Force Applied and Impulse/ Newton's Third Law

Question

How does the applied force of an object determine impulse?

How is impulse and Newton's Third Law relate to changes in momentum for colliding objects?

Answer 1

Force is change in momentum over time = d(m*v)/dt which is F = m*a if mass isn't lost... so impulse = change in momentum so force times time = F*dt = d(m*v) or F*t = m*v_f - m*v_i

Example: If i apply a constant force of 5 N for 3 seconds to a 2 kg object initially at rest how fast will it be going after the 3 seconds?

Answer: the impulse is 5*3 = 15 kg*m/s so if they had zero momentum to start with, now that I pushed it, it has 15 kg*m/s of momentum... the object has a mass of 2 kg, momentum which = m*v so 15/2 = 7.5m/s = v

For Newton's 3rd Law and Collisions:

If 2 objects collide then the force each experiences depends on the time they contact each other. If it's perfectly elastic and each one bounces back with the same momentum (but opposite signs since opposite direction) then that could result from a really high collision force but a very short time or a lesser force applied over a long time because change in momentum = F*t (or ∫F*dt if you know calculus and it's a non-constant force) which means 15 kg*m/s change in momentum could be from a 5 N force applied over 3 seconds or a 15 N force applies over 1 second.