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A disc of mass m slides with negligible friction along a flat surface with a velocity v. The disc strikes a wall head-on and bounces back in the opposite direction with a kinetic energy one fourth of its initial kinetic energy. What is the final velocity of the disc? Group of answer choices -v/2 -v/4 v/4 v/2 -v

Respuesta :

Answer:

-v/2

Explanation:

Given that:

  • a disc of mass m
  • Collides with the wall going through a sliding motion on on the plane smooth surface.
  • Upon rebounding from the wall its kinetic energy becomes one-fourth of the initial kinetic energy before collision.

We know, kinetic energy is given as:

[tex]KE_i=\frac{1}{2}. m.v^2[/tex]

consider this to be the initial kinetic energy of the body.

Now after collision:

[tex]KE_f=\frac{1}{4}\times KE_i[/tex]

[tex]KE_f=\frac{1}{4} \times \frac{1}{2}\times m.v^2[/tex]

Considering that the mass of the body remains constant before and after collision.

[tex]KE_f=\frac{1}{2}\times m.(\frac{v}{2})^2[/tex]

Therefore the velocity of the body after collision will become half of the initial velocity but its direction is also reversed which can be denoted by a negative sign.

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