10.1: Impulse-Momentum Equations for a Particle
The concepts of impulse and momentum provide a third method of solving kinetics problems in dynamics. Generally this method is called the Impulse-Momentum Method , and it can be boiled down to the idea that the impulse exerted on a body over a given time will be equal to the change in that body's momentum. The impulse is usually denoted by the variable \(J\) (not to be confused with the polar moment of inertia, which is also J) and the momentum is a body's mass times its velocity. Impulses and velocities are both vector quantities, giving us the basic equation below.
\[ \vec{J} = m \vec{v}_f - m \vec{v}_i \]
For two-dimensional problems, we can break the single vector equation down into two scalar components to solve. In this case, we simply need to break all forces and velocities into \(x\) and \(y\) components.
\[ J_x = m v_{f_x} - m v_{i_x} \]
\[ J_y = m v_{f_y} - m v_{i_y} \]
Impulse:
The concept of an impulse in its most basic form is a force integrated over a time. For a force with a constant magnitude, we can find the magnitude of the impulse by multiplying the magnitude of the force by the time that force is exerted. If the force is not constant, we simply integrate the force function over the set time period. The direction of the impulse vector will be the direction of the force vector and the units will be a force times a time (Newton-seconds or pound-seconds, for example).
\begin{align} \text{Constant Magnitude Force:} \quad &\, \vec{J} = \vec{F} * t \\ \text{Non-Constant Magnitude Force:} \quad &\, \vec{J} = \int \vec{F}(t) \, dt \end{align}
In many cases, we will discuss impulsive forces . This is an instance where we have very large forces acting over a very short time frame. In instances of impulsive forces, it is often difficult to measure the exact magnitude of the force or the time. In these cases we may only be able to deduce the magnitude of the impulse as a whole via the observed change in momentum of the body.
Momentum:
The momentum of a body will be equal to the mass of the body times its current velocity. Since velocity is a vector, the momentum will also be a vector, having both magnitude and a direction. Unlike the impulse, which happens over some set time, the momentum is captured as a snapshot of a specific instant in time (usually right before and after some impulse is exerted). The units for momentum will be mass times unit distance per unit time. This is usually kilogram-meters per second in metric, or slug feet per second in English units.
Conservation of Momentum:
In instances where there is no impulse exerted on a body, we can use the original equation to deduce that there will be no change in momentum of the body. In this instance, momentum is conserved. This will also hold for systems of bodies, where if no external impulses are exerted on the bodies in a system, the momentum will be conserved as a whole. This is the basis of analysis for many collisions, as is discussed in the following sections.
Example \(\PageIndex{1}\)
A tennis ball (0.06 kg) is served to a tennis player at a speed of 10 m/s. The player then returns the ball at a speed of 36 m/s.
- What is the impulse exerted on the ball?
- If a high-speed camera reveals the impact lasted 0.02 seconds, what is the average force exerted on the ball during the collision?
- Solution
Example \(\PageIndex{2}\)
A plane with a mass of 80,000 kg is traveling at a velocity of 200 meters per second when the engines cut out. Twenty seconds later, it’s noticed that the velocity has dropped to 190 m/s. Assuming the plane is not gaining or losing altitude, what is the average drag force on the plane?
- Solution
Example \(\PageIndex{3}\)
The plot below shows the thrust generated by the engine on a jet fighter (mass of 2500 kg) over ten seconds. If the plane is starting from rest on a runway, and friction and drag are negligible, determine the speed of the plane at the end of these ten seconds.
- Solution