The Trolley that Splits in Two
A trolley of mass $6\ \text{kg}$ rolls along a smooth track at $2\ \text{m s}^{-1}$, carrying a compressed spring inside. The spring releases and the trolley separates into two $3\ \text{kg}$ halves, so the total momentum must stay at $6 \times 2 = 12\ \text{kg m s}^{-1}$. Drag the speed of the front half: whatever it does, the back half must adjust so that the two momenta still add to 12.
front half $3v_{1}$ = 3.0 kg m s⁻¹, back half $3v_{2}$ = 9.0 kg m s⁻¹ — total 12.0 kg m s⁻¹
The split is internal, so no momentum leaves the pair: solving $3v_{1} + 3v_{2} = 12$ gives the back half's velocity $v_{2} = 4 - v_{1}$.
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. Bodies of mass $3\ \text{kg}$ and $5\ \text{kg}$ move in the same direction at $2\ \text{m s}^{-1}$ and $1\ \text{m s}^{-1}$. Their total momentum is:
R2. A body of mass $4\ \text{kg}$ moves at $3\ \text{m s}^{-1}$ to the right and another of mass $4\ \text{kg}$ moves at $1\ \text{m s}^{-1}$ to the left. Taking right as positive, the total momentum is:
R3. A body slows at a constant $0.5\ \text{m s}^{-2}$ from $3\ \text{m s}^{-1}$ to rest. How long does it take?
For any impact along a line, the total momentum is the same on both sides of the event:
| Impact | Equation |
|---|---|
| Two bodies, separate afterwards | $m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$ |
| Bodies that coalesce | $(m_{1} + m_{2})v$ replaces the right-hand side |
| A body that splits | the same equation read right to left |
Mini-example: $2\ \text{kg}$ at $5\ \text{m s}^{-1}$ strikes $3\ \text{kg}$ at rest and the two bodies move off separately at $1\ \text{m s}^{-1}$ and $v$: $10 = 2 + 3v$, so $v = \dfrac{8}{3} \approx 2.67\ \text{m s}^{-1}$.
Choose the positive direction once, mark it on a sketch and carry the sign into every velocity. The commonest lost mark is writing a rebounding velocity as positive.
| Motion | Sign |
|---|---|
| Continues in the chosen direction | positive |
| Comes back the other way | negative |
Mini-example: $3\ \text{kg}$ moving at $4\ \text{m s}^{-1}$ meets $2\ \text{kg}$ moving at $4\ \text{m s}^{-1}$ the other way. The total before is $12 - 8 = 4$. If the $3\ \text{kg}$ body returns at $1\ \text{m s}^{-1}$, then $-3 + 2v = 4$ and $v = 3.5\ \text{m s}^{-1}$ in the original direction.
No external force acts when a coupling breaks or an explosion separates a body, so the total momentum carries straight through the event; only the distribution between the pieces changes.
A truck of mass $20\,000\ \text{kg}$ rolls at $6\ \text{m s}^{-1}$ and separates into $8000\ \text{kg}$ and $12\,000\ \text{kg}$ sections. The front section leaves at $9\ \text{m s}^{-1}$, so $120\,000 = 72\,000 + 12\,000v$ and the rear section travels at $4\ \text{m s}^{-1}$.
If neither after-velocity is known, conservation alone is not enough; a second statement — a common velocity, a direction or an inequality — must be used as well.
Groups of 4: decide whether each statement about collisions, coalescing and splitting is true or false. Tap a card to pick it up, then tap the zone you want to file it in. Press Check cards to see which are misfiled, and Reset to start again. Watch the wording — "must" and "always" are usually the giveaway.
Work alone: stage 1 finds the total momentum before the impact, stage 2 uses the coalescing condition to get the common speed, and stage 3 works out one velocity after the pair is pushed apart again. A wrong option locks and reveals a hint, and a stage stays locked until the one above it is solved.
A train consists of a locomotive of mass $40\,000\ \text{kg}$ pulling four coaches, each of mass $50\,000\ \text{kg}$. The train travels at $5\ \text{m s}^{-1}$ along a straight horizontal line when the coupling between the second and third coaches breaks. The locomotive and the first two coaches continue at $7\ \text{m s}^{-1}$. The rear two coaches decelerate at a constant rate and come to rest after travelling $100\ \text{m}$. Work out how long it takes from the moment the coupling breaks until the rear two coaches come to rest.
First find the mass of each section and the total momentum before the break. The two sections share that momentum afterwards, so the rear section's speed follows at once. Then the rear section decelerates uniformly over $100\ \text{m}$ — the average speed gives the time.
Whole train: $40\,000 + 4 \times 50\,000 = 240\,000\ \text{kg}$, so $p = 240\,000 \times 5 = 1\,200\,000\ \text{kg m s}^{-1}$. Front section: $40\,000 + 2 \times 50\,000 = 140\,000\ \text{kg}$; rear section: $100\,000\ \text{kg}$.
The front section carries $140\,000 \times 7 = 980\,000\ \text{kg m s}^{-1}$, so the rear section has $1\,200\,000 - 980\,000 = 220\,000\ \text{kg m s}^{-1}$. Its speed is $\dfrac{220\,000}{100\,000} = 2.2\ \text{m s}^{-1}$.
The rear section slows uniformly from $2.2\ \text{m s}^{-1}$ to rest over $100\ \text{m}$, so its average speed is $1.1\ \text{m s}^{-1}$ and $t = \dfrac{100}{1.1} = 90.9\ \text{s}$ (3 s.f.).
