MB2 - Direct collisions | AQA A-Level Further Mathematics

A collision law describes how rapidly two bodies separate compared with how rapidly they approached. You will combine this law with momentum, model rebounds from a fixed wall and check that calculated velocities fit the physical collision.

Relative approach and separation

A faster ball catches a slower ball from behind. Must one reverse direction for them to separate afterwards? No: the front ball can simply move away faster. The collision law measures relative motion, not whether an individual velocity changes sign.

In a direct collision, the velocities lie along the line of impact. For smooth spheres this is the line joining their centres at contact. Put $A$ left of $B$ , take right as positive and let their initial velocities be $u_A,u_B$ . They approach if $u_A>u_B$ , because the gap closes at speed $u_A-u_B$ . Afterwards, the gap opens at speed $v_B-v_A$ .

Coefficient of restitution

For an approaching pair, the coefficient of restitution $e$ is the speed of separation divided by the speed of approach, measured along the line of impact. Newton's experimental law for the usual passive collision model is

v_B-v_A=e(u_A-u_B),\qquad 0\le e\le1

The approach speed must be positive for this ratio to define $e$ . The case $e=1$ has equal approach and separation speeds; $e=0$ has zero immediate relative separation speed along the line of impact.

This law is not separately supplied in the formula booklet. It describes the collision pair and conditions being modelled; it need not be one universal constant for every impact involving those materials.

[DIAGRAM: asset_name: Relative speeds in a catching collision; asset_slug: aqa7367-relative-collision; caption: Both balls can move right before and after a collision while their gap changes from closing to opening; description: Two before-and-after panels show A left of B with right positive. Before, A moves right at 6 metres per second and B right at 2, so approach speed is 4. After, A moves right at 3 and B right at 5, so separation speed is 2. Arrows begin on the balls and use one common velocity scale. The coefficient is 2 divided by 4 equals one half.]
Both balls can move right before and after a collision while their gap changes from closing to opening

Worked example 1: catching up

Question. $A$ is left of $B$ . Before collision, $u_A=6,u_B=2$ ; afterwards, $v_A=3,v_B=5$ , all in $\mathrm{m\,s}^{-1}$ with right positive. Find $e$ .

Plan. Calculate the gap's closing speed before and opening speed after.

\begin{aligned} \text{approach speed}&=6-2=4\\ \text{separation speed}&=5-3=2\\ e&=\frac{2}{4}=\frac12 \end{aligned}

Answer. $e=\tfrac12$ , a dimensionless ratio.

Check. The gap decreases by $4\,\mathrm{m}$ per second before impact and increases by $2\,\mathrm{m}$ per second afterwards. Neither ball needs to reverse.

Change only $B$ 's incoming velocity to $-2$ , leaving the listed outgoing velocities unchanged. Opposing velocities add to the approach speed through signed subtraction.

Worked example 2: approaching head-on

Question. With $A$ still left of $B$ , now $u_A=6,u_B=-2,v_A=3,v_B=5$ , in $\mathrm{m\,s}^{-1}$ . Find the restitution coefficient represented by these relative speeds.

Plan. Retain the negative sign in $u_A-u_B$ .

\begin{aligned} \text{approach speed}&=6-(-2)=8\\ \text{separation speed}&=5-3=2\\ e&=\frac{2}{8}=\frac14 \end{aligned}

Answer. $e=\tfrac14$ .

Check. In $0.1\,\mathrm{s}$ of incoming uniform motion the gap would shrink by $0.8\,\mathrm{m}$ ; with the outgoing velocities it grows by $0.2\,\mathrm{m}$ . Their ratio is again $1/4$ . This checks the relative-motion calculation; masses would also be needed to test momentum for a particular isolated pair.

Restitution is one collision equation. It cannot by itself decide whether an arbitrary set of velocities conserves momentum.

Watch out

Wrong: $e=v_B/u_A$ for two moving bodies. Use relative speeds along the line of impact: $e=(v_B-v_A)/(u_A-u_B)$ with the stated ordering and positive approach speed.

Check the relative speeds for a fresh pair.

Rebounds from a fixed smooth wall

For a direct impact on a fixed wall, the wall's velocity is zero before and after. If positive points towards the wall and incoming velocity is $u>0$ , separation requires outgoing velocity $v\le0$ . The relative-speed law becomes $-v=eu$ .

Direct impact with a fixed wall

For a direct impact with incoming velocity $u>0$ towards a fixed smooth wall,

v=-eu\qquad(0\le e\le1)

The outgoing speed is $eu$ ; the minus sign in velocity records reversal. This is an instantaneous impact model with other impulses negligible.

Not in the formula booklet: learn it.

The ball alone does not conserve momentum: the wall exerts an external impulse. Smoothness means no tangential contact impulse; in direct impact there is no tangential incoming component.

Worked example 3: a direct rebound

Question. A ball hits a fixed smooth wall directly at $4\,\mathrm{m\,s}^{-1}$ . Taking towards the wall as positive, find its outgoing velocity when $e=1/2$ .

Plan. Reverse the normal direction and multiply the incoming speed by $e$ .

v=-eu=-\frac12(4)=-2\,\mathrm{m\,s}^{-1}

Answer. Velocity is $-2\,\mathrm{m\,s}^{-1}$ , a speed of $2\,\mathrm{m\,s}^{-1}$ away from the wall.

Check. Relative separation speed divided by approach speed is $2/4=1/2$ . The ball moves into the accessible region, not through the wall.

Change only $e$ , keeping the incoming velocity and wall fixed.

Worked example 4: a more elastic rebound

Question. The same ball hits the same fixed smooth wall at $4\,\mathrm{m\,s}^{-1}$ , but now the model uses $e=3/4$ . Find its outgoing velocity.

Plan. Apply the wall law with the new restitution coefficient.

v=-\frac34(4)=-3\,\mathrm{m\,s}^{-1}

Answer. $-3\,\mathrm{m\,s}^{-1}$ , directed away from the wall.

Check. The ratio of outgoing to incoming speeds is $3/4$ , and the greater coefficient gives a greater rebound speed than example 3 while still not exceeding $4$ .

At $e=0$ , the outgoing normal velocity is zero; the model does not predict a rebound with positive normal speed. At $e=1$ , speed is preserved but velocity still reverses.

Watch out

Wrong: $v=eu$ in the same positive direction before and after. For a direct wall rebound, $v=-eu$ . A positive outgoing speed must still be accompanied by an “away from the wall” direction.

Keep that direction in your answer.

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Mechanics

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MB2 - Direct collisions

Relative approach and separation

Rebounds from a fixed smooth wall

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