A system consists of three particles with these masses and velocities: mass \(3.0 \mathrm{kg},\) moving north at \(3.0 \mathrm{m} / \mathrm{s}\) mass $4.0 \mathrm{kg},\( moving south at \)5.0 \mathrm{m} / \mathrm{s} ;\( and mass \)7.0 \mathrm{kg}\( moving north at \)2.0 \mathrm{m} / \mathrm{s} .$ What is the total momentum of the system?

We first need to calculate the momentum of each of the three particles separately. Let's represent the momentum of each particle using the formula: Momentum = mass × velocity Particle 1: Mass = \(3.0\mathrm{kg}\) Velocity = \(3.0\mathrm{m/s}\) (north) Momentum of particle 1 = \(3.0\mathrm{kg} \times 3.0\mathrm{m/s}\) = \(9\,\text{N s}\) (north) Particle 2: Mass = \(4.0\mathrm{kg}\) Velocity = \(-5.0\mathrm{m/s}\) (south, so we put a negative sign) Momentum of particle 2 = \(4.0\mathrm{kg} \times (-5.0\mathrm{m/s})\) = \(-20\,\text{N s}\) (south) Particle 3: Mass = \(7.0\mathrm{kg}\) Velocity = \(2.0\mathrm{m/s}\) (north) Momentum of particle 3 = \(7.0\mathrm{kg} \times 2.0\mathrm{m/s}\) = \(14\,\text{N s}\) (north)

Now, we need to add the momenta of the three particles to obtain the total momentum of the system. Total momentum = Momentum of particle 1 + Momentum of particle 2 + Momentum of particle 3 Total momentum = \(9\,\text{N s}\) (north) + \(-20\,\text{N s}\) (south) + \(14\,\text{N s}\) (north) Let's consider north as positive and south as negative. Total momentum = \(9\,\text{N s}\) + \((-20\,\text{N s})\) + \(14\,\text{N s}\) Total momentum = \(-2\,\text{N s}\) Since the resulting value is negative, it means the total momentum of the system is southward.

The total momentum of the system is \(2\,\text{N s}\) (south).