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Chapter 9: Problem 66

Block 1, with mass \(m_{1}\) and speed \(4.0 \mathrm{~m} / \mathrm{s}\), slides along an \(x\) axis on a frictionless floor and then undergoes a one-dimensional elastic collision with stationary block 2, with mass \(m_{2}=0.40 m_{1}\). The two blocks then slide into a region where the coefficient of kinetic friction is \(0.50 ;\) there they stop. How far into that region do (a) block 1 and \((b)\) block 2 slide?

Short Answer

Expert verified
Block 1 slides 0.30 m and Block 2 slides 1.20 m into the friction region.

Step by step solution

01

Understand Elastic Collision Exchange

First, let's understand the situation: Block 1 slides and hits Block 2 in a one-dimensional elastic collision. In such collisions, both momentum and kinetic energy are conserved.
02

Conservation of Momentum Equation

The law of conservation of momentum states that the total momentum before the collision is equal to the total momentum after. Before the collision, only Block 1 is moving, so:\[ m_1 \cdot 4.0 \mathrm{~m/s} = m_1 \cdot v_1 + m_2 \cdot v_2 \]Given that \(m_2 = 0.40 m_1\), we can simplify our variables, assuming \(v_1\) and \(v_2\) are velocities after the collision for Block 1 and Block 2, respectively.
03

Conservation of Kinetic Energy Equation

For elastic collisions, the kinetic energy is also conserved:\[ \frac{1}{2} m_1 \cdot (4.0)^2 = \frac{1}{2} m_1 \cdot v_1^2 + \frac{1}{2} m_2 \cdot v_2^2 \]Substitute \(m_2 = 0.40 m_1\) to further simplify the equation.
04

Solve System of Equations

We have two equations with variables \(v_1\) and \(v_2\). Substituting for \(m_2\) and solving them together, we find:\[ v_1 = 1.71 \mathrm{~m/s} \] and \[ v_2 = 3.43 \mathrm{~m/s} \]
05

Calculate Frictional Force and Deceleration

When the blocks enter the region with friction, they experience a force given by:\[ F_{friction} = \mu \cdot m \cdot g \]where \(\mu = 0.50\) and \(g = 9.8 \mathrm{~m/s^2}\). This frictional force causes deceleration (negative acceleration), so the deceleration is \( a = \mu \cdot g = 0.50 \cdot 9.8 \). Calculate \(a\).
06

Determine Stopping Distance for Block 1

Using the equation of motion with a final velocity of zero:\[ v_f^2 = v_1^2 - 2a d_1 \]Solve for \(d_1\) the distance Block 1 travels:\[ 0 = (1.71)^2 - 2 \cdot 4.9 \cdot d_1 \]
07

Determine Stopping Distance for Block 2

Similarly, for Block 2:\[ v_f^2 = v_2^2 - 2a d_2 \]Solve for \(d_2\):\[ 0 = (3.43)^2 - 2 \cdot 4.9 \cdot d_2 \]

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Key Concepts

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

Conservation of Momentum
In physics, the principle of conservation of momentum is a powerful tool in analyzing the motion of objects. It states that the total momentum of a closed system remains constant unless acted upon by an external force.
In the scenario where Block 1 collides with Block 2, we have a perfect example of a closed system. Before the collision, only Block 1 is moving with a speed of 4.0 m/s, giving it momentum:
  • Total initial momentum = \( m_1 \times 4.0 \)
  • Block 2 is stationary, so its initial momentum is zero.
After the collision, both blocks are moving. The total momentum before the collision equals the total momentum after.
  • \( m_1 \times 4.0 = m_1 \times v_1 + m_2 \times v_2 \)
This ensures that any change in velocity of Block 1 is exactly balanced by a change in Block 2's velocity, keeping the momentum total unchanged.
Conservation of Kinetic Energy
In elastic collisions, not only is momentum conserved, but kinetic energy is as well. This means the total kinetic energy of the system before the collision will be equal to the total kinetic energy after.
Before Block 1 hits Block 2, it has kinetic energy given by:
  • \( KE_{initial} = \frac{1}{2} m_1 (4.0)^2 \)
After the collision, both blocks have kinetic energy:
  • \( \frac{1}{2} m_1 v_1^2 + \frac{1}{2} m_2 v_2^2 \)
This equation can be used in conjunction with the conservation of momentum to solve for the velocities \( v_1 \) and \( v_2 \) post-collision. The conservation of kinetic energy ensures no energy is "lost" during the collision, which can help us accurately predict post-collision motion.
Coefficient of Kinetic Friction
When the blocks enter a region with friction, a new force comes into play, slowing them down. The coefficient of kinetic friction (\( \mu \)) describes how much frictional force acts on the blocks compared to the normal force. Here's how it works in this scenario:
  • The frictional force is \( F_{friction} = \mu \times m \times g \)
  • Given that \( \mu = 0.50 \), and \( g \) (acceleration due to gravity) is \( 9.8 \, \mathrm{m/s^2} \), the force acting against the blocks' motion can be calculated.
This friction causes deceleration, reducing the speeds of both blocks until they eventually stop. Understanding the coefficient of kinetic friction helps us appreciate how different surfaces can influence motion by affecting how quickly the blocks come to a stop.
Stopping Distance Calculation
Once the blocks enter the frictional region, their speeds begin to decrease due to the frictional force. To find out how far each block slides before stopping, we use the following equation of motion:
  • \( v_{f}^2 = v_i^2 - 2ad \)
The final velocity \( v_{f} \) is 0 m/s (since the blocks stop), \( v_i \) is the initial velocity in the friction region, and \( a \) is the deceleration caused by friction.
  • For Block 1: \( 0 = (1.71)^2 - 2 \times 4.9 \times d_1 \)
  • For Block 2: \( 0 = (3.43)^2 - 2 \times 4.9 \times d_2 \)
Solving these equations gives us the distances \( d_1 \) and \( d_2 \), representing how far each block travels before their movements are arrested by friction. A clear understanding of this concept helps in modeling the effects of friction in real-world scenarios.

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Most popular questions from this chapter

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