How Much Speed Survives a Bounce?
A ball is dropped from rest at a height of $2\ \text{m}$, so it reaches the floor at $u = \sqrt{2gh} = 6.3\ \text{m s}^{-1}$. It leaves the floor at $v = eu$, and the height it then climbs is $e^{2}h$. Drag the restitution slider and watch the rebound column: a bouncy surface keeps most of the speed, a dead one very little.
impact speed $u = 6.3\ \text{m s}^{-1}$, rebound speed $v = eu =$ 3.8 m s⁻¹, rebound height $e^{2}h =$ 0.72 m
Both the rebound speed and the rebound height are fixed by the single number $e$, which lies between 0 (the ball stays down) and 1 (the ball returns to the full height).
Try each in your head before clicking an answer — this is a test, not a reveal. See how many you get first try. The clock starts on your first answer.
R1. A ball falls from rest through $2\ \text{m}$. Its speed on reaching the ground is (take $g = 10\ \text{m s}^{-2}$):
R2. A ball reaches the floor at $8\ \text{m s}^{-1}$ and leaves it at $6\ \text{m s}^{-1}$. The ratio of the rebound speed to the impact speed is:
R3. A ball leaves the floor at $3\ \text{m s}^{-1}$. The height it reaches is (take $g = 10\ \text{m s}^{-2}$):
For two bodies that collide along a line, the speed with which they separate is a fixed fraction of the speed with which they approached:
| Statement | Meaning |
|---|---|
| $v_{2} - v_{1} = e(u_{1} - u_{2})$ | separation $= e \times$ approach |
| $0 \le e \le 1$ | $e = 0$ coalesce, $e = 1$ perfectly elastic |
Mini-example: two particles approach each other at $6\ \text{m s}^{-1}$ and separate at $3\ \text{m s}^{-1}$, so $e = \dfrac{3}{6} = 0.5$. A value greater than 1 is impossible — a collision cannot create speed of separation.
A dropped ball has one body, so the floor supplies the second half of the relation. Speeds come from the heights:
| Quantity | Relation |
|---|---|
| Impact and rebound speeds | $u = \sqrt{2gh_{1}}$, $v = \sqrt{2gh_{2}}$ |
| Coefficient of restitution | $e = \dfrac{v}{u} = \sqrt{\dfrac{h_{2}}{h_{1}}}$ |
| Rebound height | $h_{2} = e^{2}h_{1}$ |
Mini-example: a ball dropped from $2\ \text{m}$ rebounds to $0.5\ \text{m}$, so $e = \sqrt{\dfrac{0.5}{2}} = 0.5$. Notice that it is the square root of the height ratio, not the height ratio itself.
For a general impact of two free bodies, conservation of momentum gives one equation and the restitution relation gives the second, so both after-velocities can be found.
$2\ \text{kg}$ moving at $5\ \text{m s}^{-1}$ strikes $3\ \text{kg}$ at rest, and $e = 0.4$. Momentum: $10 = 2v_{1} + 3v_{2}$; restitution: $v_{2} - v_{1} = 0.4 \times 5 = 2$. Solving: $v_{1} = 0.8\ \text{m s}^{-1}$ and $v_{2} = 2.8\ \text{m s}^{-1}$.
If $e$ is the same at every bounce, the height after each bounce is multiplied by $e^{2}$ again, so the successive heights are $e^{2}h,\ e^{4}h,\ e^{6}h, \ldots$
Groups of 4: on the collision tab, fix the masses and velocities and sweep the restitution slider — the total momentum never moves, while the speed of separation locks on to e times the speed of approach. Then open the bounce tab and check that the rebound height is e² times the drop height. Three graded challenges follow.
Work alone through the four cards. Decide for each claim whether it is correct or hides an error, then read the explanation. The two errors are the classic ones — the height ratio used as e instead of its square root, and a coupling mistaken for a perfectly elastic impact.
A ball of mass $0.6\ \text{kg}$ is dropped from a height of $1.8\ \text{m}$ onto a solid floor. Each time the ball bounces on the floor it loses $10\%$ of its speed. Show that the ball first fails to reach a height of $1\ \text{m}$ after the third bounce. Take $g = 10\ \text{m s}^{-2}$.
Losing $10\%$ of its speed means the ball keeps $90\%$, so $e = 0.9$. Each bounce multiplies the height by $e^{2} = 0.81$, so build the heights one bounce at a time and compare each with $1\ \text{m}$.
The ball keeps $90\%$ of its speed at each bounce, so $e = 0.9$ and each rebound height is $e^{2} = 0.81$ times the previous one.
After the 1st bounce: $h_{1} = 0.81 \times 1.8 = 1.458\ \text{m} \ge 1$. After the 2nd bounce: $h_{2} = 0.81 \times 1.458 = 1.181\ \text{m} \ge 1$. So the ball still reaches $1\ \text{m}$ twice.
After the 3rd bounce: $h_{3} = 0.81 \times 1.181 = 0.957\ \text{m} < 1$. So the ball first fails to reach $1\ \text{m}$ after the third bounce, as required.
Particle $A$ of mass $2\ \text{kg}$ moves at $4\ \text{m s}^{-1}$ along a smooth horizontal surface towards particle $B$ of mass $6\ \text{kg}$, which is at rest. The coefficient of restitution between the particles is $0.5$. (i) Show that the velocity of $A$ after the impact is $-0.5\ \text{m s}^{-1}$. (ii) Find the velocity of $B$ after the impact.
Neither after-velocity is known, so momentum conservation alone is not enough — two equations are needed. Momentum gives $2 \times 4 = 2v_{A} + 6v_{B}$; restitution gives $v_{B} - v_{A} = 0.5 \times 4$. Solve them together: part (i) asks for $A$'s velocity, and a negative answer means $A$ has rebounded.
