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Chapter 8: Problem 72

A small wooden block with mass 0.800 kg is suspended from the lower end of a light cord that is 1.60 m long. The block is initially at rest. A bullet with mass 12.0 \(g\) is fired at the block with a horizontal velocity \(v_0\). The bullet strikes the block and becomes embedded in it. After the collision the combined object swings on the end of the cord. When the block has risen a vertical height of 0.800 m, the tension in the cord is 4.80 N. What was the initial speed \(v_0\) of the bullet?

Short Answer

Expert verified
The initial speed of the bullet \( v_0 \) is approximately 232 m/s.

Step by step solution

01

Understand the Problem

We're dealing with a conservation of momentum problem followed by energy conservation. First, the bullet embeds itself in the block, and then the combined mass rises to a certain height due to the momentum transferred by the bullet. Our task is to find the initial speed \( v_0 \) of the bullet.
02

Apply Conservation of Energy

First, we need to determine the velocity of the block and bullet immediately after the collision. When the block rises to 0.800 m, the kinetic energy has transformed into potential energy and some tension energy. Use: \[ mgh = \frac{1}{2} mv^2 \] with \( h = 0.800 \text{ m} \). Calculate the velocity \( v \) of the block when it reaches this height.
03

Calculate Tension Contribution to Energy

The tension in the cord at the highest point is 4.80 N. Since the tension does not affect the vertical position directly, we relate it through forces in a circular motion. The tension also contributes as \( T - mg = \frac{mv^2}{r} \), where \( v \) is the velocity just before reaching the 0.8 m height and \( r \) is the radius here equal to \( 1.6 - 0.8 = 0.8 \). Solve for \( v \).
04

Solve for Initial Velocity After Collision

Apply the conservation of energy at the top point: \( mgh = \frac{1}{2} mv^2 \). Calculate \( v \) using \( g = 9.8 \text{ m/s}^2 \).
05

Apply Conservation of Momentum

Using the found velocity \( v \) after collision, apply the conservation of momentum. The initial momentum \( mv_0 \) equals the final momentum \( (m_1 + m_2)v \): \( 0.012 v_0 = (0.800 + 0.012)v \). Solve for the initial velocity \( v_0 \) of the bullet.

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Key Concepts

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

Conservation of Momentum
In any closed system with no external forces, the momentum before an event is equal to the momentum after. This principle is crucial in understanding collisions like the one between the bullet and the block in our exercise. Momentum is the product of mass and velocity, and in this scenario, we calculate the momentum of both bullet and block as they interact.
  • Momentum before collision is given by the bullet: \( \text{momentum} = m_{bullet} \times v_0 \).
  • Momentum after the collision includes both the bullet and block: \( (m_{bullet} + m_{block}) \times v \).
We utilize these equations to link the pre and post-collision movements, enabling us to find the bullet's initial velocity. This step is essential as it helps in transforming the system's kinetic profile without any energy being lost, just changing forms.
Conservation of Energy
The principle of conservation of energy states that energy cannot be created or destroyed, only transformed from one form to another. In this exercise, initial kinetic energy converts to potential and rotational energy eventually.
  • Initially, the bullet's kinetic energy: \( \frac{1}{2} m_{bullet} v_0^2 \).
  • As the bullet embeds into the block, the system's kinetic energy gradually transforms to potential energy: \( m_{total}gh \).
Understanding this transition is crucial to determining how high the block swings after absorbing the bullet's impact. The energy transformation helps explain why the system behaves as observed, providing a robust technique to calculate unknown quantities such as velocities.
Circular Motion
Circular motion comes into play once the bullet and block move as a single entity. Post-collision, the block and bullet swing on a cord, presenting a classic example of circular dynamics.
  • In circular motion, forces ensure the rotating paths are consistently followed.
  • Tension in the cord is pivotal, contributing alongside gravitational force to maintain circular motion.
  • The centripetal force required is provided by tension, calculated using: \( F = \frac{mv^2}{r} \), where \(r\) is the effective length they swing through.
Centripetal and tension forces are crucial in keeping everything in motion. By analyzing these forces, you understand the balance needed for sustained circular trajectories.
Kinetic Energy
Kinetic energy is a form of energy associated with motion. Initially, the bullet possesses kinetic energy, which is partially transferred to the wood block upon impact, causing both to move jointly.
  • Kinetic energy at collision onset is \( \frac{1}{2} mv^2 \).
  • As the block rises, kinetic energy decreases and converts to potential energy.
After the merger of bullet and block, their composite kinetic energy is less than that of the lone bullet but is pivotal in translating the motion into potential energy, explaining how high they rise in the absence of external work.
Potential Energy
Potential energy relates to the position of an object within a force field, here primarily gravity. As the block and bullet rise, their system gains potential energy at a loss of kinetic energy.
  • Potential energy is given by: \( PE = mgh \).
  • Higher elevation translates to more potential energy stored.
  • This energy increase implies work done against gravity.
Understanding potential energy allows us to see how energy conservation remains intact throughout the motion. As kinetic energy diminishes, the corresponding rise in potential energy reflects the height reached, making it a key aspect of analyzing the post-collision system behavior.

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Most popular questions from this chapter

You and your friends are doing physics experiments on a frozen pond that serves as a frictionless, horizontal surface. Sam, with mass 80.0 kg, is given a push and slides eastward. Abigail, with mass 50.0 kg, is sent sliding northward. They collide, and after the collision Sam is moving at 37.0\(^\circ\) north of east with a speed of 6.00 m/s and Abigail is moving at 23.0\(^\circ\) south of east with a speed of 9.00 m/s. (a) What was the speed of each person before the collision? (b) By how much did the total kinetic energy of the two people decrease during the collision?

Friends Burt and Ernie stand at opposite ends of a uniform log that is floating in a lake. The log is 3.0 m long and has mass 20.0 kg. Burt has mass 30.0 kg; Ernie has mass 40.0 kg. Initially, the log and the two friends are at rest relative to the shore. Burt then offers Ernie a cookie, and Ernie walks to Burt's end of the log to get it. Relative to the shore, what distance has the log moved by the time Ernie reaches Burt? Ignore any horizontal force that the water exerts on the log, and assume that neither friend falls off the log.

The mass of a regulation tennis ball is 57 g (although it can vary slightly), and tests have shown that the ball is in contact with the tennis racket for 30 ms. (This number can also vary, depending on the racket and swing.) We shall assume a 30.0-ms contact time. The fastest-known served tennis ball was served by "Big Bill" Tilden in 1931, and its speed was measured to be 73 m/s. (a) What impulse and what force did Big Bill exert on the tennis ball in his record serve? (b) If Big Bill's opponent returned his serve with a speed of 55 m/s, what force and what impulse did he exert on the ball, assuming only horizontal motion?

Obviously, we can make rockets to go very fast, but what is a reasonable top speed? Assume that a rocket is fired from rest at a space station in deep space, where gravity is negligible. (a) If the rocket ejects gas at a relative speed of 2000 m/s and you want the rocket's speed eventually to be 1.00\(\times\) 10\(^{-3}c\), where \(c\) is the speed of light in vacuum, what fraction of the initial mass of the rocket and fuel is \(not\) fuel? (b) What is this fraction if the final speed is to be 3000 m/s?

Two skaters collide and grab on to each other on frictionless ice. One of them, of mass 70.0 kg, is moving to the right at 4.00 m/s, while the other, of mass 65.0 kg, is moving to the left at 2.50 m/s. What are the magnitude and direction of the velocity of these skaters just after they collide?
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