Friction First, Momentum Second
A block of mass $2\ \text{kg}$ is sent sliding at $8\ \text{m s}^{-1}$ across rough ground with coefficient of friction $0.5$. Friction removes kinetic energy metre by metre, so the block slows before anything else can happen. Drag the distance travelled and watch the speed fall — the speed at the moment of an impact is the input every momentum calculation needs.
speed now $v = \sqrt{64 - 10d} =$ 6.6 m s⁻¹, friction force $\mu mg = 10\ \text{N}$
The work done against friction, $10d$ joules, comes straight out of the kinetic energy $\dfrac{1}{2}(2)(8^{2}) = 64\ \text{J}$ — the block stops after $6.4\ \text{m}$.
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 body of mass $3\ \text{kg}$ slides $5\ \text{m}$ on rough ground with coefficient of friction $0.4$. The work done against friction is (take $g = 10\ \text{m s}^{-2}$):
R2. A body of mass $4\ \text{kg}$ moves at $5\ \text{m s}^{-1}$. Its kinetic energy is:
R3. A body decelerates steadily from $5\ \text{m s}^{-1}$ to rest over $2\ \text{m}$. The magnitude of its deceleration is:
A full momentum question is usually three stages joined together, and each stage has its own tool:
| Stage | Tool |
|---|---|
| Speed just before an impact on rough ground | work–energy: kinetic energy lost $= \mu mg \times d$ |
| Speeds just after the impact | conservation of momentum along the line |
| Distance each body then slides | work–energy again: $\dfrac{1}{2}mv^{2} = \mu mg \times s$ |
Notice that $m$ cancels in the sliding stage, so a body leaving an impact at speed $v$ slides $s = \dfrac{v^{2}}{2\mu g}$ whatever its mass.
Energy removed over a distance $d$:
| Form | Equation |
|---|---|
| Work–energy | $\dfrac{1}{2}mu^{2} - \dfrac{1}{2}mv^{2} = \mu mg\,d$ |
| Simplified | $v^{2} = u^{2} - 2\mu g\,d$ |
Mini-example: a block sent off at $8\ \text{m s}^{-1}$ on ground with $\mu = 0.5$ is travelling at $v = \sqrt{64 - 2(5)(3)} = \sqrt{34} \approx 5.8\ \text{m s}^{-1}$ after $3\ \text{m}$.
Apply conservation across the impact to get each body's speed, then hand each speed back to the sliding formula to find the distances and any final gap between the bodies.
Two bodies travelling in the same direction collide again only if the one behind is moving faster. Compare the after-speeds before answering, as in the three-ball questions of this week.
A body leaves an impact at $4\ \text{m s}^{-1}$ on ground with $\mu = 0.4$; it slides $s = \dfrac{16}{2 \times 0.4 \times 10} = 2\ \text{m}$ before stopping.
Groups of 4: each card on the left is a quantity a full question might ask for; each card on the right is the method that starts it off. Tap one card on each side to link them, then press Check to score. Agree on the opening line of each method out loud before you link — that is the habit a full question rewards.
Work alone. One slider sets the force and one sets the time it acts; the readout reports the impulse I = Ft. The two tables double and treble each factor in turn, so the same pattern that built the momentum in Lesson 1 now appears as the momentum change a force delivers. Three graded challenges follow.
A particle $P$ of mass $4\ \text{kg}$ is projected with speed $9\ \text{m s}^{-1}$ along rough horizontal ground. The coefficient of friction between $P$ and the ground is $0.4$. After $7\ \text{m}$ it strikes a particle $Q$ of mass $3\ \text{kg}$. The coefficient of friction between $Q$ and the ground is also $0.4$. $Q$ comes to rest after $2\ \text{m}$. Show that the speed of $P$ immediately before the collision is $5\ \text{m s}^{-1}$ and find the speed of $Q$ immediately after the collision. Take $g = 10\ \text{m s}^{-2}$.
Two separate energy audits. Before the impact, the kinetic energy $P$ loses equals the friction work $\mu mg \times 7$. After the impact, $Q$'s kinetic energy at the moment of the collision is exactly the friction work it takes to slide $2\ \text{m}$.
$F_{P} = \mu m_{P}g = 0.4 \times 4 \times 10 = 16\ \text{N}$, so over $7\ \text{m}$ friction removes $16 \times 7 = 112\ \text{J}$.
$\dfrac{1}{2}(4)(9^{2}) - \dfrac{1}{2}(4)v^{2} = 112$, so $162 - 2v^{2} = 112$, giving $v^{2} = 25$ and $v = 5\ \text{m s}^{-1}$, as required.
$Q$ slides $2\ \text{m}$ with friction $0.4 \times 3 \times 10 = 12\ \text{N}$, so $\dfrac{1}{2}(3)v_{Q}^{2} = 12 \times 2 = 24$ and $v_{Q}^{2} = 16$: $Q$ leaves the collision at $4\ \text{m s}^{-1}$.
