The Momentum Balance
Particle A of mass $3\ \text{kg}$ slides at $4\ \text{m s}^{-1}$ towards particle B of mass $5\ \text{kg}$, which is at rest. The blue bar is the total momentum of the pair before the collision: $3 \times 4 = 12\ \text{kg m s}^{-1}$. If the collision brings A to rest, B must carry all of that momentum away. Drag B's speed until the second bar matches the first — the collision cannot change the total.
momentum after $p_{B} = 5v =$ 10.0 kg m s⁻¹ — still different from the 12.0 before the collision
Particle A is brought to rest, so it carries no momentum away. Every kilogram metre per second that leaves A must arrive at B.
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 released from rest and falls freely. After 2 s its speed is (take $g = 10\ \text{m s}^{-2}$):
R2. A body of mass $5\ \text{kg}$ is acted on by a resultant force of $15\ \text{N}$. Its acceleration is:
R3. A ball travelling at $4\ \text{m s}^{-1}$ rebounds and travels at $4\ \text{m s}^{-1}$ in the opposite direction. The magnitude of its change in velocity is:
Every moving body carries momentum, defined as the product of its mass and its velocity:
| Quantity | Statement |
|---|---|
| Momentum | $p = mv$ |
| Unit | $\text{kg m s}^{-1}$, the same as the newton second $\text{N s}$ |
Momentum is a vector: along a line it is positive in one direction and negative in the other. Mini-example: a trolley of mass $3\ \text{kg}$ rolling at $4\ \text{m s}^{-1}$ has momentum $3 \times 4 = 12\ \text{kg m s}^{-1}$; a $6\ \text{kg}$ trolley at $2\ \text{m s}^{-1}$ has the same momentum, so either can be stopped by the same impulse.
During a collision the two bodies push on each other with equal and opposite forces, so the total momentum of the pair cannot change:
| Situation | Equation |
|---|---|
| General collision along a line | $m_{1}u_{1} + m_{2}u_{2} = m_{1}v_{1} + m_{2}v_{2}$ |
| Bodies that coalesce | $m_{1}u_{1} + m_{2}u_{2} = (m_{1} + m_{2})v$ |
Mini-example: a $2\ \text{kg}$ body moving at $5\ \text{m s}^{-1}$ strikes a $5\ \text{kg}$ body at rest and is brought to rest itself. Then $10 = 5v$, so the second body leaves at $2\ \text{m s}^{-1}$.
An explosion or a split is the same law run backwards: if the total momentum is zero before, the pieces afterwards have equal and opposite momenta.
A stationary shell bursts into a $1\ \text{kg}$ piece and a $3\ \text{kg}$ piece. If the lighter piece leaves at $6\ \text{m s}^{-1}$, then $0 = 1 \times 6 - 3v$, so the heavier piece leaves at $2\ \text{m s}^{-1}$ the other way.
Choose one direction as positive before writing anything down, and keep it: a body travelling the other way has negative momentum, whether it is before or after the impact.
Groups of 4: for each card, decide whether the total momentum of the system described is conserved. 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. Two of the six are the traps — an external force acting on the system changes its momentum.
Work alone. One slider sets the mass and one sets the speed; the readout reports the momentum p = mv. Watch the two tables: in the first, the speed is held fixed while the mass is doubled and trebled; in the second, the mass is held fixed while the speed is doubled and trebled. Read the pattern off the numbers before you touch the sliders, then answer the three graded challenges.
Particle $A$ of mass $4\ \text{kg}$ moves in a straight line across a smooth horizontal surface at $3\ \text{m s}^{-1}$. Particle $B$ of mass $6\ \text{kg}$ is at rest on the surface. $A$ collides with $B$. After the collision $A$ is at rest and $B$ moves away from $A$ with speed $u\ \text{m s}^{-1}$. Find the value of $u$.
Take the direction in which $A$ is travelling as positive. The total momentum before the collision is $4 \times 3 + 6 \times 0$, and after the collision only $B$ is moving. Set the two totals equal.
Take the direction of $A$'s motion as positive. Then $p_{\text{before}} = m_{A}u_{A} + m_{B}u_{B} = 4 \times 3 + 6 \times 0 = 12\ \text{kg m s}^{-1}$.
