Two Balls, One Meeting Point
Ball X is released from rest at a height of $21\ \text{m}$ at the same instant that ball Y is thrown vertically upwards from the ground at $15\ \text{m s}^{-1}$. Slide the clock and watch both heights: X falls faster and faster, Y slows as it climbs, and the gap closes — they meet when the two heights are equal. Momentum will decide what happens next.
height of X 16.0 m, height of Y 10.0 m — gap = 6.0 m
Both heights are given by the same equation of motion, so the meeting time comes from setting $21 - 5t^{2} = 15t - 5t^{2}$.
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 is thrown vertically upwards at $15\ \text{m s}^{-1}$. After $1\ \text{s}$ its velocity is (take $g = 10\ \text{m s}^{-2}$):
R2. The ball thrown upwards at $15\ \text{m s}^{-1}$ is at a height after $1\ \text{s}$ of:
R3. A ball is dropped from rest. After $1.5\ \text{s}$ its speed is (take $g = 10\ \text{m s}^{-2}$):
Momentum questions in a vertical line are ordinary momentum questions with a sign chosen once and kept:
| Choice | Consequence |
|---|---|
| Downwards positive | a falling body has positive momentum, a rising body negative momentum |
| Upwards positive | the signs are reversed — but they must not be mixed mid-question |
Mini-example: with downwards positive, a $0.05\ \text{kg}$ ball rising at $4\ \text{m s}^{-1}$ has downward momentum $0.05 \times (-4) = -0.2\ \text{kg m s}^{-1}$.
A floor bounce reverses the velocity, and the bounce speed is tied to the height reached:
| Stage | Relation |
|---|---|
| Arrival at the floor | $u = \sqrt{2gh_{1}}$ |
| Leaving the floor | $v = eu = \sqrt{2gh_{2}}$ |
| Momentum change of the ball | $m(u + v)$, directed away from the floor |
Mini-example: a ball dropped from $1.8\ \text{m}$ arrives at $6\ \text{m s}^{-1}$; leaving at $4\ \text{m s}^{-1}$ it rises to $\dfrac{16}{20} = 0.8\ \text{m}$.
The collision itself needs the momenta just before contact, so the meeting time and height come first from the equations of motion: set the two heights equal.
At the meeting, add the momenta with their signs and divide by the combined mass to get the velocity of the coalesced body: mini-example — a $0.05\ \text{kg}$ ball falling at $10\ \text{m s}^{-1}$ meets a $0.05\ \text{kg}$ ball rising at $4\ \text{m s}^{-1}$; with downwards positive, $0.05(10) + 0.05(-4) = 0.3$, so the $0.1\ \text{kg}$ combined body falls at $3\ \text{m s}^{-1}$.
Once the coalesced speed is known, the rest of the question is a constant-acceleration stage: $s = u_{0}T + \dfrac{1}{2}gT^{2}$ gives the remaining time to the ground.
Groups of 4: each card on the left describes an impact; each card on the right gives a rebound quantity. Tap one card on each side to link them — a numbered badge appears on both. Press Check to score the links, and Reset to shuffle and start again. Talk through the direction each answer belongs to before you link.
Work alone on the bounce tab. Set the drop height and the coefficient of restitution, then read the impact speed, the rebound speed, the rebound height and the momentum absorbed by the floor. Try a heavier ball with the same bounce to see which readouts change and which do not, then answer the three graded challenges.
Ball $X$ has mass $0.03\ \text{kg}$. It falls vertically from rest from a window $30\ \text{m}$ above the ground. Ball $Y$ has mass $0.01\ \text{kg}$. At the same time that ball $X$ starts to fall, ball $Y$ is projected vertically upwards from ground level directly towards ball $X$ with initial speed $20\ \text{m s}^{-1}$. Find the downward momentum of each ball just before they meet. Take $g = 10\ \text{m s}^{-2}$.
They meet when their heights above the ground are equal. Write both heights in terms of the same time $t$, solve for $t$, then use $v = u + at$ for each ball. Take downwards as positive for the momenta.
After time $t$: height of $X$ is $30 - 5t^{2}$ and height of $Y$ is $20t - 5t^{2}$. Setting them equal gives $30 = 20t$, so $t = 1.5\ \text{s}$.
$X$ is falling: $v_{X} = 10 \times 1.5 = 15\ \text{m s}^{-1}$ downwards. $Y$ is still rising: $v_{Y} = 20 - 10 \times 1.5 = 5\ \text{m s}^{-1}$ upwards.
With downwards positive: $p_{X} = 0.03 \times 15 = 0.45\ \text{kg m s}^{-1}$ downwards, and $p_{Y} = 0.01 \times (-5) = -0.05\ \text{kg m s}^{-1}$, that is $0.05\ \text{kg m s}^{-1}$ upwards.
A ball of mass $0.2\ \text{kg}$ is dropped from rest from a height of $5\ \text{m}$ onto a fixed horizontal floor. The coefficient of restitution between the ball and the floor is $0.6$. Find (i) the speed with which the ball reaches the floor (ii) the speed with which it leaves the floor (iii) the height it reaches afterwards (iv) the momentum absorbed by the floor. Take $g = 10\ \text{m s}^{-2}$.
