Whenever the force is not easy to calculate. Impulse is $$J=\Delta p= \int\limits_0^{\delta t}F\mathrm dt$$. This comes directly from $F=\frac{\mathrm dp}{\mathrm dt}$
For most collisions, the force isn't easily defined. It's usually an extremely large force $F$ operating in an extremely short time $\delta t$. It need not be, though--most spring-mass collisions have a well-defined force. But for collisions like clapping your hands together, throwing a wooden block at a wall, etc, we can't easily calculate the force since we don't know a whole host of things, like the moduli of the object and material. So what we do is this: We safely neglect $\delta t$. During the colission, we have little or no change in displacement. The only thing that changes is velocity, so we can just code it as a change in momentum $J=\Delta p$. This way, we can work with the equations of conservation of momentum etc without having to calculate the collision force.
Some examples
Lets say you have a pulley system. A simple one: A person A is hanging from one end, and a block B of equal mass is on the other end. The system is at rest. An object C approaches A at a speed $v$. A catches it. Find the final velocities of the system.
Here, we cannot directly conserve momentum as the pulley hangs from a ceiling and that ceiling can take away momentum. Neither can we conserve energy as the collision is inelastic. But, using impulses, we can conserve momentum: Let $J_T=\int Tdt$ where $T$ is the tension. Since tension is same in both strings attached to A and B, this impulse will act on both. It is also upwards for both, as tension is upwards. Now, we can actually conserve momentum individually for the two bodies. Instead of using $\Delta p=0$, we use $J= \Delta p$. Your $J$ can be eliminated easily (two equations, two unknowns--$J$ and final velocity)
Another example is when you have a ball bouncing off a wall; when the wall has friction. We can apply normal coefficient of restitution equations to find the final perpendicular velocity $v_\perp$, but we can't do the same for the parallel direction, as friction is there. But, we know that $f=\mu N$. Using this, we get $J_f=\mu J_N$ (plug it into $\int Fdt$, $\mu$ is a constant). Now, since we already calculated the change in velocity, we can calculate $J_N=J_\perp=m\Delta v_\perp$. Using this, we can find $J_f$. Since $J_f=J_\parallel=\Delta v_\parallel$, we can find the change in parallel velocity. That's where we can get these formulae from.
If you want elaboration on the examples, please let me know. I've assumed a bit while using them.