$A$ is at rest and $B$ moves away at $u$, so $p_{\text{after}} = 4 \times 0 + 6 \times u = 6u$.
The surface is smooth, so no external horizontal force acts on the pair and momentum is conserved: $6u = 12$, so $u = 2\ \text{m s}^{-1}$.
A gun of mass $3\ \text{kg}$ rests on a smooth horizontal surface and fires a bullet of mass $20\ \text{g}$ horizontally with speed $300\ \text{m s}^{-1}$. Find the speed of recoil of the gun.
An explosion is conservation of momentum starting from zero: the gun and the bullet are together at rest before the shot, so their momenta afterwards must be equal and opposite. Take the bullet's direction as positive and give the gun's velocity a letter.
Both the gun and the bullet are at rest before the shot, so $p_{\text{before}} = 3 \times 0 + 0.02 \times 0 = 0$.
The bullet's mass in kilograms is $20\ \text{g} = 0.02\ \text{kg}$. Take the bullet's direction as positive. The bullet carries $0.02 \times 300 = 6\ \text{kg m s}^{-1}$ and the gun $3v$, so $p_{\text{after}} = 6 + 3v$.
Conservation gives $6 + 3v = 0$, so $v = -2\ \text{m s}^{-1}$: the gun moves at $2\ \text{m s}^{-1}$ in the direction opposite to the bullet.
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. A body of mass $4\ \text{kg}$ moves at $3\ \text{m s}^{-1}$. Its momentum is:
Q2. Particle A of mass $2\ \text{kg}$ moves at $3\ \text{m s}^{-1}$ towards particle B of mass $4\ \text{kg}$ moving at $1\ \text{m s}^{-1}$ in the opposite direction. Taking A's direction as positive, the total momentum of A and B is:
Q3. Two bodies of equal mass move directly towards each other, each at $3\ \text{m s}^{-1}$, and coalesce. The speed of the combined body is:
Q4. A body of mass $2\ \text{kg}$ moving at $6\ \text{m s}^{-1}$ splits into two $1\ \text{kg}$ pieces. One piece continues in the original direction at $4\ \text{m s}^{-1}$. The velocity of the other piece is:
Work each on paper, then reveal the solution to check your working.
Two particles, $A$ and $B$, have masses $3\ \text{kg}$ and $2\ \text{kg}$. They move along a straight horizontal line towards each other, each with speed $4\ \text{m s}^{-1}$, and collide. The particles coalesce to form a single particle. Find the speed of the combined particle.
Reveal Solution
Take the direction of $A$'s motion as positive. The total momentum before is $3 \times 4 + 2 \times (-4) = 12 - 8 = 4\ \text{kg m s}^{-1}$. After the collision the mass is $5\ \text{kg}$, so $5v = 4$ and $v = 0.8\ \text{m s}^{-1}$, in the direction $A$ was originally moving.
Two particles, $A$ and $B$, move in a straight line on a smooth horizontal surface, both in the same direction. $A$ has mass $m\ \text{kg}$ and moves at $5\ \text{m s}^{-1}$; $B$ has mass $0.2\ \text{kg}$ and moves at $2\ \text{m s}^{-1}$ ahead of $A$. Find, in terms of $m$, an expression for the total momentum of $A$ and $B$.
Reveal Solution
Both travel the same way, so the momenta add: $m \times 5 + 0.2 \times 2 = 5m + 0.4$, in $\text{kg m s}^{-1}$.
Two particles, $A$ and $B$, move in a straight line on a smooth horizontal surface, both in the same direction. $A$ has mass $m\ \text{kg}$ and moves at $5\ \text{m s}^{-1}$; $B$ has mass $0.2\ \text{kg}$ and moves at $2\ \text{m s}^{-1}$ ahead of $A$. $A$ collides with $B$ and they coalesce to form a single particle $C$, which moves at $3\ \text{m s}^{-1}$. Find the value of $m$.
Reveal Solution
Before: $5m + 0.4$. After: the combined mass is $(m + 0.2)\ \text{kg}$ moving at $3\ \text{m s}^{-1}$, so $5m + 0.4 = 3(m + 0.2) = 3m + 0.6$. Hence $2m = 0.2$ and $m = 0.1\ \text{kg}$.
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