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blondinia [14]
3 years ago
5

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?​
Physics
1 answer:
Marina CMI [18]3 years ago
4 0

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}

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