A particle of mass 0.25 kg is moving with velocity (3i + 7j) m s–1, when it receives the impulse (5i – 3j) N s. Find the speed of the particle immediately after the impulse.

In order to answer this question, we must first understand the concept of impulse. When a force acts on an object, its momentum is being changed, and this change is defined as the impulse. Put simply, Impulse = final momentum - initial momentum. In the same way that forces can be resolved, impulse can be seen as acting in two perpendicular directions, allowing us to work with the 'i' and 'j' components separately. Looking at the 'i' components in the question, the initial velocity is given as 3 m/s, thus the momentum (mass * velocity) is 0.25 * 3 = 0.75 kgm/s. The impulse, or change in momentum, is given as 5 Ns, thus the final momentum must be 5 + 0.75 = 5.75 kgm/s. The final velocity in the 'i' direction is therefore 5.75/0/25 = 23m/s. We may do the same thing in the j direction, where initial momentum is 1.75 kgm/s, the impulse is -3 Ns (meaning it is acting in the opposite direction of the particle's travel), so the final momentum is -1.25kgm/s , and its final velocity is -5 kgm/s. The final velocity is therefore 23i - 5j m/s. The question asks for speed, so we must find the magnitude of the velocity, which is 23.537 m/s.