An impulse of $10 \text{ kg m/s}$ could be the result of a force of $1 \text{ N}$ applied over $10 \text{ s}$ or of a force of $10 \text{ N}$ applied over $1 \text{ s}$.
If mass and force are constant during the period of impact, the impulse is also directly proportional to the force and the change in velocity: $$\vecphy{J}=\Delta \vecphy{p}=\vecphy{F}t$$ This can be shown as follows: \begin{align*} \vecphy{J}&=\Delta\vecphy{p}&\quad\quad \vecphy{F}&=\frac{d\vecphy{p}}{dt}&\\ &=\vecphy{p}_1-\vecphy{p}_0&\quad\quad &=\frac{\Delta\vecphy{p}}{t}&\\ &=m(\vecphy{v}_1-\vecphy{v}_0)\\ &=m\Delta\vecphy{v}\\ \Rightarrow\vecphy{J}&=\Delta\vecphy{p}\\ &=\vecphy{F}t \end{align*}
The magnitude of impulse is given by the area under the graph of force $(\vecphy{F})$ against time $(t)$