{~m} / \mathrm{s}\). Calculate the magnitude and direction of the velocity of the block after each of the following forces has been applied for \(5.00 \mathrm{~s}\) : (a) a force of \(5.00 \mathrm{~N}\) directed to the right; (b) a force of \(7.00 \mathrm{~N}\) directed to the left.

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