Answer :
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 .
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 .