Block $P$ of mass $2\ \text{kg}$ is projected at $10\ \text{m s}^{-1}$ along rough horizontal ground with coefficient of friction $0.4$. After travelling $8\ \text{m}$ it strikes block $Q$ of mass $3\ \text{kg}$, which is at rest on the same ground. The blocks coalesce. Find (i) the speed of $P$ immediately before the impact (ii) their common speed immediately after it (iii) the distance the pair then slides before coming to rest. Take $g = 10\ \text{m s}^{-2}$.
Three stages, three tools. Before the impact, friction removes $\mu mg \times 8$ joules from $P$'s kinetic energy. Across the impact, momentum is conserved in the coalescence. Afterwards, the pair's kinetic energy is spent again against friction — and note that the mass cancels in that last stage.
Friction is $\mu mg = 0.4 \times 2 \times 10 = 8\ \text{N}$, so over $8\ \text{m}$ it removes $8 \times 8 = 64\ \text{J}$. Starting energy $\dfrac{1}{2}(2)(10^{2}) = 100\ \text{J}$, so at the impact $\dfrac{1}{2}(2)v^{2} = 36$ and $v = 6\ \text{m s}^{-1}$.
$2 \times 6 = (2 + 3)v$, so $5v = 12$ and $v = 2.4\ \text{m s}^{-1}$.
The friction on the pair is $\mu Mg = 0.4 \times 5 \times 10 = 20\ \text{N}$. Equating the kinetic energy to the work done against friction, $\dfrac{1}{2}(5)(2.4^{2}) = 20s$, so $14.4 = 20s$ and $s = 0.72\ \text{m}$.
Answer, then check. Options are shuffled every load. The clock starts on your first answer.
Q1. A body of mass $2\ \text{kg}$ is sent sliding at $8\ \text{m s}^{-1}$ across ground with coefficient of friction $0.5$. After $3\ \text{m}$ its speed is (take $g = 10\ \text{m s}^{-2}$):
Q2. A body of mass $4\ \text{kg}$ moving at $5\ \text{m s}^{-1}$ collides with, and coalesces with, a body of mass $1\ \text{kg}$ at rest. Their common speed is:
Q3. A body leaves an impact at $4\ \text{m s}^{-1}$ on ground with coefficient of friction $0.4$. Before stopping it slides:
Q4. After an impact, object $A$ travels at $2\ \text{m s}^{-1}$ and object $B$, which is behind $A$, travels at $2.5\ \text{m s}^{-1}$ in the same direction. Then:
Work each on paper, then reveal the solution to check your working.
A particle of mass $3\ \text{kg}$ is projected with speed $4\ \text{m s}^{-1}$ towards a stationary particle of mass $5\ \text{kg}$. The particles coalesce on impact. Find the speed at which the particles move after the collision.
Reveal Solution
$3 \times 4 = 8v$, so $v = 1.5\ \text{m s}^{-1}$ in the original direction of motion.
A particle of mass $3\ \text{kg}$ is projected with speed $4\ \text{m s}^{-1}$ towards a stationary particle of mass $5\ \text{kg}$, and the two coalesce. The coalesced particles then move towards a particle of mass $2\ \text{kg}$. After this second collision the coalesced particles remain stationary and the $2\ \text{kg}$ particle moves with speed $3\ \text{m s}^{-1}$. Find the speed and direction of motion of the $2\ \text{kg}$ particle before the collision.
Reveal Solution
First collision: $3 \times 4 = 8v$, so the coalesced mass moves at $1.5\ \text{m s}^{-1}$. For the second collision, take its direction as positive: $8(1.5) + 2u = 0 + 2(3)$, so $12 + 2u = 6$ and $u = -3\ \text{m s}^{-1}$. The $2\ \text{kg}$ particle was moving at $3\ \text{m s}^{-1}$ towards the coalesced particles.
A particle $P$ of mass $4\ \text{kg}$ is projected with speed $9\ \text{m s}^{-1}$ along rough horizontal ground with coefficient of friction $0.4$. After $7\ \text{m}$ it strikes a particle $Q$ of mass $3\ \text{kg}$, also on ground with coefficient of friction $0.4$. The speed of $P$ immediately before the collision is $5\ \text{m s}^{-1}$ and the speed of $Q$ immediately after it is $4\ \text{m s}^{-1}$. Find the distance between $Q$ and $P$ when both have come to rest. Take $g = 10\ \text{m s}^{-2}$.
Reveal Solution
Momentum across the impact: $4 \times 5 = 4v_{P} + 3 \times 4$, so $4v_{P} = 8$ and $P$ moves on at $2\ \text{m s}^{-1}$.
Sliding distances after the impact: $P$ slides $\dfrac{2^{2}}{2 \times 0.4 \times 10} = 0.5\ \text{m}$, while $Q$ slides the stated $2\ \text{m}$.
Measuring from the point where $P$ started: $P$ finishes at $7 + 0.5 = 7.5\ \text{m}$ and $Q$ at $7 + 2 = 9\ \text{m}$, so the distance between them is $1.5\ \text{m}$.
Exit Ticket
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