Calculating momentum of an object
Everybody knows that it is dangerous to drive in front of a big truck on the highway because of how long it takes the big truck to slow down, even though it is going the same speed as all the small cars. Likewise, if a little kid going very quickly crashes into a slow moving adult on an ice skating rink, it is a very different outcome than if a fast adult crashes into a slow kid. The reason for these things is the connection between force and momentum. In fact, we can see this connection from Newton's second law.
The momentum of an object is defined as its mass (the kind defined by \(m = F/a\)) times its velocity. Like the velocity, it has a magnitude as well as a direction. Practically, momentum can be thought of as the tendency for an object to stay along its current path through space. The more momentum an object has, the bigger the force required to change its velocity.
Writing it out in the form \(F=ma\), we see that force is equal to the mass times the acceleration. Acceleration is, by definition, the rate at which the velocity changes. If we assume that the mass of an object is constant, then we can reinterpret the second law in a useful way.
Calculating the momentum of an object
Change in momentum
Suppose that before and after the application of a force \(F\), the velocity of an object is \(v_i\), and \(v_f\), respectively.
Because the mass \(m\) is the same, before and after, the momentum before and after is \(p_i = mv_i\) and \(p_f = mv_f\), respectively.
The change in momentum is then \(\Delta p = p_f - p_i = m(v_f-v_i) = m\Delta v\).
The rate at which the momentum changed is then \(\displaystyle\Delta p / \Delta t = m\Delta V/ \Delta t\). But, \(\Delta v/\Delta t\) is just the acceleration, therefore, the rate of change in momentum is \(\displaystyle ma = F\)!
So, we can see clearly that force is an exact measure of how quickly the momentum of an object will change. Whereas with velocity, we have to divide by \(m\) to find how much \(v\) will be changed by \(F\), with momentum the relationship is direct. This suggests that momentum is the more natural quantity to work with in classical mechanics. In fact, as you progress through mechanics, quantum physics, and on to the frontiers of theoretical physics, one hardly talks about velocities, but of momenta. But that's for later.
Impulse and the change in momentum
As we showed, we can reinterpret the second law \(F=ma\) as \(F = \Delta p / \Delta t\), meaning that force is equal to the rate of change of momentum. Therefore, if we have a known force that acts on an object for some known amount of time \(\Delta t\), we can use the second law to calculate the change in momentum: \(\Delta p = F\Delta t\).
For example, if a backcountry skier is pulled by a rope with a constant force of tension \(T=\)10 N, for \(t=\)10 s, their momentum must increase by \(F\Delta t\) = 100 kg m/s.
This is fine is we have a known force. However, in many cases momentum is relatively easy to measure by experiment, while force is more difficult to measure. For example, if we throw a ball at a wall, it bounces off in the opposite direction, a change in momentum. It is clear that the momentum has changed and we can measure it by timing the motion of the ball, but how can we possibly measure the force between the wall and the ball?
Happily, we can turn the second law on its head and use the change in momentum to find the unknown force.
Suppose we throw a ball of mass \(m\) at a wall, and it bounces off elastically meaning that it bounces back at the same speed. If we record the compression of the ball against the wall, we can figure out the approximate amount of time it spent interacting with the wall, \(\Delta t\). Using the second law, we can then infer the strength (on average) of the force with which the wall pushed on the ball!
Impulse of a variable force
It states that= Net external impulse = change in momentum in the direction of F. This is very useful when you have to get the net external impulse of the object. This is widely used in collision mechanics, generally where friction comes into play. This theorem is always valid where as the Conservation of momentum is not valid when there is a net external force.