To find the kinetic energy of the center of mass for the given system, we need to first understand the concept of the center of mass and how kinetic energy is calculated. In this scenario, we have a 1 kg mass colliding elastically with a stationary 5 kg mass. After the collision, the 1 kg mass moves in the opposite direction at a speed of 2 m/s. Let’s break this down step by step.
Understanding the System
We have two masses:
- Mass \( m_1 = 1 \) kg, which moves with a speed \( v_1' = -2 \) m/s after the collision (the negative sign indicates the reversal of direction).
- Mass \( m_2 = 5 \) kg, which is initially stationary (thus its speed is \( v_2 = 0 \) m/s).
Calculating the Center of Mass Velocity
The velocity of the center of mass (\( v_{cm} \)) for a two-body system can be calculated using the formula:
v_{cm} = \frac{m_1 v_1' + m_2 v_2}{m_1 + m_2}
Plugging in the values:
v_{cm} = \frac{(1 \, \text{kg} \cdot -2 \, \text{m/s}) + (5 \, \text{kg} \cdot 0 \, \text{m/s})}{1 \, \text{kg} + 5 \, \text{kg}} = \frac{-2 \, \text{kg m/s}}{6 \, \text{kg}} = -\frac{1}{3} \, \text{m/s}
Calculating the Kinetic Energy of the Center of Mass
The kinetic energy (\( KE \)) of the center of mass is given by the formula:
KE_{cm} = \frac{1}{2} (m_1 + m_2) v_{cm}^2
Now substituting the values we have:
KE_{cm} = \frac{1}{2} (1 \, \text{kg} + 5 \, \text{kg}) \left(-\frac{1}{3} \, \text{m/s}\right)^2
KE_{cm} = \frac{1}{2} (6 \, \text{kg}) \left(\frac{1}{9} \, \text{m}^2/\text{s}^2\right)
KE_{cm} = 3 \, \text{kg} \cdot \frac{1}{9} \, \text{m}^2/\text{s}^2 = \frac{1}{3} \, \text{J}
Final Result
Therefore, the kinetic energy of the center of mass for this system is 1/3 Joule. This reflects the energy associated with the motion of the combined mass system, taking into account how they move relative to each other after the collision.
