Question

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

Answer 1

-v/2

Step-by-step 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:

`KE_i=(1)/(2). m.v^2`

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

Now after collision:

`KE_f=(1)/(4)* KE_i`

`KE_f=(1)/(4) * (1)/(2)* m.v^2`

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

`KE_f=(1)/(2)* m.((v)/(2))^2`

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.