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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?.

Respuesta :

The kinetic energy of the two-block system before the collision will be [tex]\frac{9}{2} MV_S^2[/tex]. Momentum get conserved during the collision.

What is collision?

The abrupt, violent coming together in direct contact of two bodies is known as the collision.a falling object and a floor are examples of collision.

According to the law of conservation of momentum, the momentum of the body before the collision is always equal to the momentum of the body after the collision.

According to the law of conservation of momentum;

Momentum before collision =Momentum after collision

[tex]\rm MV+2m(0)=(3M)V_S\\\\ V= \frac{3MV_S}{M} \\\\ V= 3V_S[/tex]

Velocity of the block before collision is [tex]\rm 3V_S[/tex].

The kinetic energy of the two-block system before the collision is found as;

[tex]\rm KE = \frac{1}{2} mv^2 \\\\\ \rm KE = \frac{1}{2} m(3V_S)^2 \\\\\ \rm KE = \frac{9}{2} MV_S^2[/tex]

Hence, the kinetic energy of the two-block system before the collision will be [tex]\frac{9}{2} MV_S^2[/tex].

To learn more about the collision, refer to the link;

brainly.com/question/13138178

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