Direct impact and Newton's law of restitutionEdexcel International A Level Further Maths: Revision notes

A direct impact is a collision between two particles (or smooth spheres) whose velocities before and after the collision lie along the line of centres, so everything happens in one dimension. Draw a diagram showing the velocities before and after, choose one direction as positive and write every velocity as a signed value. During the impact the only horizontal forces are the equal and opposite impulses between the particles, so total momentum is conserved whatever the value of $e$: $m_1u_1+m_2u_2=m_1v_1+m_2v_2.$ Conservation of momentum alone gives one equation in two unknowns $v_1,v_2$, so a second equation is always needed.

Newton's law of restitution gives the second equation. For a direct impact between two particles: $\text{speed of separation}=e\times\text{speed of approach},\qquad\text{i.e. }v_2-v_1=e\,(u_1-u_2),$ where particle 2 is the one ahead in the positive direction. The coefficient of restitution $e$ depends only on the materials of the two particles. Check the sign: after the impact, particle 1 cannot be ahead of particle 2, so $v_2-v_1\ge0$. Solving the momentum and restitution equations simultaneously gives $v_1$ and $v_2$. A quick check: substitute both answers back into both equations.

The coefficient of restitution satisfies $0\le e\le1$.

• $e=1$: perfectly elastic. Speed of separation equals speed of approach and no kinetic energy is lost.
• $e=0$: inelastic. The particles do not separate; they move together with a common velocity (they coalesce).
• $0<e<1$: partially elastic. Kinetic energy is lost. A value of $e$ above $1$ would mean kinetic energy is gained in a collision, which cannot happen, so a calculated $e>1$ shows an error. If a quadratic such as $e^2=\frac14$ appears, reject the negative root and take $e=\frac12$. Imposing $0\le e\le1$ on a result such as $v_B=\frac{23+15e}{4}$ gives the range of possible final speeds.

Momentum is always conserved, but kinetic energy is lost whenever $e<1$ (sound, heat and deformation). On a smooth horizontal surface there is no change in potential energy, so the loss of mechanical energy is $\Delta E=\left(\tfrac12m_1u_1^2+\tfrac12m_2u_2^2\right)-\left(\tfrac12m_1v_1^2+\tfrac12m_2v_2^2\right).$ Calculate the kinetic energy of each particle before and after, then subtract; the answer must be positive (or zero if $e=1$). A negative answer means a calculation error. The impulse on a particle is its change of momentum. In the worked example below, the impulse on $B$ is $3(3)-3(0)=9$ N s.

$A$ ($2$ kg, $5$ m s$^{-1}$) hits $B$ ($3$ kg, at rest) with $e=0.5$. Momentum: $10=2v_A+3v_B$. Restitution: $v_B-v_A=0.5(5)=2.5$. Substituting $v_A=v_B-2.5$: $10=5v_B-5$, so $v_B=3$ and $v_A=0.5$ m s$^{-1}$, both in the original direction. Kinetic energy before $=25$ J, after $=0.25+13.5=13.75$ J, so $11.25$ J is lost. Method: (1) diagram with a positive direction; (2) momentum equation; (3) restitution equation; (4) solve; (5) check $v_B\ge v_A$ and that the energy loss is positive.