One object is traveling northeast with a momentum of 12 kg-m/s, while a second object is traveling southwest on the same path as the first object with a momentum of 4 kg-m/s. When the two objects collide, they stick together. What is the momentum after their collision?
In order to solve this problem, we need to break vectors down into their components.
The first object has a momentum of 12 kg-m/s, which can be broken down into two components: one in the northeast direction and one in the southeast direction.
Let's call the vector in the northeast direction NE and the vector in the southeast direction SE. We can calculate the magnitude of these vectors by using the Pythagorean theorem:
NE = sqrt(12^2/2) = sqrt(72) = 6sqrt(2) kg-m/s
SE = sqrt(12^2/2) = sqrt(72) = 6sqrt(2) kg-m/s
Similarly, the second object has a momentum of 4 kg-m/s, which can also be broken down into NE and SE components:
NE = sqrt(4^2/2) = sqrt(8) = 2sqrt(2) kg-m/s
SE = sqrt(4^2/2) = sqrt(8) = 2sqrt(2) kg-m/s
Since the two objects are traveling in opposite directions along the same path, their momenta have opposite signs:
The NE components add up to: 6sqrt(2) - 2sqrt(2) = 4sqrt(2) kg-m/s
The SE components add up to: 6sqrt(2) + 2sqrt(2) = 8sqrt(2) kg-m/s
Now, when the two objects collide and stick together, their momenta are added together:
Total momentum = 4sqrt(2) kg-m/s + 8sqrt(2) kg-m/s = 12sqrt(2) kg-m/s
Therefore, the momentum after their collision is 12sqrt(2) kg-m/s.