Let’s begin by examining the momentum form of Euler’s first law by defining the momentum for a system of particles. Consider a system composed of N particles, where each particle i has a mass m_i and a velocity \mathbf{v}_i. The linear momentum \mathbf{p}_i of each particle is given by the product of its mass and velocity:
\mathbf{p}_i = m_i \mathbf{v}_i
The total linear momentum \mathbf{P} of the system is the sum of the momenta of all individual particles:
\mathbf{P} = \sum_{i=1}^{N} \mathbf{p}_i = \sum_{i=1}^{N} m_i \mathbf{v}_i
Next, consider the position vectors from a fixed origin O in an inertial frame to each of the particles, denoted by \mathbf{r}_i. Taking the time derivative of these position vectors yields the velocities:
\frac{d\mathbf{r}_i}{dt} = \mathbf{v}_i
Previously, it was established that the sum of the position vectors weighted by their respective masses for a system of particles is equal to the total mass M multiplied by the position vector of the center of mass \mathbf{R}:
\sum_{i=1}^{N} m_i \mathbf{r}_i = M \mathbf{R}
Extending this concept to a continuously distributed mass, the linear momentum \mathbf{P} of the body is represented as:
\mathbf{P} = M \mathbf{V}
where \mathbf{V} is the velocity of the center of mass, and M remains constant. Taking the time derivative of the linear momentum gives:
\frac{d\mathbf{P}}{dt} = M \frac{d\mathbf{V}}{dt} = M \mathbf{A}
Here, \mathbf{A} is the acceleration of the center of mass. According to Euler’s first law, the sum of the forces acting on the system is equal to the mass of the center of mass multiplied by its acceleration:
\sum \mathbf{F} = M \mathbf{A}
This equation shows that the total external force on a continuously distributed mass is equal to the time derivative of its linear momentum, mirroring the relationship observed for a single particle.
We can derive the impulse-momentum relationship by integrating the momentum form of Euler’s first law over a time interval from t_1 to t_2:
\int_{t_1}^{t_2} \sum \mathbf{F} \, dt = \mathbf{P}(t_2) - \mathbf{P}(t_1)
The left-hand side of this equation represents the impulse, which is the integral of the force over the specified time period. The right-hand side represents the change in linear momentum \Delta \mathbf{P} of the system during that interval.
A specific case of impulse is known as an impact, which occurs during a collision of bodies over a very short time period, effectively making \Delta t approach zero. In such scenarios, if the sum of the external forces in the direction of the momentum becomes negligible or vanishes, the impulse also vanishes. Consequently, the change in momentum is zero, leading to the conservation of momentum:
\Delta \mathbf{P} = 0 \quad \Rightarrow \quad \mathbf{P}(t_2) = \mathbf{P}(t_1)
In practical problems involving large impact forces, smaller forces such as gravity are often ignored to simplify the analysis, allowing the conservation of momentum to be applied effectively.
The momentum form of Euler’s first law establishes that the total external force acting on a system is equal to the rate of change of its linear momentum. Integrating this relationship over time yields the impulse-momentum theorem, which is particularly useful in analyzing collisions and impacts where momentum is conserved in the absence of significant external forces.
Let’s analyze an example using the principle of conservation of momentum. We have two blocks: the first block is initially at rest on a smooth, frictionless horizontal surface, and the second block is sliding toward it with an initial speed v. After the collision, the two blocks stick together and move as a single unit.
We assume no external forces act in the horizontal direction, so the total linear momentum of the system is conserved. Let the mass of each block be m, and let the velocity of the combined blocks after the collision be v_1.
The total momentum before the collision is the momentum of the moving block, as the block at rest has no initial velocity. Thus, the initial momentum p_0 is:
p_0 = m v + m \cdot 0 = m v
After the collision, the two blocks stick together and move as a single object with mass 2m and velocity v_1. The total momentum p_1 is:
p_1 = (m + m)v_1 = 2m v_1
By the law of conservation of momentum, the total momentum before the collision equals the total momentum after the collision:
p_0 = p_1
Substituting the expressions:
m v = 2m v_1
Canceling m from both sides:
v = 2 v_1
The velocity of the two blocks after the collision is:
v_1 = \frac{v}{2}
This means that after the collision, the two blocks, which now move together, travel at half the initial speed of the sliding block. This result is consistent with the conservation of momentum and the symmetry of the system.
In most collisions, the objects do not stick together; instead, they rebound with different velocities. To analyze such scenarios, an additional parameter called the Coefficient of Restitution (COR) is required.
The Coefficient of Restitution is defined as the ratio of the relative velocity of separation of the bodies after impact to their relative velocity of approach before impact. Mathematically, it is expressed as:
e = \frac{v_B' - v_A'}{v_A - v_B}
where:
When e = 1, the collision is perfectly elastic, meaning no kinetic energy is lost. When e = 0, the collision is perfectly inelastic, meaning the objects stick together.
The derivation relies on Newton’s impact law and conservation of linear momentum.
Newton’s law states that the relative velocity of separation after the collision is proportional to the relative velocity of approach before the collision:
v_B' - v_A' = -e (v_A - v_B)
The total momentum before and after the collision must be conserved. For two bodies of masses m_A and m_B, this gives:
m_A v_A + m_B v_B = m_A v_A' + m_B v_B'
Rewriting these equations, we solve for v_A' and v_B':
From the restitution equation:
v_B' = v_A' - e (v_A - v_B)
Substitute this expression for v_B' into the momentum equation:
m_A v_A + m_B v_B = m_A v_A' + m_B \left(v_A' - e (v_A - v_B)\right)
Expanding and simplifying:
m_A v_A + m_B v_B = v_A'(m_A + m_B) - m_B e (v_A - v_B)
Solve for v_A':
v_A' = \frac{m_A v_A + m_B v_B + m_B e (v_A - v_B)}{m_A + m_B}
Now substitute v_A' back into the expression for v_B':
v_B' = \frac{m_A v_A + m_B v_B + m_B e (v_A - v_B)}{m_A + m_B} - e (v_A - v_B)
Simplify to find v_B':
v_B' = \frac{m_A v_A + m_B v_B - m_A e (v_A - v_B)}{m_A + m_B}
These equations provide the final velocities v_A' and v_B' of the two bodies after the collision. The coefficient of restitution e determines the elasticity of the collision, with e = 1 corresponding to no kinetic energy loss (perfectly elastic collision) and e = 0 (perfectly plastic collision) corresponding to maximum energy dissipation.
The line of action is defined as the common normal or the perpendicular line to the impacting surfaces of two colliding bodies. This line is normal to the contact surfaces at the point of impact. When we analyze collisions, the behavior of the bodies along this line determines the nature of the impact.
If the mass centers of the colliding bodies lie on the line of action, the impact is classified as a central impact. When, in addition to the mass centers lying on the line of action, the velocities of the colliding bodies are also aligned with the line of action, the impact is further classified as a direct central impact.
However, if the velocities of one or both bodies are not aligned with the line of action but the mass centers remain on it, the collision is referred to as an oblique central impact. In this case, the velocities have components both along and perpendicular to the line of action.
To simplify the analysis of collisions, the following assumptions are made:
These assumptions allow us to focus solely on the changes in velocity along the line of action, where the impulse acts during the collision. By applying conservation of momentum and, if needed, the coefficient of restitution, the post-collision velocities along the line of action can be determined. The velocity components perpendicular to the line of action remain constant throughout the collision, simplifying the overall problem.