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A \(5.00 \mathrm{~g}\) bullet is fired horizontally into a \(1.20 \mathrm{~kg}\) wooden block resting on a horizontal surface. The coefficient of kinetic friction between block and surface is \(0.20 .\) The bullet remains embedded in the block, which is observed to slide \(0.230 \mathrm{~m}\) along the surface before stopping. What was the initial speed of the bullet?

Short Answer

Expert verified
The initial speed of the bullet is approximately \(392 \mathrm{~m/s}\).

Step by step solution

01

Understand the Problem

We have a bullet and a block involved in an inelastic collision where the bullet embeds in the block. We need to find the initial speed of the bullet before impact. Given are the mass of the bullet (\(5.00 \mathrm{~g}\)), the mass of the block (\(1.20 \mathrm{~kg}\)), the friction coefficient (\(0.20\)), and the distance the block slides (\(0.230 \mathrm{~m}\)).
02

Calculate Work Done by Friction

Calculate the work done by friction as the block slides to a stop using the formula: \( W = f_k \times d \). The kinetic friction force \( f_k = \mu_k \times m_t \times g \) where \(\mu_k = 0.20\), \(m_t = 1.205 kg\) (total mass: block + bullet), and \(g = 9.8 \mathrm{~m/s^2}\). Substitute these values to find work done.
03

Use Work-Energy Principle

The work done by friction is equal to the initial kinetic energy of the bullet-block system (work-energy principle). Set \( W = \frac{1}{2} m_t v^2 \) and solve for \(v\), the speed of the combined system right after the collision.
04

Use Conservation of Momentum

Use conservation of momentum for the collision to relate the initial speed of the bullet to the speed of the bullet-block system just after collision. The formula is \(m_b v_b = m_t v\), where \(m_b = 0.005 \mathrm{~kg}\) (mass of bullet), \(v_b\) is initial speed of bullet, and \(v\) is the speed calculated from Step 3.
05

Calculate Initial Speed of Bullet

Rearrange the momentum conservation equation to find \(v_b\): \(v_b = \frac{m_t}{m_b} v\). Substitute the values of \(v\), \(m_t\), and \(m_b\) to find \(v_b\) - the initial speed of the bullet.

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Key Concepts

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

Kinetic Energy
Kinetic energy is the energy of motion. When an object moves, it has kinetic energy, which is given by the formula \( KE = \frac{1}{2} mv^2 \), where \( m \) is the mass and \( v \) is the velocity of the object. In the exercise, the bullet and block system acquires kinetic energy right after the collision. This energy comes from the initial motion of the bullet.
In an inelastic collision, like the one described in the exercise, not all kinetic energy is conserved. While the total energy remains constant, some kinetic energy turns into other energy forms, such as sound or thermal energy due to deformation or heat. Hence, the kinetic energy calculated initially will help determine the speed of the bullet-block system immediately after the collision. Understanding kinetic energy is crucial in solving motion and energy-related physics problems.
Conservation of Momentum
The conservation of momentum is a fundamental principle in physics stating that in an isolated system of colliding bodies, the total momentum remains constant before and after the collision. It applies to the exercise where a bullet embeds itself into a block, forming a single moving system.
We use the momentum conservation formula \( m_b v_b = m_t v \), where \( m_b \) is the mass of the bullet, \( v_b \) is its initial speed, \( m_t \) is the total mass post-collision, and \( v \) is their combined speed.
  • Momentum before collision = Momentum after collision.
  • The bullet's momentum is transferred to the block upon impact.
  • Solving this equation will allow us to determine the bullet's initial velocity.
By keeping the momentum constant, we can solve complex physics problems, like determining unknown velocities in collision scenarios.
Work-Energy Principle
The work-energy principle states that the work done by the forces acting on an object equals the change in its kinetic energy. In the context of the exercise, after the inelastic collision, the block slides and the work done by friction diminishes the system's kinetic energy until it comes to rest.
The principle is mathematically expressed as \( W = \Delta KE \), where \( W \) is the work done, and \( \Delta KE \) is the change in kinetic energy.
To calculate the work done by friction, we use \( W = f_k \times d \), where \( d \) is the distance the block slides, and \( f_k \) is the kinetic friction force.
  • Friction does negative work, reducing the system's kinetic energy.
  • The reduction in kinetic energy is equal to the initial kinetic energy post-collision.
Applying this principle enables us to understand how forces like friction affect the motion and energy of objects.
Kinetic Friction
Kinetic friction is the force that opposes the motion of two surfaces sliding against each other. In the exercise, it acts between the block and the surface, bringing the block-bullet system to a stop.
Kinetic friction force \( f_k \) is calculated by \( f_k = \mu_k \times m_t \times g \), where \( \mu_k \) is the coefficient of kinetic friction, \( m_t \) is the total mass, and \( g \) is the acceleration due to gravity.
  • This force works against the sliding object, causing it to decelerate.
  • Kinetic friction converts kinetic energy into heat.
  • Different surfaces have different coefficients of kinetic friction.
Understanding kinetic friction is vital for solving physics problems that involve motion and determining stopping distances.
Physics Problem Solving
Solving physics problems requires a methodical approach to identify knowns and unknowns, apply relevant principles, and compute results. For the exercise at hand, we addressed an inelastic collision scenario followed by energy transformation due to friction.
Here are key steps in effective problem-solving:
  • Comprehend the problem: Gather all given data and understand the system dynamics.
  • Identify applicable principles: Use conservation laws and the work-energy principle.
  • Verify calculations: Double-check computations to ensure accuracy.
  • Reflect on solutions: Consider whether results make physical sense in the given context.
By following these steps, anyone can systematically approach complex problems and achieve reliable solutions. Practicing this approach in various scenarios builds a more profound understanding of physics concepts.

