Question

Blocks \(A\) (mass 2.00 kg) and \(B\) (mass 10.00 kg, to the right of \(A\)) move on a frictionless, horizontal surface. Initially, block \(B\) is moving to the left at 0.500 m/s and block \(A\) is moving to the right at 2.00 m/s. The blocks are equipped with ideal spring bumpers, as in Example 8.10 (Section 8.4). The collision is headon, so all motion before and after it is along a straight line. Find (a) the maximum energy stored in the spring bumpers and the velocity of each block at that time; (b) the velocity of each block after they have moved apart.

Short answer

(a) Maximum spring energy is 5.0 J, and both blocks have velocity -1/12 m/s. (b) After separation, \( v_{A,f} = -1.75 \, \text{m/s} \) and \( v_{B,f} = 0.33 \, \text{m/s} \).

Step-by-step solution

Understand the Conservation of Momentum

The initial momentum of the system must equal the momentum at the point of maximum compression as momentum is conserved in elastic collisions. Given that \( v_{A,i} = 2.00 \, \text{m/s} \) and \( v_{B,i} = -0.500 \, \text{m/s} \) (since \(B\) is moving left), we write:\[ m_A v_{A,i} + m_B v_{B,i} = m_A v_{A} + m_B v_{B} \]Substituting the known values: \[ 2.00 \cdot 2.00 + 10.00 \cdot (-0.500) = 2.00 \cdot v_{A} + 10.00 \cdot v_{B} \] Simplifying gives us: \[ 4 - 5 = 2v_A + 10v_B \] \[ -1 = 2v_A + 10v_B \] (equation 1).

Use Energy Conservation at Maximum Compression

At maximum compression, all of the kinetic energy that is lost is converted into potential energy stored in the spring. The initial kinetic energy is:\[ KE_{initial} = \frac{1}{2} m_A v_{A,i}^2 + \frac{1}{2} m_B v_{B,i}^2 \]Substitute the known values:\[ KE_{initial} = \frac{1}{2} \cdot 2.00 \cdot (2.00)^2 + \frac{1}{2} \cdot 10.00 \cdot (-0.500)^2 \]\[ KE_{initial} = 4 + 1.25 = 5.25 \, \text{J} \]This converts to kinetic energy at maximum compression \(KE_{compression}\) and spring energy \(E_{spring}\):\[ KE_{initial} = KE_{compression} + E_{spring} \]\[ E_{spring} = KE_{initial} - KE_{compression} \]

Express Velocities at Maximum Compression

At maximum compression, velocities of both blocks are equal; let this common velocity be \(v\):\[ 2v_A + 10v_B = 2v + 10v = 12v \]From equation 1: \[ -1 = 12v \] \[ v = -\frac{1}{12} \, \text{m/s} \] So \( v_A = v_B = -\frac{1}{12} \, \text{m/s} \) at maximum compression.

Calculate Maximum Spring Energy

Substitute velocities into stored energy equation:\[ KE_{compression} = \frac{1}{2} \cdot 2.00 \cdot (-\frac{1}{12})^2 + \frac{1}{2} \cdot 10.00 \cdot (-\frac{1}{12})^2 \]\[ KE_{compression} = \frac{1}{24} + \frac{5}{24} = \frac{6}{24} = 0.25 \, \text{J} \]\[ E_{spring} = KE_{initial} - KE_{compression} = 5.25 - 0.25 = 5.0 \, \text{J} \]

Determine Velocities After Separation

For velocities post-collision, use the elastic collision properties where kinetic energy is conserved:Post-collision velocities:\[ v_{A,f} = \frac{(m_A - m_B)}{(m_A + m_B)}v_{A,i} + \frac{2m_B}{(m_A + m_B)}v_{B,i} \]\[ v_{A,f} = \frac{(2 - 10)}{12} \times 2 + \frac{20}{12} \times (-0.5) \]\[ v_{A,f} = \frac{-8}{12} \times 2 + \frac{10}{12}(-0.5) \]\[ v_{A,f} = -1.33 \text{ m/s} + (-0.42 \text{ m/s}) = -1.75 \text{ m/s} \]Similarly for \(B\):\[ v_{B,f} = \frac{2m_A}{(m_A + m_B)}v_{A,i} + \frac{(m_B - m_A)}{(m_A + m_B)}v_{B,i} \]\[ v_{B,f} = \frac{4}{12} \times 2 + \frac{8}{12} \times (-0.5) \]\[ v_{B,f} = \frac{8}{12} - rac{4}{12} = 0.33 \, \text{m/s} \]

Key concepts

Conservation of Momentum

In physics, the conservation of momentum is a fundamental principle, especially during collisions. Momentum is defined as the product of an object's mass and velocity. In an elastic collision, such as the one in the problem, total momentum before the collision equals total momentum after the collision. This holds true because no external forces are acting on the system in the horizontal direction.

For instance, with blocks A and B colliding, the formula for conservation of momentum can be set up as follows:

\[ m_A v_{A,i} + m_B v_{B,i} = m_A v_{A,f} + m_B v_{B,f} \]

where:

• \( m_A \) and \( m_B \) are the masses of blocks A and B respectively.
• \( v_{A,i} \) and \( v_{B,i} \) are their initial velocities.
• \( v_{A,f} \) and \( v_{B,f} \) are their velocities after collision.

By applying this principle, one can solve for the velocities of both blocks at various points during the collision.

Kinetic Energy

Kinetic energy is the energy possessed by an object due to its motion and is given by the formula \( KE = \frac{1}{2}mv^2 \). In elastic collisions, like the one between the two blocks, kinetic energy is conserved.

This means that the total kinetic energy before and after the collision remains the same. This principle helps us understand how energy transfers between objects during the collision. It's important to note that some kinetic energy is temporarily converted to potential energy at the point of maximum spring compression.

If we consider both blocks before the collision, their initial kinetic energies can be calculated using:

\[KE_{initial} = \frac{1}{2} m_A (v_{A,i})^2 + \frac{1}{2} m_B (v_{B,i})^2\]At maximum spring compression, the potential energy stored in the springs is maximum, which in this scenario, modifies kinetic energies temporarily.

Spring Potential Energy

Spring potential energy is a form of potential energy stored in springs when they are compressed or stretched. In the collision of blocks equipped with spring bumpers, at the moment of maximum compression, the kinetic energy of the blocks is converted into spring potential energy.

The potential energy stored in a spring can be calculated using Hooke’s law: \( E_{spring} = \frac{1}{2} k x^{2} \), where \(k\) is the spring constant and \(x\) is the displacement from the equilibrium position.

In this problem, the concept is applied by equating the initial kinetic energy to the sum of the kinetic energy at moment of maximum compression and the potential energy stored in the spring, allowing us to solve for the stored spring energy:

\[E_{spring} = KE_{initial} - KE_{compression}\] This principle helps us determine how much energy is stored and temporarily taken from the kinetic energy of the blocks during the collision.

Frictionless Surfaces

Frictionless surfaces are idealized surfaces the exercises assume to have no friction between objects and the surfaces on which they move. This simplification is crucial to solve problems where we are interested in observing only the fundamental interactions between colliding objects without additional forces altering their motion.

When an object moves on a frictionless surface, it experiences no force slowing it down due to surface resistance. Therefore, only the forces applied by collisions or springs affect their momentum and energy.

In this given exercise, both blocks A and B are moving on a frictionless horizontal surface. This means that apart from the internal forces during the collision (i.e., spring force), no external forces will alter their speeds or directions. Understanding this concept helps us focus solely on conservation laws of momentum and energy without worrying about friction complicating the calculations.