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

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:

$KE_i=\frac{1}{2}. m.v^2$

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

Now after collision:

$KE_f=\frac{1}{4}\times KE_i$

$KE_f=\frac{1}{4} \times \frac{1}{2}\times m.v^2$

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

$KE_f=\frac{1}{2}\times m.(\frac{v}{2})^2$

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.