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Most popular questions from this chapter

A rifle bullet with mass \(8.00 \mathrm{~g}\) strikes and embeds itself in a block with a mass of \(0.992 \mathrm{~kg}\) that rests on a frictionless, horizontal surface and is attached to a coil spring. (See Figure \(8.50 .\) ) The impact compresses the spring \(15.0 \mathrm{~cm}\). Calibration of the spring shows that a force of \(0.750 \mathrm{~N}\) is required to compress the spring \(0.250 \mathrm{~cm} .\) (a) Find the magnitude of the block's velocity just after impact. (b) What was the initial speed of the bullet?

To warm up for a match, a tennis player hits the \(57.0 \mathrm{~g}\) ball vertically with her racket. If the ball is stationary just before it is hit and goes \(5.50 \mathrm{~m}\) high, what impulse did she impart to it?

Archerfish are tropical fish that hunt by shooting drops of water from their mouths at insects above the water's surface to knock them into the water, where the fish can eat them. \(\mathrm{A} 65 \mathrm{~g}\) fish at rest just on the water's surface can expel a \(0.30 \mathrm{~g}\) drop of water in a short burst of \(5.0 \mathrm{~ms}\). High-speed measurements show that the water has a speed of \(2.5 \mathrm{~m} / \mathrm{s}\) just after the archerfish expels it. What is the average force the fish exerts on the drop of water? A. \(0.00015 \mathrm{~N}\) B. \(0.00075 \mathrm{~N}\) C. \(0.075 \mathrm{~N}\) D. \(0.15 \mathrm{~N}\)

You (mass \(55 \mathrm{~kg}\) ) are riding your frictionless skateboard (mass \(5.0 \mathrm{~kg}\) ) in a straight line at a speed of \(4.5 \mathrm{~m} / \mathrm{s}\) when a friend standing on a balcony above you drops a \(2.5 \mathrm{~kg}\) sack of flour straight down into your arms. (a) What is your new speed, while holding the flour sack? (b) since the sack was dropped vertically, how can it affect your horizontal motion? Explain. (c) Suppose you now try to rid yourself of the extra weight by throwing the flour sack straight up. What will be your speed while the sack is in the air? Explain.

Just as it has reached an upward speed of \(5.0 \mathrm{~m} / \mathrm{s}\) during a vertical launch, a rocket explodes into two pieces. Photographs of the explosion reveal that the lower piece, with a mass one-fourth that of the upper piece, was moving downward at \(3.0 \mathrm{~m} / \mathrm{s}\) the instant after the explosion. (a) Find the speed of the upper piece just after the explosion. (b) How high does the upper piece go above the point where the explosion occurred?

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