Question

Two large blocks of wood are sliding toward each other on the frictionless surface of a frozen pond. Block a has mass 4. 00 kg and is initially sliding east at 2. 00 m/s. Block b has mass 6. 00 kg and is initially sliding west at 2. 50 m/s. The blocks collide head-on. After the collision block b is sliding east at 0. 50 m/s.

Answer 1

Two large blocks of wood are sliding towards each other on the frictionless surface of a frozen pond. The blocks collide head-on. Then the decrease in total kinetic energy after the collision is 13.18 J.

From the law of conservation of linear momentum, we know, Pi = Pf

where, Pi is total initial momentum

Pf is total final momentum

m₁u₁ + m₂u₂ = M₁V₁ + M₂V₂

4*2 + 6*(-2.5) = 4* V₁ + 6*(0.5)

4V₁ = 8 - 15 -3

V₁ = -2.5 m/s

Now, let us calculate the initial and final kinetic energies.

The formula to find out kinetic energy is K.E = 1/2 mv²

Initial kinetic energy = 1/2* 4* 2² + 1/2* 6* (-2.5)² = 26.75 J

Final kinetic energy = 1/2* 4* 2.5² + 1/2* 6* (-0.5)² = 13.57 J

The difference in total kinetic energy is K₂ - K₁ = (13.57 - 26.75) = -13.18 J

Thus, the decrease in total kinetic energy of the given blocks after collision is 13.18 J.

To know more about kinetic energy:

brainly.com/question/22174271

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