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Chapter 9: Problem 52

a \(10 \mathrm{~g}\) bullet moving directly upward at \(1000 \mathrm{~m} / \mathrm{s}\) strikes and passes through the center of mass of a \(5.0 \mathrm{~kg}\) block initially at rest. 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?

Short Answer

Expert verified
The block rises approximately 0.073 meters above its initial position.

Step by step solution

01

- Understand the Scenario

We have a 10 g bullet moving upward at 1000 m/s that passes through a 5 kg block at rest. After passing through, the bullet slows down to 400 m/s. We need to find the maximum height the block gains due to this interaction.
02

- Convert Units

Convert the bullet's mass from grams to kilograms for easier calculations. The mass of the bullet is 10 g, which is 0.01 kg.
03

- Apply Conservation of Momentum

Since the system is closed and isolated, we apply the conservation of momentum before and after the collision. The initial momentum of the system is the momentum of the bullet:Initial momentum: \[ p_i = m_b imes v_{b_i} = 0.01 imes 1000 = 10 \, \text{kg m/s} \]Final momentum: \[ p_f = m_b imes v_{b_f} + m_B imes v_B = 0.01 imes 400 + 5.0 imes v_B \]Set the initial and final momentum equal:\[ 10 = 0.01 imes 400 + 5.0 imes v_B \]Solving for \(v_B\), the velocity of the block after the collision.
04

- Solve for Block's Velocity

Solve the equation:\[ 10 = 0.01 imes 400 + 5.0 imes v_B \]\[ 10 = 4 + 5.0 imes v_B \]\[ 6 = 5.0 imes v_B \]\[ v_B = \frac{6}{5.0} = 1.2 \, \text{m/s} \]
05

- Calculate Maximum Height

Use the block's velocity after the collision to determine the maximum height it rises using energy conservation. Potential energy at the maximum height equals the initial kinetic energy.Kinetic energy:\[ KE = \frac{1}{2} m_B v_B^2 = \frac{1}{2} \times 5.0 \times (1.2)^2 = 3.6 \, \text{J} \]Potential energy at maximum height:\[ PE = m_B g h \]Equating kinetic energy to potential energy gives:\[ 3.6 = 5.0 \times 9.81 \times h \]Solve for \(h\):\[ h = \frac{3.6}{5.0 \times 9.81} = \frac{3.6}{49.05} \approx 0.073 \text{ meters} \]

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Key Concepts

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

Energy Conservation
Energy conservation is a fundamental principle in physics that states energy cannot be created or destroyed, only transferred or converted from one form to another. In the context of our exercise, after the bullet passes through the block, the block moves upwards due to the energy transferred to it.
  • The kinetic energy of the block after the collision is crucial. It's calculated using the formula \( KE = \frac{1}{2} m v^2 \), where \( m \) is the mass and \( v \) is the velocity of the block.
  • The maximum height the block reaches relates to potential energy. At the peak, all kinetic energy converts to potential energy.
  • We equate the initial kinetic energy to the potential energy at the highest point to find the maximum height.
Energy conservation allows us to solve for unknown variables, like maximum height, by equating energy forms. This ensures no energy is lost to the universe, adhering to this vital conservation law.
Collision Dynamics
Collision dynamics help us understand how bodies interact during collisions. Each type of collision, elastic or inelastic, affects speed and energy differently. In our scenario, the bullet and block experience a type of collision.
  • Initially, the bullet moves upward and impacts the stationary block.
  • The bullet loses some of its velocity to the block, imparting energy to the block.
The outcome reveals a momentum transfer since it is neither a purely elastic nor fully inelastic collision. Both objects still retain some motion afterward. The dynamics showcase the conservation of momentum principle, crucial in determining post-collision behavior.
Kinematics
Kinematics focuses on describing motion without regard to the forces that cause it. In this exercise, once the block gains motion from the bullet, kinematic equations help predict its path.
  • After the bullet collision, the block's velocity is calculated using momentum conservation.
  • This velocity helps in determining further motion aspects, like how high it can rise.
While kinematics doesn't account for the forces, it provides a way to measure the movement. Calculating how high the block rises falls into the domain of kinematics, using derived motion equations from initial velocity data.
Maximum Height Calculations
Maximum height calculations hinge on understanding how energy converts as an object moves. Once the block is in motion, its initial kinetic energy determines how high it will rise.
  • Identify the initial kinetic energy post-collision. This step involves knowing the block's mass and velocity.
  • Use the formula for potential energy \( PE = mgh \) to express the maximum height, where \( m \) is mass, \( g \) is gravitational acceleration, and \( h \) is height.
  • The maximum height formula comes from setting kinetic energy equal to potential energy since energy shifts forms as the block climbs.
This calculation is an application of the conservation of energy principle. It allows us to find how high the block will rise without additional forces acting upon it except gravity.

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Most popular questions from this chapter

A \(4.0 \mathrm{~kg}\) mess kit sliding on a frictionless surface explodes into two \(2.0 \mathrm{~kg}\) parts: \(3.0 \mathrm{~m} / \mathrm{s},\) due north, and \(5.0 \mathrm{~m} / \mathrm{s}, 30^{\circ}\) north of cast. What is the original speed of the mess kit?

A \(0.25 \mathrm{~kg}\) puck is initially stationary on an ice surface with negligible friction. At time \(t=0,\) a horizontal force begins to move the puck. The force is given by \(\vec{F}=\left(12.0-3.00 t^{2}\right) \mathrm{i},\) with \(\vec{F}\) in newtons and \(t\) in seconds, and it acts until its magnitude is zero. (a) What is the magnitude of the impulse on the puck from the force between \(t=0.500 \mathrm{~s}\) and \(t=1.25 \mathrm{~s} ?\) (b) What is the change in momentum of the puck between \(t=0\) and the instant at which \(F=0 ?\)

block \(A\) (mass \(1.6 \mathrm{~kg}\) ) slides into block \(B\) (mass \(2.4 \mathrm{~kg}\) ), along a frictionless surface. The directions of three velocities before \((i)\) and after \((f)\) the collision are indicated; the corresponding speeds are \(v_{A i}=5.5 \mathrm{~m} / \mathrm{s}\) \(v_{B i}=2.5 \mathrm{~m} / \mathrm{s},\) and \(v_{B f}=4.9 \mathrm{~m} / \mathrm{s} .\) What are the (a) speed and (b) direction (left or right) of velocity \(\vec{v}_{A f} ?\) (c) Is the collision elastic?

A bullet of mass \(10 \mathrm{~g}\) strikes a ballistic pendulum of mass \(2.0 \mathrm{~kg} .\) The center of mass of the pendulum rises a vertical distance of \(12 \mathrm{~cm}\). Assuming that the bullet remains embedded in the pendulum, calculate the bullet's initial speed.

A collision occurs between a \(2.00 \mathrm{~kg}\) particle traveling with velocity \(\vec{v}_{1}=(-4.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{i}}+(-5.00 \mathrm{~m} / \mathrm{s}) \hat{\mathrm{j}}\) and a \(4.00 \mathrm{~kg}\) particle traveling with velocity \(\vec{v}_{2}=(6.00 \mathrm{~m} / \mathrm{s}) \mathrm{i}+(-2.00 \mathrm{~m} / \mathrm{s}) \mathrm{j} .\) The colli- sion connects the two particles. What then is their velocity in (a) unit-vector notation and as a (b) magnitude and (c) angle?
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