Question: How much damage would a space probe cause if it can get about 4 lightyears away in 10 years and doesn't have any brakes when it arrives? Could it cause a global extinction event on the planet it was sent to? Do actual concept probes think about this?

Answer: 
Calculating the damage done due to an impact is an imprecise business, but we could use the [KT extinction event](https://en.wikipedia.org/wiki/Cretaceous%E2%80%93Paleogene_extinction_event) (that finished off the dinosaurs) as a benchmark. This impact had [an energy of around $4.2\times 10^{23}$ joules](https://en.wikipedia.org/wiki/Chicxulub_crater#Impact_specifics). The question is then how fast our probe would have to be going to deliver this much energy.

For a relativistic projectile the total energy is given by:

$$ E^2 = \frac{m^2v^2c^2}{1 - v^2/c^2} + m^2c^4 $$

and the kinetic energy is just this energy minus the rest mass energy $mc^2$.

Let's assume our probe has a mass of a ton ($10^3$ kg), which is around the mass of the [Voyager probes](https://en.wikipedia.org/wiki/Voyager_program). In that case to match the energy of the Chicxulub impact would need a speed of around $0.99999998c$. You're suggesting a speed of $0.4c$, and at that speed the kinetic energy would be around $8\times 10^{18}$ joules, which is a factor 50000 lower than the Chicxulub impact energy. Even so I wouldn't like to be standing underneath it when it hit.

But there are a couple of points that need to be made.

Firstly space is big - really big - and it's mostly empty. The chance of a probe hitting anything is so ridiculously small that no-one is going to take it seriously.

Secondly NASA does care what happens to its probes and this is part of the mission planning. For example [the Cassini probe is going to be deliberately crashed into Saturn](https://en.wikipedia.org/wiki/Cassini%E2%80%93Huygens#Spacecraft_disposal) where the heat of reentry will destroy it (and any Earth life forms that might be hiding on it).
The space probe won't do any damage as it will be vaporized in the exosphere. When the probe enters the interplanetary medium at a distance of the order of 100 AU from the star, it will collide with ions at a density of the order of a 5 ions per $\text{cm}^3$. The collisions with protons happen at a speed of 0.4 c, therefore at an energy of the order of 100 MeV, so this is quite similar to being exposed to intense alpha radiation. The temperature of the heat shield will be raised to about 600 K. The heat shield may tolerate temperatures up to about 6000 K before it will vaporize. This temperature will be reached when the density is $10^4$ times larger, so 50,000 ions per $\text{cm}^3$. But at 1000 km altitude above Earth you'll typically already have $\sim 10^{6}$ atoms per $\text{cm}^3$.

So, as the probe approaches an Earth like planet, the temperature of its heat shield gradually rises due to the increasingly intense bombardment of the ions and atoms it scoops up. At 10,000 km the shield will be red hot, 0.02 seconds later when it is at 4000 km it will most likely have vaporized, exposing the interior to temperatures of the order of 6000 K. The entire probe then simply vaporizes completely before even reaching 1000 km distance.

The atoms the probe consist of will keep on moving into the atmosphere until they collide with atoms. Because of the disintegration, each atom is now "on its own" while in a solid only the atoms at the surface would suffer collisions. All the atoms will have collided with atoms from the upper atmosphere already at a height of a few hundred kilometers. In these high energy collisions, the atoms get ionized and will then experience the strong Lorentz force due to the Earth's magnetic field. These processes will cause the energy of the probe to get dispersed well above 100 km altitude.