When a ball bounces off a wall, it is reflected with the same speed but travels in the opposite direction: v → -v The kinetic energy is constant, and the net work is done on the ball by the wall is zero. Since **Δv = v _{f} – v_{i} = -2v = a_{ave }Δt**, we define the impulse as force * time,

*J = F Δt ….. ……….. ……(1)*

So the momentum of an object with mass in as mass * velocity is

*p = mv ……. …….. ……(2)*

Newton’s second law, ** F = ma** , can then be written,

*J = Δp ……. …….. ……(3)*

This relationship is somewhat analogous to the work-kinetic energy theorem. It allows us to describe changes in the velocity of an object without knowing details of the forces [the magnitudes and direction and for how long they last]. For the ball bouncing off the wall, we do not know the magnitude of the force exerted by the wall on the ball or the duration of the force; however, the impulse of the force has a known value, ** J = – 2mv**. The unit of momentum and impulse are

We have two objects [pucks] sliding on (idealized) ice and traveling at velocities v_{1,i} and v_{2,i} respectively. They collide and exert forces F_{12} and F_{21} on each other. These two forces are an action-reaction pair so that F_{12} = – F_{21}. After the collision, they travel at velocities v_{1f} and v_{2f}. Note that there no external forces, for example the gravitational force in the case when the surface is inclined.

We have for the impulses,

**F _{12} Δt = m Δv_{1} = m_{1} (v_{1,f} — v_{1,i})**

**F _{21} Δt = m Δv_{2} = m_{2} (v_{2,f} — v_{2,i})**

Newton’s third law states F_{12} = -F_{21}, or F_{12} + F_{21} = 0, so that the addition of the two equations yields

**(F _{12} + F_{21}) Δt = m_{1} (v_{1,f} — v_{1,i}) + m_{2} (v_{2,f} — v_{2,i}) = 0 or, **

**m _{1} v_{1,i} + m_{2} v_{2,i} = m_{1} v_{1,f} + m_{2} v_{2,f}**

We define the total momentum of the system [mass m_{1} and mass m_{2}],

**p _{tot} = m_{1} v_{1} + m_{2} v_{2}**

We thus see that the total momentum is the same before and after the collision. We say that the total momentum is conserved. No detailed knowledge of the forces is necessary.