A bullet of mass 188 grams was fired at 797 meters per second into a wooden target of mass three kilograms which was at rest. The bullet embedded into the target, and they started to move together. Find the speed of the bullet and target after the impact, given that the total momentum of the system did not change.
Since the momentum of the system did not change, we can use the conservation of momentum. This tells us that the momentum before is equal to the momentum after such that 𝑚 one 𝑢 one plus 𝑚 two 𝑢 two is equal to 𝑚 one 𝑣 one plus 𝑚 two 𝑣 two, where 𝑚 one and 𝑚 two are the masses of the two objects, 𝑢 one and 𝑢 two are the velocities before impact, and 𝑣 one and 𝑣 two are the velocities after impact.
In this question, a bullet is being fired into a wooden target. We are told that the bullet has a mass of 188 grams and the target has a mass of three kilograms. As these units are different, we will use the fact that there are 1000 grams in a kilogram to find the mass of the bullet in kilograms. 188 divided by 1000 is 0.188. The mass of the bullet is 0.188 kilograms.
We also know from the question that the bullet was fired at 797 meters per second, whereas the wooden target was at rest. The momentum before is therefore equal to 0.188 multiplied by 797 plus 3000 multiplied by zero. This is equal to 149.836.
We are told that after the impact the bullet becomes embedded into the target. This means that this has a total mass of 3.188 kilograms. We need to find the speed of this object, which we will call 𝑣 meters per second. The momentum after impact is therefore equal to 3.188 multiplied by 𝑣. And we know this must be equal to 149.836. Dividing through by 3.188 gives us 𝑣 is equal to 47. The speed of the bullet and target after the impact is therefore equal to 47 meters per second.