M1.4 - Dynamics of a particle moving in a straight line or plane

A force is a vector interaction that can change a particle's velocity. Draw or list only the forces acting on the chosen particle, combine them as a resultant, then apply Newton's second law in a fixed coordinate system.

Common forces include weight $m\mathbf g$ vertically downward, a normal reaction perpendicular to a contact surface, tension along a taut string, friction along a rough contact, and stated driving or resistance forces. Resolve each force along the chosen axes before adding; never insert a force merely because an object is moving.

A particle of mass $2$ kg is acted on by forces $(-2\mathbf i+3\mathbf j)$ N and $(4\mathbf i+5\mathbf j)$ N. Hence$$\sum\mathbf F=2\mathbf i+8\mathbf j\text{ N},\qquad \mathbf a=\mathbf i+4\mathbf j\text{ m s}^{-2}.$$Its acceleration magnitude is $\sqrt{17}\,\mathrm{m\,s^{-2}}$; dividing the force magnitude by mass gives the same value.

A connected-particle problem becomes manageable when the body or system is chosen before equations are written. For every stage: identify the active connections and contacts, choose a positive direction, resolve forces, apply $\sum F=ma$, then use kinematics only after the acceleration is known.

Masses $3$ kg and $2$ kg hang on opposite sides of a smooth pulley, with the $3$ kg mass moving downward. For the separate particles,$$3g-T=3a,\qquad T-2g=2a.$$Adding eliminates the internal tension: $g=5a$, so $a=g/5$. Substitution gives $T=12g/5$ N. The same $a$ must be used because the taut inextensible string constrains both motions.

Use a whole-system equation to find acceleration efficiently, then a single-particle equation to recover an internal tension or coupling force. On an incline, resolve perpendicular first when a reaction or friction is needed, then resolve parallel to the motion. Every term in an equation must be a force on the selected body and every $ma$ term must use its mass.

When a particle hits the ground, a string becomes slack or breaks, contact is lost, or an applied force changes, the force model changes immediately. End the first stage, carry forward the boundary velocity and position, draw a new force set, find the new acceleration, and begin a new constant-acceleration interval.

Linear momentum is signed in one dimension: for mass $m$ moving with velocity $v$, $p=mv$. Choose one positive direction before a collision and give every initial and final velocity its sign. Speed alone is insufficient when a particle reverses.

Impulse is the change in momentum of one particle and has units $\mathrm{N\,s}=\mathrm{kg\,m\,s^{-1}}$. The impulses two colliding particles exert on each other are equal in magnitude and opposite in direction. Report $|I|$ when a magnitude is requested.

During a short direct collision, the two internal impulses cancel when both particles are treated as one system. If external impulse is negligible over that interval, total linear momentum is conserved. This is a signed equation, not an equation between total speeds or kinetic energies.

A $2$ kg particle moving at $4\,\mathrm{m\,s^{-1}}$ collides directly with a $3$ kg particle moving at $-1\,\mathrm{m\,s^{-1}}$. If the second particle leaves at $1\,\mathrm{m\,s^{-1}}$, conservation gives$$2(4)+3(-1)=2v+3(1),$$so $v=1\,\mathrm{m\,s^{-1}}$. The impulse on the first particle is $2(1-4)=-6\,\mathrm{N\,s}$, hence magnitude $6\,\mathrm{N\,s}$.

For a particle moving along a rough surface in this syllabus, friction acts along the contact surface opposite the relative motion. Its magnitude is the coefficient of friction times the normal reaction.

The coefficient $\mu$ is dimensionless. The reaction $R$ must be found from forces perpendicular to the surface; it equals $mg$ only on a horizontal surface with no other force having a perpendicular component. A pull angled upward reduces $R$, while a push angled downward increases it.

A $5$ kg block is pulled across a horizontal floor by a $20$ N force at $30^\circ$ above the horizontal. If $\mu=0.2$, perpendicular balance gives$$R=5g-20\sin30^\circ=39\text{ N}.$$Thus $F=0.2(39)=7.8$ N. Horizontally,$$20\cos30^\circ-7.8=5a,$$so $a=1.90\,\mathrm{m\,s^{-2}}$ to three significant figures.

On an incline with no other perpendicular force, $R=mg\cos\theta$. Friction still opposes motion: it acts down the plane while the particle moves up, and up the plane while it moves down. Recalculate $R$ whenever the applied-force geometry changes.