Further Mechanics 1

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• Momentum is a vector: $\mathbf{p}=m\mathbf{v}$. Use conservation of momentum only when the total external impulse in the chosen direction is zero; use $\mathbf{J}=m(\mathbf{v}-\mathbf{u})$ when an impulse changes one particle's velocity.
• For direct impact, choose one positive direction and keep signed velocities. When A catches B from behind in the positive direction, solve momentum with $v_B-v_A=e(u_A-u_B)$: the right-hand side is the speed of approach and the left-hand side is the speed of separation. For ordinary spheres, $0\le e\le1$.
• For successive impacts, finish one impact completely before starting the next. Update the particles' positions and signed velocities, decide whether they are approaching again, and then write a new momentum-restitution pair.
• For a smooth oblique impact, resolve parallel and perpendicular to the line of centres, or normal and tangential to a wall. Only the normal components change; the tangential components are unchanged because a smooth contact gives no tangential impulse.
• Work-energy is usually shorter than resolving forces over a distance: $\Delta KE=$ total work done. Include signs explicitly for gravity, resistance and elastic forces. Use conservation of mechanical energy only when no non-conservative force does work.
• Power is the rate of doing work. Instantaneously, $P=\mathbf{F}\mathbin{\cdot}\mathbf{v}$; in one-dimensional motion this is $P=Fv$. At constant speed, the driving force balances the total resistance before you use $P=Fv$.
• For an elastic string or spring of natural length $l$, modulus $\lambda$ and extension $x$, Hooke's law gives $T=\dfrac{\lambda x}{l}$ and the stored elastic energy is $\dfrac{\lambda x^2}{2l}$. A string has zero tension at $x=0$ and is slack below natural length; unlike a string, a spring can also be compressed.
• Worked example - vector impulse: a $2$ kg particle has velocity $(3\mathbf{i}-2\mathbf{j})\text{ m s}^{-1}$ and receives impulse $(-4\mathbf{i}+6\mathbf{j})\text{ N s}$. Since $\mathbf{v}=\mathbf{u}+\mathbf{J}/m$, its new velocity is $(\mathbf{i}+\mathbf{j})\text{ m s}^{-1}$.
• Worked example - direct impact: masses $2$ kg and $3$ kg move in the same direction at $5$ and $1\text{ m s}^{-1}$, with $e=\tfrac12$. The equations $2v_A+3v_B=13$ and $v_B-v_A=2$ give $v_A=\tfrac75$ and $v_B=\tfrac{17}{5}\text{ m s}^{-1}$. The kinetic-energy loss is $26.5-19.3=7.2$ J.
• Worked example - oblique spheres: equal smooth spheres have line of centres parallel to $\mathbf{i}$. Sphere A approaches with velocity $6\mathbf{i}+4\mathbf{j}$, B is at rest and $e=\tfrac12$. The $\mathbf{j}$ component of A stays $4$; the one-dimensional normal calculation gives $\mathbf{v}_A=1.5\mathbf{i}+4\mathbf{j}$ and $\mathbf{v}_B=4.5\mathbf{i}$.
• Worked example - elastic energy: a $2$ kg particle is attached to a horizontal spring with $l=1$ m and $\lambda=50$ N. Released from rest at extension $0.2$ m, it has elastic energy $1$ J, so at natural length $\tfrac12(2)v^2=1$ and $v=1\text{ m s}^{-1}$.

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