$2 \times 4 + 6 \times 0 = 2v_{A} + 6v_{B}$, so $2v_{A} + 6v_{B} = 8$, or $v_{A} + 3v_{B} = 4$.
Speed of approach $= 4 - 0 = 4$, so the speed of separation is $v_{B} - v_{A} = 0.5 \times 4 = 2$.
From step 2, $v_{B} = v_{A} + 2$. Substituting into step 1: $v_{A} + 3(v_{A} + 2) = 4$, so $4v_{A} = -2$ and $v_{A} = -0.5\ \text{m s}^{-1}$, as required. Particle $A$ therefore rebounds at $0.5\ \text{m s}^{-1}$.
Back-substituting, $v_{B} = v_{A} + 2 = 1.5\ \text{m s}^{-1}$. Check the separation: $1.5 - (-0.5) = 2$, which is $0.5 \times 4$ ✓.
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. Two bodies collide with a speed of approach of $5\ \text{m s}^{-1}$ and leave the impact with a speed of separation of $3\ \text{m s}^{-1}$. The coefficient of restitution is:
Q2. A ball strikes the ground at $8\ \text{m s}^{-1}$. If the coefficient of restitution is $0.5$, the rebound speed is:
Q3. A ball is dropped from a height of $3\ \text{m}$ onto a floor with coefficient of restitution $0.5$. It rebounds to a height of:
Q4. A $2\ \text{kg}$ particle moving at $4\ \text{m s}^{-1}$ strikes an identical particle at rest, and the impact is perfectly elastic ($e = 1$). After the impact:
Work each on paper, then reveal the solution to check your working.
A ball of mass $0.6\ \text{kg}$ is dropped from a height of $1.8\ \text{m}$ onto a solid floor. Each time the ball bounces on the floor it loses $10\%$ of its speed. Work out how much momentum was absorbed by the floor in the first bounce. Take $g = 10\ \text{m s}^{-2}$.
Reveal Solution
Impact speed: $u = \sqrt{2 \times 10 \times 1.8} = 6\ \text{m s}^{-1}$ downwards. The ball leaves at $v = 0.9 \times 6 = 5.4\ \text{m s}^{-1}$ upwards, so its momentum changes by $0.6 \times (6 + 5.4) = 6.84\ \text{kg m s}^{-1}$ — the rebound adds the two speeds. That is the momentum absorbed by the floor: $6.84\ \text{N s}$.
A ball of mass $0.6\ \text{kg}$ is dropped from a height of $1.8\ \text{m}$ onto a solid floor. Each time the ball bounces on the floor it loses $10\%$ of its speed. State the modelling assumptions you have made in working out the motion of the ball.
Reveal Solution
Assumptions: the ball behaves as a particle; the motion stays vertical, with no sideways component; air resistance is neglected, so the ball falls freely at $g = 10\ \text{m s}^{-2}$; the floor is fixed and does not move or deform; the coefficient of restitution is the same at every bounce, so the ball keeps exactly $10\%$ of its speed each time; and the contact with the floor is instantaneous, so no time is spent in contact.
Balls $X$, $Y$ and $Z$ lie at rest on a smooth horizontal surface, with $Y$ between $X$ and $Z$. Balls $X$ and $Z$ each have mass $2\ \text{kg}$ and ball $Y$ has mass $1\ \text{kg}$. Ball $X$ is given a velocity of $1\ \text{m s}^{-1}$ towards ball $Y$. Balls $X$ and $Y$ collide; after this collision the speed of $Y$ is three times the speed of $X$. Ball $Y$ goes on to collide with ball $Z$; after this collision the speed of $Y$ is the same as the speed of $X$, and the speed of $Z$ is twice the speed of $Y$. Finally ball $Y$ collides with ball $X$ again; after this collision the speed of $Y$ is twice the speed of $X$, and the speed of $Z$ is four times the speed of $Y$. Show that the balls are now all travelling in the same direction and that no further collisions occur.
Reveal Solution
Take the positive direction as the way $X$ was given. First impact: $2 \times 1 = 2v_{X} + v_{Y}$ with $v_{Y} = 3v_{X}$, so $2 = 5v_{X}$ and $v_{X} = 0.4$, $v_{Y} = 1.2\ \text{m s}^{-1}$ (both positive — the tested alternative $v_{X} = -2$ would create kinetic energy, so it is rejected).
Second impact ($Y$ on $Z$): momentum $1.2 = v_{Y} + 2v_{Z}$, with $|v_{Y}| = 0.4$ and $|v_{Z}| = 0.8$: only $v_{Y} = -0.4$, $v_{Z} = 0.8$ fits, so $Y$ rebounds and $Z$ moves away at $0.8\ \text{m s}^{-1}$.
Third impact ($Y$ on $X$): the momentum of the $X$–$Y$ pair is $2(0.4) + 1(-0.4) = 0.4\ \text{kg m s}^{-1}$. $Z$ is untouched, so its speed stays $0.8$; the statement that the speed of $Z$ is four times the speed of $Y$ gives $v_{Y} = 0.2$ and then the speed of $Y$ being twice the speed of $X$ gives $v_{X} = 0.1\ \text{m s}^{-1}$. Check: $1(0.2) + 2(0.1) = 0.4$ ✓.
Final velocities: $X$ at $+0.1$, $Y$ at $+0.2$ and $Z$ at $+0.8\ \text{m s}^{-1}$ — all in the same direction, and with $Z$ ahead of $Y$ ahead of $X$ and each moving faster than the one behind it. Every pair is therefore separating, so no further collisions occur.
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