Two uses of $v^{2} = 2gh$ surround the bounce: one to find the speed before it and one to find the height after it. In between, restitution multiplies the speed by $e$, and the momentum absorbed is the change in the ball's momentum — a rebound, so the two speeds add.
$u^{2} = 2gh = 2 \times 10 \times 5 = 100$, so $u = 10\ \text{m s}^{-1}$ downwards.
$v = eu = 0.6 \times 10 = 6\ \text{m s}^{-1}$ upwards.
$h_{2} = \dfrac{v^{2}}{2g} = \dfrac{36}{20} = 1.8\ \text{m}$, which is also $e^{2}h = 0.36 \times 5$ ✓.
With upwards positive the ball's velocity changes from $-10$ to $+6$, so $\Delta p = 0.2\,(6 + 10) = 3.2\ \text{kg m s}^{-1}$: the floor absorbs $3.2\ \text{N s}$.
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. A ball is projected vertically upwards at $12\ \text{m s}^{-1}$. After $2\ \text{s}$ its velocity is (take $g = 10\ \text{m s}^{-2}$):
Q2. Taking downwards as positive, the momentum of a $0.5\ \text{kg}$ ball moving upwards at $4\ \text{m s}^{-1}$ is:
Q3. A $0.2\ \text{kg}$ ball falling at $6\ \text{m s}^{-1}$ and a $0.1\ \text{kg}$ ball rising at $3\ \text{m s}^{-1}$ meet and coalesce. Taking downwards as positive, the velocity of the combined body is:
Q4. A ball leaves the floor at $6\ \text{m s}^{-1}$. The height it reaches is (take $g = 10\ \text{m s}^{-2}$):
Work each on paper, then reveal the solution to check your working.
Particle $A$ has mass $4\ \text{kg}$. It moves with speed $3\ \text{m s}^{-1}$ in a straight line on a smooth horizontal surface. Particle $B$ has mass $6\ \text{kg}$ and is at rest on the surface. Particle $A$ collides with particle $B$. After the collision, $A$ and $B$ move away from each other, $A$ rebounding along its original line with speed $v\ \text{m s}^{-1}$ and $B$ moving forward with speed $4v\ \text{m s}^{-1}$. Find the value of $v$.
Reveal Solution
Taking $A$'s direction as positive: $4 \times 3 = -4v + 6(4v)$, so $12 = 20v$ and $v = 0.6\ \text{m s}^{-1}$ ($B$ leaves at $2.4\ \text{m s}^{-1}$). The rebound makes $A$'s momentum negative.
Ball $X$ of mass $0.03\ \text{kg}$ falls from rest from a window $30\ \text{m}$ above the ground. At the same instant ball $Y$ of mass $0.01\ \text{kg}$ is projected vertically upwards from ground level at $20\ \text{m s}^{-1}$ directly towards $X$. The balls coalesce and the combined object falls to the ground. Show that the combined object reaches the ground $2.68\ \text{s}$ after ball $X$ started to fall. Take $g = 10\ \text{m s}^{-2}$.
Reveal Solution
The balls meet when $30 - 5t^{2} = 20t - 5t^{2}$, so $t = 1.5\ \text{s}$, at a height $30 - 5(1.5)^{2} = 18.75\ \text{m}$.
Just before they meet: $X$ falls at $15\ \text{m s}^{-1}$ and $Y$ rises at $5\ \text{m s}^{-1}$, so the total downward momentum is $0.03(15) + 0.01(-5) = 0.4\ \text{kg m s}^{-1}$. The combined mass is $0.04\ \text{kg}$, so the object leaves the meeting at $10\ \text{m s}^{-1}$ downwards.
It then falls $18.75\ \text{m}$ with initial speed $10\ \text{m s}^{-1}$: $18.75 = 10T + 5T^{2}$, so $T^{2} + 2T - 3.75 = 0$ and $T = \dfrac{-2 + \sqrt{19}}{2} = 1.18\ \text{s}$. Total time $= 1.5 + 1.18 = 2.68\ \text{s}$, as required.
Object $A$ has mass $3\ \text{kg}$ and moves with velocity $10\ \text{m s}^{-1}$ towards object $B$, which has mass $6\ \text{kg}$ and is stationary. After they collide, object $A$ bounces back with speed $2\ \text{m s}^{-1}$. Object $B$ then collides with object $C$, which has mass $17\ \text{kg}$ and is stationary. Object $C$ moves at $3\ \text{m s}^{-1}$ after this collision. Deduce whether or not there will be a third collision and explain your reasoning.
Reveal Solution
First collision: $3 \times 10 = 3(-2) + 6v_{B}$, so $v_{B} = 6\ \text{m s}^{-1}$ forwards.
Second collision: $6 \times 6 = 6v_{B}' + 17 \times 3$, so $36 = 6v_{B}' + 51$ and $v_{B}' = -2.5\ \text{m s}^{-1}$: $B$ rebounds at $2.5\ \text{m s}^{-1}$.
Now $A$ travels backwards at $2\ \text{m s}^{-1}$ and $B$ travels the same way at $2.5\ \text{m s}^{-1}$. $B$ is behind $A$ but moving faster in the same direction, so $B$ will catch $A$: there will be a third collision. ($C$ moves away at $3\ \text{m s}^{-1}$ and is not involved again.)
Exit Ticket
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