Ball $A$ of mass $2\ \text{kg}$ moves along a smooth horizontal surface at $5\ \text{m s}^{-1}$ towards ball $B$ of mass $3\ \text{kg}$, which is moving towards $A$ at $2\ \text{m s}^{-1}$. After the collision $A$ has reversed and moves at $1\ \text{m s}^{-1}$. Find the velocity of $B$ after the collision.
Take $A$'s direction of travel as positive, so $B$'s velocity before the collision is negative. After the collision $A$ has reversed, so its velocity is $-1$: write the total momentum before and after and set them equal.
With $A$'s direction positive, $p_{\text{before}} = 2 \times 5 + 3 \times (-2) = 10 - 6 = 4\ \text{kg m s}^{-1}$.
$A$ rebounds at $1\ \text{m s}^{-1}$, so $p_{\text{after}} = 2 \times (-1) + 3v = -2 + 3v$.
Conservation gives $-2 + 3v = 4$, so $3v = 6$ and $v = 2\ \text{m s}^{-1}$: $B$ travels at $2\ \text{m s}^{-1}$ in the direction in which $A$ was originally moving. Both bodies therefore end up travelling the same way, with $A$ behind $B$.
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. A body of mass $3\ \text{kg}$ moves at $4\ \text{m s}^{-1}$ and a body of mass $2\ \text{kg}$ moves at $5\ \text{m s}^{-1}$ in the opposite direction. Taking the first direction as positive, the total momentum is:
Q2. A ball of mass $0.5\ \text{kg}$ moving at $4\ \text{m s}^{-1}$ hits a stationary ball of mass $0.5\ \text{kg}$ and they coalesce. Their common speed is:
Q3. A body of mass $2\ \text{kg}$ moving at $6\ \text{m s}^{-1}$ collides with a $4\ \text{kg}$ body at rest. Afterwards the $2\ \text{kg}$ body moves at $1\ \text{m s}^{-1}$ in the opposite direction. The speed of the $4\ \text{kg}$ body is:
Q4. A body of mass $5\ \text{kg}$ moving at $3\ \text{m s}^{-1}$ splits into a $3\ \text{kg}$ piece and a $2\ \text{kg}$ piece. The $3\ \text{kg}$ piece continues at $5\ \text{m s}^{-1}$. The velocity of the $2\ \text{kg}$ piece is:
Work each on paper, then reveal the solution to check your working.
Two balls travel towards one another along the $x$-axis. The first ball has mass $2\ \text{kg}$ and travels at $3\ \text{m s}^{-1}$ in the positive $x$-direction; the second ball has mass $5\ \text{kg}$ and travels at $1\ \text{m s}^{-1}$ in the negative $x$-direction. After the collision the balls travel at the same speed but in opposite directions. Work out the speed of the balls after the collision.
Reveal Solution
Total momentum before: $2 \times 3 + 5 \times (-1) = 1\ \text{kg m s}^{-1}$. If the common speed is $s$, the $5\ \text{kg}$ ball continues in the positive direction and the $2\ \text{kg}$ ball moves the other way: $5s - 2s = 1$, so $s = \dfrac{1}{3}\ \text{m s}^{-1}$.
Two particles, $A$ and $B$, of masses $3\ \text{kg}$ and $2\ \text{kg}$, move along a straight horizontal line towards each other, each with speed $4\ \text{m s}^{-1}$. After the collision particle $A$ moves in the same direction as before but with speed $0.4\ \text{m s}^{-1}$. Find the speed of $B$ after the collision.
Reveal Solution
Taking $A$'s original direction as positive, the total momentum before is $3 \times 4 + 2 \times (-4) = 4\ \text{kg m s}^{-1}$. After the collision: $3 \times 0.4 + 2v = 4$, so $2v = 2.8$ and the speed of $B$ is $1.4\ \text{m s}^{-1}$ (it has reversed and travels in $A$'s original direction).
Three balls, $A$, $B$ and $C$, lie in that order in a straight line on a smooth horizontal surface. $A$ has mass $0.4\ \text{kg}$ and moves at $4\ \text{m s}^{-1}$ towards $B$, which has mass $m\ \text{kg}$ and is stationary. $C$ has mass $0.25\ \text{kg}$ and moves at $0.8\ \text{m s}^{-1}$ away from $B$. $A$ hits $B$ and then $B$ hits $C$, and there are no further impacts. $A$ and $C$ now each have speed $1\ \text{m s}^{-1}$ and are both moving in directions away from $B$. Find the range of possible values of $m$.
Reveal Solution
Take the direction from $A$ towards $C$ as positive. After the impacts $A$ moves at $-1$ and $C$ at $+1$. First impact: $0.4 \times 4 = 0.4(-1) + mv_{B}$, so $v_{B} = \dfrac{2}{m}$.
For $B$ to catch $C$: $\dfrac{2}{m} > 0.8$, hence $m < 2.5$.
Second impact: $m \times \dfrac{2}{m} + 0.25 \times 0.8 = mv_{B}' + 0.25 \times 1$, so $2.2 = mv_{B}' + 0.25$ and $v_{B}' = \dfrac{1.95}{m}$. For no further impact, $B$ must not catch $C$ again: $\dfrac{1.95}{m} \le 1$, hence $m \ge 1.95$.
So $1.95 \le m < 2.5$.
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