What is the percent change in momentum of a proton that accelerates from 0.41c to 0.85c?

At these speeds, we can't ignore the relativistic masses.

At 0.41c:  m = m₀ / √(1-.41²) = m₀ / √0.8319 = m₀ / 0.9121 = 1.096 m₀ 

At 0.85c:  m = m₀ /  √(1-.85²) = m₀ / √0.2775 = m₀ / 0.5268 = 1.898 m₀

The change in momentum is  (1.898m₀ x 0.85c) - (1.096m₀ x 0.41c)

                                        = (1.613 - 0.449) m₀ c  =  1.164 m₀ c .

As a fraction of the original momentum, that's  (1.164 / 0.449)

                                                                 = (2.592) = 159% increase .

the simple machine has three main features. they are:

loadfulcrumeffort

fulcrum is the center point or pivot point of the machine. effort is input force/ applied force. load is external force/ output.

the keisha 's machine labels are

x : load

y: fulcrum

z: lever