M2 4.3 - Successive collisions

You already know

For $P$ left of $Q$, use

These equations apply separately at each direct collision with negligible external impulse.

Worked example 1: handing velocities to the next impact

Question. Equal-mass particles $P$ and $Q$ move on a smooth horizontal line. Initially $P$ moves right at $6\,\mathrm{m\,s^{-1}}$ towards stationary $Q$. Their coefficient of restitution is $1/2$. After their first collision, $Q$ reaches a barrier and returns left at $3\,\mathrm{m\,s^{-1}}$. It then collides with $P$ again, before $P$ reaches the barrier. Find the velocities after each particle-particle collision.

Plan. Solve the first impact, then replace only $Q$'s velocity by its stated return velocity before solving the second.

At the second collision the incoming pair is $(1.5,-3)$, not $(6,0)$:

Answer. First impact: $(1.5,4.5)\,\mathrm{m\,s^{-1}}$. Second impact: $(-1.875,0.375)\,\mathrm{m\,s^{-1}}$, right positive.

Check. The second pair sums to $-1.5$, and $0.375-(-1.875)=2.25$; both second-impact equations hold.

Worked example 2: a slower return

Question. The same first collision occurs, but $Q$ returns left at $1\,\mathrm{m\,s^{-1}}$. Find the velocities after the second particle-particle impact.

Plan. Use incoming velocities $1.5$ and $-1$ at the second collision.

Adding the equations gives $2w_Q=1.75$, then $w_P=0.5-w_Q$.

Answer. $w_P=-0.375\,\mathrm{m\,s^{-1}}$ and $w_Q=0.875\,\mathrm{m\,s^{-1}}$.

Check. Their sum is $0.5$ and separation speed is $1.25$, half the approach speed $2.5$.

Watch out

Momentum is conserved during each particle-particle impact, but the barrier supplies an external impulse between them. Do not conserve the particles' combined momentum across the entire sequence.

Worked example 3: two successive pairs

Question. Three particles, each of mass $1\,\mathrm{kg}$, lie in order $P,Q,R$ on a smooth horizontal line. $P$ moves right at $6\,\mathrm{m\,s^{-1}}$; $Q$ and $R$ are at rest. The coefficient of restitution is $1/2$ for both $P$-$Q$ and $Q$-$R$. Find their velocities immediately after the first $Q$-$R$ collision.

Plan. Use the previous equal-mass first impact, then solve a new pair with incoming velocities $4.5$ and zero.

For $Q$-$R$,

giving $w_Q=1.125$ and $w_R=3.375$.

Answer. $(1.5,1.125,3.375)\,\mathrm{m\,s^{-1}}$, right positive.

Check. Total momentum is $1.5+1.125+3.375=6\,\mathrm{kg\,m\,s^{-1}}$. The $Q$-$R$ separation speed is $2.25$, as required.

Worked example 4: a heavier third particle

Question. Repeat the previous sequence with $R$ of mass $2\,\mathrm{kg}$; all other data are unchanged.

Plan. Retain $P$'s velocity and use the new mass in the $Q$-$R$ momentum equation.

Substituting $w_Q=w_R-2.25$ gives $3w_R=6.75$.

Answer. $(v_P,w_Q,w_R)=(1.5,0,2.25)\,\mathrm{m\,s^{-1}}$.

Check. Total momentum is $1.5+0+2(2.25)=6\,\mathrm{kg\,m\,s^{-1}}$; the stationary $Q$ has transferred more momentum to the heavier $R$.