Force as Rate of Momentum Change; Impulse

It is common knowledge that during a collision, the total momentum of the system is conserved. However, for individuals, its momentum can change. This change in momentum is called impulse, or $\vec{j}$. By Newton's 3rd Law, during a collision between 2 objects (object A and B), the normal force exerted by B onto A is the same as the normal force exerted by A onto B. By Newton's 2nd Law, whenever there's force, there's a corresponding acceleration. Whenever there's acceleration, there's a change in velocity, and thus a change in momentum. From kinematics we know: $$\triangle\vec{v} = \vec{a}_{ave} \cdot t$$ Now multiply by m on both sides: $$m\cdot\triangle\vec{v} = m\cdot\vec{a}_{ave}\cdot t$$ $m\cdot\triangle\vec{v}$ is change in momentum, or $\vec{j}$, while $m\cdot\vec{a}_{ave}$ is the average normal force during collision. So we get: $$\vec{j} = \vec{F}_{ave}\cdot t$$ or more commonly expressed as: $$\frac{\vec{j}}{t} = \vec{F}_{ave}$$ which is the equation form of the impulse-momentum theorem.

An example problem: 2 blocks (A and B) are moving toward each other at speeds of 5 m/s and 3 m/s respectively. Block A has mass of 3 kg while block B has mass of 5 kg. The blocks came to a stop upon collision (Took place within .02s), calculate the average normal force exerted by either block.

The change of momentum for block A is: $$(3 kg)(5 m/s) - (3kg)(0m/s) = 15 Ns$$ This change took place within .02s, so we get that the average normal force is: $$\vec{F}_{ave} = \frac{15 Ns}{.02s} = 750N$$