Published by Patrick Mutisya · 14 days ago
State and apply each of Newton’s laws of motion to solve quantitative problems involving momentum and impulse.
A body remains at rest, or moves with constant velocity, unless acted upon by a net external force.
The net external force on a body equals the rate of change of its momentum.
\$\mathbf{F}_{\text{net}} = \frac{d\mathbf{p}}{dt} = \frac{d}{dt}(m\mathbf{v})\$
For constant mass this reduces to the familiar form:
\$\mathbf{F}_{\text{net}} = m\mathbf{a}\$
For every action force there is an equal and opposite reaction force.
\$\mathbf{F}{AB} = -\mathbf{F}{BA}\$
Linear momentum of a particle of mass \$m\$ moving with velocity \$\mathbf{v}\$ is defined as:
\$\mathbf{p} = m\mathbf{v}\$
Momentum is a conserved vector quantity in isolated systems.
Impulse is the change in momentum produced by a force acting over a time interval \$\Delta t\$:
\$\mathbf{J} = \int{t1}^{t_2} \mathbf{F}\,dt = \Delta\mathbf{p}\$
For a constant force:
\$\mathbf{J} = \mathbf{F}\,\Delta t\$
In the absence of external forces, the total momentum of a system remains constant:
\$\sum \mathbf{p}{\text{initial}} = \sum \mathbf{p}{\text{final}}\$
This principle underlies the analysis of collisions and explosions.
| Collision Type | Momentum | Kinetic Energy | Typical Example |
|---|---|---|---|
| Elastic | Conserved | Conserved | Billiard balls |
| Inelastic | Conserved | Not conserved (some lost as heat, deformation) | Car crash |
| Perfectly Inelastic | Conserved | Maximum loss (objects stick together) | Two carts coupling |
Two spheres, \$m1 = 0.5\ \text{kg}\$ moving at \$u1 = 4\ \text{m·s}^{-1}\$ and \$m_2 = 0.8\ \text{kg}\$ initially at rest, collide elastically. Find their velocities after the collision.
Solution steps:
\$m1u1 + m2u2 = m1v1 + m2v2\$
Since \$u_2 = 0\$:
\$0.5(4) = 0.5v1 + 0.8v2\$
\$\frac{1}{2}m1u1^{2} + \frac{1}{2}m2u2^{2} = \frac{1}{2}m1v1^{2} + \frac{1}{2}m2v2^{2}\$
\$\frac{1}{2}(0.5)(4^{2}) = \frac{1}{2}(0.5)v1^{2} + \frac{1}{2}(0.8)v2^{2}\$
\$v1 = \frac{m1 - m2}{m1 + m2}\,u1 = \frac{0.5 - 0.8}{1.3}\times4 = -0.923\ \text{m·s}^{-1}\$
\$v2 = \frac{2m1}{m1 + m2}\,u_1 = \frac{2(0.5)}{1.3}\times4 = 3.08\ \text{m·s}^{-1}\$
| Law | Statement | Mathematical Form | Typical Application |
|---|---|---|---|
| First | A body at rest or in uniform motion remains so unless acted upon by a net external force. | \$\mathbf{F}_{\text{net}} = 0 \;\Rightarrow\; \mathbf{p} = \text{constant}\$ | Spacecraft drift, frictionless tracks |
| Second | Net external force equals the rate of change of momentum. | \$\mathbf{F}_{\text{net}} = \dfrac{d\mathbf{p}}{dt}\$ | Calculating acceleration, impulse problems |
| Third | For every action there is an equal and opposite reaction. | \$\mathbf{F}{AB} = -\mathbf{F}{BA}\$ | Collision analysis, rocket propulsion |
Consult the Cambridge International AS & A Level Physics (9702) syllabus sections on “Momentum and Newton’s laws of motion” for detailed exam specifications and additional worked examples.