The kinetic energy of the two-block system with the given masses before the collision is determined as ³/₂Mvs².
The kinetic energy of the blocks before the collision is calculated as follows;
For a negligible friction effect, the kinetic energy of the two - block system after the collision is equal to the kinetic energy before the collision.
K.E(initial) = K.E(after)
K(after) = ¹/₂(m₁ + m₂)vs²
K(after) = ¹/₂(m + 2m)vs²
K(after) = ¹/₂(3m)vs²
[tex]K.E(after) = \frac{3}{2} Mv_s^2[/tex]
Thus, the kinetic energy of the two-block system with the given masses before the collision is determined as ³/₂Mvs².
The complete question is below:
block x of mass m slides across a horizontal surface where friction is negligible. block x collides with block y of mass 2m that is initially at rest, as shown in figure 1. after the collision, both blocks slide together with a speed vs , as shown in figure 2.
What is the kinetic energy of the two-block system before the collision?
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