Question

Object 1 moves at 30° angle with momentum of 30 kg m/s collides inelastically with object 2 moving at a 250° angle with a momentum of 50 kg m/s. What is the resulting magnitude and direction of the vector?

Answer 1

The resulting momentum is the sum of the momenta before the collision (according to the momentum conservation law):

$$\mathbf{p}_1 + \mathbf{p}_2 = \mathbf{p}'$$

where $\mathbf{p}_1 = 30\cdot (\cos30\degree, \sin30\degree)$ is the momentum of Object 1 (in coordinate form), and $\mathbf{p}_2 =50\cdot (\cos250\degree, \sin250\degree)$ is the momentum of Object 2.

Thus, obtain:

$$\mathbf{p}' = 30\cdot (\cos30\degree, \sin30\degree) + 50\cdot (\cos250\degree, \sin250\degree)=\\ =(30\cos30\degree + 50\cos250\degree, 30\sin30\degree + 50\sin250\degree) \approx (8.9, -32)$$

The magnitude of this vector is:

$$p = \sqrt{8.9^2 + (-32)^2} \approx 33\space kg\cdot m/s$$

The direction is:

$$\theta = \arctan\dfrac{-32}{8.9} \approx 286\degree$$

Answer. Magnitude: $33\space kg\cdot m/s$, direction: $286\degree$.