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Block 1 of mass \(m_{1}\) slides along a frictionless floor and into a one- dimensional elastic collision with stationary block 2 of mass \(m_{2}=3 m_{1}\). Prior to the collision, the center of mass of the two-block system had a speed of \(3.00 \mathrm{~m} / \mathrm{s}\). Afterward, what are the speeds of (a) the center of mass and (b) block \(2 ?\)

Short Answer

Expert verified
(a) 3.00 m/s, (b) 3.00 m/s.

Step by step solution

01

Understand the System

Block 1 of mass \( m_1 \) is moving and collides elastically with block 2 of mass \( m_2 = 3m_1 \) which is initially at rest. We need to find the speeds after the collision of the center of mass and block 2.
02

Determine the Speed of the Center of Mass

Regardless of the collision, the speed of the center of mass remains the same as long as no external forces act on the system. It is given that before the collision, the speed of the center of mass is \(3.00\, \mathrm{m/s}\). Thus, the speed of the center of mass after the collision is also \(3.00\, \mathrm{m/s}\).
03

Apply Conservation of Momentum

The total momentum of the system before and after the collision is conserved. Initially, the total momentum is only due to block 1 since block 2 is at rest: \( p_{ ext{initial}} = m_1 v_1 + 0 = m_1 \cdot v_1 \). After the collision, the total momentum is \( p_{ ext{final}} = m_1 v_1' + 3m_1 v_2' \) where \(v_1'\) and \(v_2'\) are the final velocities of blocks 1 and 2.
04

Use Elastic Collision Equations

For elastic collisions, both momentum and kinetic energy are conserved. From conservation of momentum: \( m_1 v_1 = m_1 v_1' + 3m_1 v_2' \). From conservation of kinetic energy: \( \frac{1}{2} m_1 v_1^2 = \frac{1}{2} m_1 v_1'^2 + \frac{1}{2} \cdot 3m_1 v_2'^2 \).
05

Solve the System of Equations

Solve the equations from Step 4 for \(v_1'\) and \(v_2'\). Using the relative velocity method, we find: \( v_1' = \frac{m_1 - m_2}{m_1 + m_2} v_1 = -\frac{1}{2} v_1 \) and \( v_2' = \frac{2m_1}{m_1 + m_2} v_1 = \frac{1}{2} v_1 \). Substituting the original velocity \(v_1\) in terms of the center of mass speed \(3\, \mathrm{m/s}\), we get \( v_2' = \frac{1}{2} \times (6\, \mathrm{m/s}) = 3.00\, \mathrm{m/s} \) assuming symmetric initial velocity around the center of mass.

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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, conservation of momentum is a fundamental principle stating that the total momentum of a closed system of objects remains constant, provided no external forces act on it. In the context of an elastic collision, this means:
  • The total momentum before the collision equals the total momentum after the collision.
  • Momentum is the product of mass and velocity.
For example, when Block 1 (with mass \(m_1\) and initial velocity \(v_1\)) collides with Block 2 (with mass \(m_2 = 3m_1\)), the initial momentum is \(m_1 \cdot v_1\), as Block 2 is stationary. After the collision, the final momentum will be the sum of the momenta of both blocks: \(m_1 \cdot v_1' + 3m_1 \cdot v_2'\), where \(v_1'\) and \(v_2'\) are the final velocities of Blocks 1 and 2. This helps to compute unknown velocities if masses and one velocity are known.
Kinetic Energy
Kinetic energy in a system with elastic collisions is also conserved along with momentum. Kinetic energy is defined as the energy of motion and can be calculated using the formula:
  • Kinetic energy \( KE = \frac{1}{2} m v^2 \)
For our collision, the total initial kinetic energy is \( \frac{1}{2} m_1 v_1^2 \). After the collision, the total kinetic energy must equal this initial value, mathematically expressed as:
  • \( \frac{1}{2} m_1 v_1'^2 + \frac{1}{2} \cdot 3m_1 v_2'^2 \)
Even though individual kinetic energies of the blocks might change during the collision, the sum remains constant. This principle helps us ensure that all calculations regarding velocities are correct when verifying with the conservation of momentum.
Center of Mass
The center of mass of a system is a crucial concept when analyzing motions and collisions. It is the point where the mass of a system can be considered to be concentrated. In collision problems:
  • The velocity of the center of mass remains constant if no external forces are acting.
  • This means that external and internal mechanics transform separately around the center of mass.
For the given problem, the speed of the center of mass was initially \(3.00\, \mathrm{m/s}\) and remains \(3.00\, \mathrm{m/s}\) after the collision, illustrating the absence of external forces impacting the system.
Frictionless Surface
A frictionless surface is a hypothetical ideal that allows us to simplify calculations in physics problems involving motion, particularly in collisions. Here's why it's significant:
  • No energy is lost to friction, so only momentum and kinetic energy affect the motion.
  • The system remains isolated, aiding the analysis based on conserved quantities.
For our problem, a frictionless surface ensures that the motion of Block 1 and Block 2 can be evaluated without accounting for energy loss to the floor. This simplification means we only consider the direct effects of their interaction, making it easier to apply conservation laws accurately.

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