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M 3

Momentum

Mechanics · Paper 4

🎯What you need to be able to do

  • Calculate the linear momentum of a moving body, with correct units.
  • Apply the principle of conservation of momentum to a collision.
  • Handle bodies that coalesce, and bodies that separate.
  • Use signs consistently to represent direction along a line.
  • Interpret a negative result as motion in the opposite direction.

📚The mathematics

Linear momentum

\[ \text{momentum} = mv \]

Mass in kilograms and velocity in metres per second give momentum in kg m s\(^{-1}\) (equivalently N s). Momentum is a vector: along a straight line it carries a sign, and that sign is the whole content of this topic.

Conservation of momentum

When two bodies collide and no external force acts along the line of motion, the total momentum before equals the total momentum after:

\[ m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2 \]

with \(u\) for velocities before and \(v\) for after. Note that momentum is conserved in every collision, but kinetic energy generally is not — some is lost to sound, heat and deformation. That energy loss is examined in M 5.

Getting the signs right

Choose a positive direction, mark it on your diagram, and give every velocity a sign in that frame. A body moving the other way has a negative velocity, and its momentum is negative. This is the single largest source of error on this topic, because the arithmetic works perfectly well with the wrong signs and produces a confident wrong answer.

Two practical habits:

  • Draw the situation twice — before and after — with an arrow and a value for every velocity. Unknown directions can be drawn either way; the algebra corrects you.
  • If an unknown velocity comes out negative, that is not an error. It means the body actually moves opposite to the direction you assumed. Say so explicitly in the answer: it is usually a mark.

Coalescence and separation

When bodies coalesce — stick together, couple, or one embeds in the other — they move afterwards with a common velocity \(v\), and the right-hand side collapses:

\[ m_1u_1 + m_2u_2 = (m_1 + m_2)v \]

This is the easiest case, because there is only one unknown.

When a body separates into parts — an explosion, a gun recoiling, a person jumping from a stationary trolley — the same principle runs in reverse. If the system starts at rest, the total momentum is zero, so the parts must carry equal and opposite momenta:

\[ 0 = m_1v_1 + m_2v_2 \;\Longrightarrow\; m_1v_1 = -m_2v_2 \]

The lighter part moves faster, which is why a rifle recoils slowly and the bullet leaves quickly.

A sanity check worth running

After a collision on a straight line, the two bodies cannot pass through each other. So if body 1 was behind body 2 and moving faster, then afterwards \( v_1 \le v_2 \). An answer in which the rear body ends up faster than the front one is physically impossible and signals a sign or algebra error.

✏️Worked example

Two particles \(A\) and \(B\), of masses 3 kg and 2 kg, move towards each other along the same straight line. \(A\) has speed 6 m s\(^{-1}\) and \(B\) has speed 4 m s\(^{-1}\). (a) They collide and coalesce. Find the velocity of the combined particle. (b) Instead, suppose they collide and \(A\) is brought to rest. Find the velocity of \(B\) after the collision.

Take the direction of \(A\)’s motion as positive. Then \(A\) has velocity \( +6 \) and \(B\), moving towards \(A\), has velocity \( -4 \).

(a) Total momentum before:

\[ (3)(6) + (2)(-4) = 18 - 8 = 10\ \text{kg m s}^{-1} \]

Coalescing gives a combined mass of 5 kg with common velocity \(v\):

\[ 5v = 10 \;\Longrightarrow\; v = 2\ \text{m s}^{-1} \]

The positive sign means the combined particle moves in \(A\)’s original direction, which makes sense since \(A\) carried the larger momentum.

(b) Momentum is still conserved and still totals \( 10 \). With \(A\) at rest afterwards, let \(B\) have velocity \( v_B \):

\[ 10 = (3)(0) + (2)v_B \;\Longrightarrow\; v_B = 5\ \text{m s}^{-1} \]

The positive sign means \(B\) has reversed and now travels at 5 m s\(^{-1}\) in \(A\)’s original direction — which is the only way the system can keep its forward momentum once \(A\) has stopped.

