Question #f3f56

1 Answer
Aug 1, 2016

Answer:

#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.