Published by Patrick Mutisya · 14 days ago
Define and use linear momentum as the product of mass and velocity.
The linear momentum \$\mathbf{p}\$ of a particle is defined as the product of its mass \$m\$ and its velocity \$\mathbf{v}\$:
\$\mathbf{p}=m\mathbf{v}\$
Momentum is a vector quantity; it has the same direction as the velocity.
| Symbol | Quantity | SI Unit | Expression |
|---|---|---|---|
| \$m\$ | Mass | kilogram (kg) | given |
| \$\mathbf{v}\$ | Velocity | metre per second (m·s⁻¹) | given or measured |
| \$\mathbf{p}\$ | Linear momentum | kilogram metre per second (kg·m·s⁻¹) | \$m\mathbf{v}\$ |
Newton’s second law can be written in terms of momentum:
\$\mathbf{F}_{\text{net}}=\frac{d\mathbf{p}}{dt}\$
If the mass of the object is constant, this reduces to the familiar form:
\$\mathbf{F}_{\text{net}}=m\frac{d\mathbf{v}}{dt}=m\mathbf{a}\$
Thus, a net external force changes the momentum of a body.
In the absence of external forces, the total linear momentum of a closed system remains constant:
\$\sum \mathbf{p}{\text{initial}} = \sum \mathbf{p}{\text{final}}\$
This principle is especially useful for analysing collisions and explosions.
\$mA vA + mB vB = mA vA' + mB vB'\$
(where primed quantities are after the collision).
\$\frac{1}{2}mA vA^2 + \frac{1}{2}mB vB^2 = \frac{1}{2}mA {vA'}^2 + \frac{1}{2}mB {vB'}^2\$
\$vA' = -0.57\ \text{m·s}^{-1},\qquad vB' = 1.43\ \text{m·s}^{-1}\$
\$0.5(2.0) + 0.8(0) = 0.5(-0.57) + 0.8(1.43) \approx 1.0\ \text{kg·m·s}^{-1}\$
Linear momentum \$\mathbf{p}=m\mathbf{v}\$ is a fundamental vector quantity linking mass and velocity. Newton’s second law states that the net external force equals the time rate of change of momentum. In isolated systems, total momentum is conserved, providing a powerful tool for solving collision and explosion problems.