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Grade 12AIPMT

a point mass 1 kg collides elastically with stationary mass of 5 kg. after the collison the 1 kg mass reverses its direction and moves with a speed of 2 m/s.For the system of these masses the kinetic energy of the center of masses is what?

Profile image of RAGINI SEIWAL
8 Years agoGrade 12
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1 Answer

Profile image of Gaurav Gupta
8 Years ago

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.