Check it. In (a), the combined speed of 2 m s\(^{-1}\) lies between \( -4 \) and \( +6 \), as a coalesced velocity always must. In (b), recompute the total momentum afterwards: \( 0 + 2(5) = 10 \) ✓, matching the value before. Recomputing the total after the collision is a ten-second check that catches every sign error in this topic.
“Towards each other” means opposite signs. Using \( +6 \) and \( +4 \) because both speeds are quoted as positive numbers gives a total momentum of 26 and a combined velocity of 5.2 m s\(^{-1}\) — a wholly plausible-looking answer that is simply wrong. Speed is never negative; velocity is. Convert the words into signed velocities before touching the equation.

📝Practise

Work through these, then reveal the answer.

1. A truck of mass 2000 kg moving at 5 m s\(^{-1}\) collides with a stationary truck of mass 3000 kg and they couple together. Find their common speed.
Momentum before: \( 2000(5) + 3000(0) = 10\,000 \) kg m s\(^{-1}\). After, the combined mass is 5000 kg, so \( 5000v = 10\,000 \) and \( v = 2 \) m s\(^{-1}\) in the original direction. The combined body is always slower than the moving one was — a useful check.
2. A ball of mass 0.5 kg moving at 8 m s\(^{-1}\) strikes a wall and rebounds at 6 m s\(^{-1}\). Find the change in momentum.
Taking the initial direction as positive: momentum before \( = 0.5(8) = 4 \); after \( = 0.5(-6) = -3 \). Change \( = -3 - 4 = -7 \) kg m s\(^{-1}\), so the magnitude is 7 kg m s\(^{-1}\), directed away from the wall. Treating the rebound as \( +6 \) gives a change of 1, which badly understates it — the sign is doing the essential work.
3. Particles of mass 4 kg and 1 kg move in the same direction at 3 m s\(^{-1}\) and 1 m s\(^{-1}\) respectively. The 4 kg particle catches up and collides. Afterwards the 1 kg particle moves at 4 m s\(^{-1}\). Find the velocity of the 4 kg particle.
Momentum before: \( 4(3) + 1(1) = 13 \). After: \( 4v + 1(4) = 13 \), so \( 4v = 9 \) and \( v = 2.25 \) m s\(^{-1}\). Check the physical constraint: the rear particle now moves at 2.25 m s\(^{-1}\) and the front one at 4 m s\(^{-1}\), so the rear is slower — they do not pass through each other ✓.
4. A gun of mass 5 kg fires a bullet of mass 0.02 kg at 400 m s\(^{-1}\). Find the recoil speed of the gun.
The system starts at rest, so the total momentum is zero: \( 0 = 0.02(400) + 5v \), giving \( 5v = -8 \) and \( v = -1.6 \) m s\(^{-1}\). The negative sign means the gun moves opposite to the bullet, at a speed of 1.6 m s\(^{-1}\) — stating that interpretation is part of the answer.
5. Two particles of masses 2 kg and 5 kg approach each other at 7 m s\(^{-1}\) and 2 m s\(^{-1}\). After the collision the 2 kg particle moves backwards at 3 m s\(^{-1}\). Find the velocity of the 5 kg particle.
Take the 2 kg particle's initial direction as positive. Before: \( 2(7) + 5(-2) = 14 - 10 = 4 \). After: \( 2(-3) + 5v = 4 \), so \( -6 + 5v = 4 \), giving \( 5v = 10 \) and \( v = 2 \) m s\(^{-1}\). The positive sign means the 5 kg particle has reversed and now moves in the original direction of the 2 kg particle.
6. A person of mass 60 kg steps off a stationary raft of mass 240 kg at 1.5 m s\(^{-1}\). Find the raft's velocity, assuming no resistance from the water.
Total momentum starts at zero: \( 0 = 60(1.5) + 240v \), so \( 240v = -90 \) and \( v = -0.375 \) m s\(^{-1}\). The raft moves at 0.375 m s\(^{-1}\) in the opposite direction to the person. Note the raft, being four times heavier, moves at a quarter of the speed — momentum shares out inversely with mass.

🔗Go deeper — other people’s work

These are external resources, not mine. If one stops working, tell me and everything above it on this page still stands.

  • PhET — Collision Lab, where you can set masses and velocities and watch momentum stay constant
  • The Physics Classroom — momentum and collisions
  • Khan Academy — conservation of momentum