[Physics] Calculate force of impact on running dog


I'm going to hold a dog protection tournament. I'd like to give the audience information about the force of the impact of the running dog when it hit(attacks) the decoy on a huge screen. I'm a software developer student so I have the neccesary skills to development the software/electronics for the project. However, i'm not a physicist and I'm getting quite confused looking on all the formulas that I find online.

My question is therefore. Is it possible to calculate the force of impact in kilograms of the dog hitting the decoy (seen as a stationary object)?

As an example how should I calculate the force if say the dog weights 40kg and is running at a speed of 30 km/t?

Best Answer

The main thing that you need here is the notion of impulse. Consider Newton's second law:

$$\sum{\vec F} = m{\vec a}$$

In our case, the only relevant force is the contact force between our dog and the decoy. Before we move on, let's rephrase this force a little. Acceleration is the rate of change of velocity, and we know that momentum is given by ${\vec p} = m {\vec v}$, so we have:

$$\begin{align} m{\vec a} &= m\frac{d{\vec v}}{dt}\\ &= \frac{d}{dt}\left(m{\vec v}\right)\\ &= \frac{d{\vec p}}{dt} \end{align}$$

So, remembering from calculus that we can say, if $F = \frac{dp}{dt}$, then we can also say $F_{avg} = \frac{\Delta p}{\Delta t}$, we only need to know how much our dog changes velocity and how much time this takes. Let's assume our decoy stops the dog completely. Then, we have $v_{f} = p_{f} = 0$. Initially, we have our dog moving at $30 {\rm kmh} = 8.3 {\rm m/s}$, so we have $p_{i} = 250 {\,\,\rm kg\cdot m/s}$, so $\Delta p = -250{\,\,\rm kg\cdot m/s}$. Now, all you need to do is to estimate the amount of time it takes for your dog to come to a stop. I will contend that 10 seconds is too long, and 1 millisecond is too short. You can reason about this some more to come up with a range of estimates, but it's going to depend on the composition of your decoy and of your dog to get anything much more precise than an estimate.