Question

two blocks are held together with a compressed spring between them on the surface of a slippery table .one block has three times the inertia of the other .when the blocks are released ,the spring pushes them away from each other .what is the ratio of their kinetic energies after the release?​

Answer 1

Explanation:

The initial kinetic energy $KE_0$ for both blocks is zero. Let $m_1= m$ and $m_2 =3m$. So using the conservation law of linear momentum, we can write

$0 = m_1v_1 - m_2v_2$

or

$v_2 = \dfrac{m_1}{m_2}v_1 = \dfrac{m}{3m}v_1 = \dfrac{1}{3}v_1$

The final kinetic energies for the two masses are

$KE_1 = \frac{1}{2}m_1v_1^2 = \frac{1}{2}mv_1^2$

$KE_2 = \frac{1}{2}m_2v_2^2 = \frac{1}{2}(3m)(\frac{1}{3}v_1)^2 = \frac{1}{2}m(\frac{1}{3}v_1^2)$

Therefore, the ratio of their kinetic energies is

$\dfrac{\Delta KE_2}{\Delta KE_1} = \dfrac{\frac{1}{2}(\frac{1}{3}v_1^2)}{\frac{1}{2}v_1^2} = \dfrac{1}{3}$