Question

Question #f3f56

Answer 1

`v = 2 pm 2 sqrt 2 \ m/s`, where `v` is the velocity before the 2m/s increase is applied

Explanation:

i think you're saying than a 2m/s increase in velocity doubles KE?

if that is correct, then in math terms:

`(1/2 m (v+ 2)^2)/(1/2 m v^2) = 2`, where `v` is the velocity before the increase is applied

So cancelling some stuff

`( (v+ 2)^2)/( v^2) = 2`

and then solving the quadratic

`v^2 + 4v + 4 = 2v^2`

`v^2 - 4v - 4 = 0`

completing the square

`(v-2)^2 - 4 - 4 = 0`

`(v-2)^2 = 8`

`v = 2 pm 2 sqrt 2 \ m/s`

both of these solutions make sense.

if the particle is travelling to the right at `v = 2 + 2 sqrt 2 \ m/s` and increase its velocity to `v = 4 + 2 sqrt 2 \ m/s`, the KE should double

Similarly, if the particle is travelling to the left, ie at `v = 2 - 2 sqrt 2 \ m/s`, and it increases its velocity to `v = 4 - 2 sqrt 2 \ m/s`, now moving to the right, the KE should also double.