Suppose that we are working with a horizontal string with two masses attached. There is no friction on the floor and we move those two attached masses away from the equilibrium position by an undefined distance.
Once the string with those two attached masses passes through the equilibrium possition, we remove one of the attached masses.
As there are no forces present in that moment (as we are in the equilibrium position), we should apply the conservation of the linear momentum such that:
(m1 + m2 ) * v_slow = m1 *v_fast (as we remove the m2 with no speed)
So, we move from an initial Kinetic energy = ((m1+m2)*v_slow^2)/2
To a final Kinetic energy = (m1*v_fast^2)/2 from which we can derive using the conservation of the linear momentum --> ((m1+m2)^2 * v_slow^2)/(2*m1) and as ((m1+m2)^2)/m1 > (m1+m2), we have essentialy given energy to the system.
Can someone explain me where does this "extra" energy come from?
Thanks
Once the string with those two attached masses passes through the equilibrium possition, we remove one of the attached masses.
As there are no forces present in that moment (as we are in the equilibrium position), we should apply the conservation of the linear momentum such that:
(m1 + m2 ) * v_slow = m1 *v_fast (as we remove the m2 with no speed)
So, we move from an initial Kinetic energy = ((m1+m2)*v_slow^2)/2
To a final Kinetic energy = (m1*v_fast^2)/2 from which we can derive using the conservation of the linear momentum --> ((m1+m2)^2 * v_slow^2)/(2*m1) and as ((m1+m2)^2)/m1 > (m1+m2), we have essentialy given energy to the system.
Can someone explain me where does this "extra" energy come from?
Thanks