Successive impactsEdexcel International A Level Further Maths: Revision notes

In successive impacts several collisions happen one after another along a straight line: between particles, or between a particle and a fixed plane (a wall) hit at right angles. Impacts with a plane surface are never oblique in this topic. The method never changes. For each collision in turn:

• draw a before-and-after diagram with one positive direction for the whole problem;
• write the momentum equation $m_1u_1+m_2u_2=m_1v_1+m_2v_2$;
• write the restitution equation, speed of separation $=e\times$ speed of approach;
• solve, and carry the new velocities forward as the starting velocities for the next collision. After the last collision, compare velocities to decide whether another collision can happen.

A fixed wall does not move, so conservation of momentum is not used for the particle (the wall supplies an impulse); Newton's law of restitution alone gives the result. If a particle hits the wall at right angles with speed $u$, it rebounds with speed $v=eu,$ in the opposite direction. The speed is multiplied by $e$ and the direction is reversed. The impulse of the wall on the particle is $m(eu+u)=m(1+e)u$, directed away from the wall. The kinetic energy lost is $\frac12mu^2(1-e^2)$.

Typical set-up: $A$ hits $B$, then $B$ hits the wall and rebounds, and $A$ and $B$ may collide again. Work through in order: (1) $A$ and $B$ collide, giving two new velocities; (2) $B$ hits the wall, so its speed becomes $e_{\text{wall}}$ times its speed and its direction reverses; (3) compare $A$ and $B$. Example: $A$ ($2$ kg) at $6$ m s$^{-1}$ hits $B$ ($1$ kg) with $e=\frac12$: momentum $12=2v_A+v_B$ and $v_B-v_A=3$, so $v_A=3$ and $v_B=6$. $B$ hits the wall with $e=\frac13$ and rebounds at $2$ m s$^{-1}$ towards $A$, which is still moving towards the wall at $3$ m s$^{-1}$: they approach each other, so a second collision occurs. Take care: a particle cannot reach the wall while another particle is between it and the wall.

With particles $A$, $B$, $C$ in a line, collisions happen between neighbouring particles only, and only when the rear particle is moving faster than the one in front (in the same direction), or the two are moving towards each other. After each collision, list the velocity of every particle. A further collision occurs when two neighbours satisfy: the one behind has a greater velocity than the one in front. No further collision is possible when the velocities, from the back, are in increasing order (and no particle can return to a wall). Example: $A$ ($4$ kg, $6$ m s$^{-1}$) hits $B$ ($2$ kg) at rest, $e=\frac12$: $v_A=3$, $v_B=6$. $B$ hits $C$ ($3$ kg) at rest: $12=2v_B'+3v_C$ and $v_C-v_B'=3$, so $v_B'=0.6$ and $v_C=3.6$. Since $A$ at $3$ m s$^{-1}$ is faster than $B$ at $0.6$ m s$^{-1}$, $A$ collides with $B$ again.

Every collision with $e<1$ loses kinetic energy. For a total over several collisions, subtract the final total kinetic energy from the initial total, or add the loss at each collision. A wall impact loses $\frac12mu^2(1-e^2)$. Problems often involve a symbolic mass or speed ($m$, $3m$, $u$), so work in fractions and keep $m$ and $u$ throughout; they usually cancel. For a 'show that there is a second collision' question, finish with a sentence that compares the two velocities (which is faster, which is behind, which direction). Always check that each value of $e$ used is the one for that pair of bodies.