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Chapter 7: Problem 34

A \(10 \mathrm{~g}\) bullet moving directly upward at \(1000 \mathrm{~m} / \mathrm{s}\) strikes and passes through the center of a \(5.0 \mathrm{~kg}\) block initially at rest (Fig. 7-29). The bullet emerges from the block moving directly upward at \(400 \mathrm{~m} / \mathrm{s}\). To what maximum height does the block then rise above its initial position? (Hint: Use free-fall equations from Chapter 3.)

Short Answer

Expert verified
0.26 meters

Step by step solution

01

- Understand the initial momentum

Calculate the initial momentum of the bullet before collision. Use the formula for linear momentum: \( p = mv \), where \( m \) is mass and \( v \) is velocity. \( p_{bullet, initial} = 0.01 \text{ kg} \times 1000 \text{ m/s} \)
02

- Calculate the final momentum of the bullet

Determine the momentum of the bullet after it exits the block. \( p_{bullet, final} = 0.01 \text{ kg} \times 400 \text{ m/s} \)
03

- Determine the momentum transferred to the block

Calculate the change in momentum of the bullet, which is imparted to the block: \( \text{Change in momentum } = p_{bullet, initial} - p_{bullet, final} \)
04

- Find the velocity of the block

Use the change in momentum to determine the block's velocity just after the collision: \( p_{block} = (5.0 \text{ kg}) \times v_{block} \). Set this equal to the momentum transferred to solve for \( v_{block} \).
05

- Calculate the maximum height using the kinematic equation

Convert the block's upward velocity to the maximum height reached, using the kinematic equation: \( v^2 = u^2 + 2as \). Here, final velocity \( v \) is 0 m/s at the maximum height, initial velocity \( u \) is the velocity obtained in Step 4, acceleration \( a \) is -9.8 m/s虏 (due to gravity), and \( s \) is the height.
06

- Solve for the height

Rearrange the kinematic equation to solve for \( s \): \( s = \frac{u^2}{2g} \)
07

- Plug in the values

Input the block's velocity from Step 4 into the equation from Step 6 to find the maximum height.

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Key Concepts

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

Linear Momentum
Momentum is a fundamental concept in physics, representing the quantity of motion an object has. Defined as the product of mass and velocity, it is expressed by the formula: \( p = mv \). Linear momentum is particularly useful in studying collisions, where the total momentum of a system is conserved. In this problem, we first calculate the initial and final momentum of the bullet. Before the collision, the bullet's momentum is \( 0.01 \text{ kg} \times 1000 \text{ m/s} \). After it exits the block, its momentum is \( 0.01 \text{ kg} \times 400 \text{ m/s} \). The difference between these values gives the momentum transferred to the block.
Kinematic Equations
Kinematic equations describe the motion of objects without considering the forces that cause this motion. They are particularly useful when dealing with constant acceleration, such as the free-fall motion under gravity. In this problem, we use the kinematic equation: \( v^2 = u^2 + 2as \). Here, \( v \) is the final velocity (0 m/s at the maximum height), \( u \) is the initial velocity (obtained from the block's velocity just after the collision), \( a \) is the acceleration due to gravity (-9.8 m/s虏), and \( s \) is the height we need to find.
Free-Fall Motion
Free-fall is a special case of motion where gravity is the only force acting on the object. Under free-fall, all objects experience the same acceleration due to gravity, \( g = 9.8 \text{ m/s}^2 \). For the block in this problem, we can use kinematic equations to determine the maximum height it reaches after the bullet passes through it. Since the block is moving upwards against gravity, the final velocity at the highest point is 0 m/s, and we can calculate the distance it travels upwards using initial velocity.
Collision Dynamics
Collisions are interactions where objects come into contact with each other and exchange momentum and energy. There are different types of collisions, such as elastic and inelastic. In this exercise, the collision between the bullet and the block involves a transfer of momentum. By understanding the change in the bullet's momentum, we can find the momentum imparted to the block. Using this information, we can continue to solve for the block鈥檚 subsequent motion. This scenario emphasizes the conservation of momentum principle, which states that the total momentum before the collision equals the total momentum after the collision, assuming no external forces are involved.
Energy Conservation
The principle of energy conservation states that energy cannot be created or destroyed, only converted from one form to another. In this problem, kinetic energy of the bulky and the bullet before collision gets partly converted into gravitational potential energy of the block as it rises. The initial kinetic energy of the block can be found using: \( KE = \frac{1}{2}mv^2 \), where \( m \) is the block's mass and \( v \) is its velocity just after collision. That kinetic energy then translates into the gravitational potential energy at its peak height: \( PE = mgh \), where \( h \) is the height.

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Most popular questions from this chapter

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