Conservation of Linear Momentum to solve problems

The key idea. During a collision the two spheres push on each other with equal and opposite forces (Newton's third law). These internal forces cancel, so the total linear momentum of the system is unchanged:

$m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$

where $u$ are velocities before and $v$ are velocities after, and momentum is mass × velocity.

Momentum is a vector — signs matter. Choose one positive direction for the whole problem and keep it. A sphere moving the other way has a negative velocity. Getting the signs consistent is the entire skill.

One equation, two unknowns. A direct impact leaves you with two unknown final velocities $v_1, v_2$ but CLM gives only one equation. You need a second relation — Newton's experimental law — to pin both down.

Cambridge tip. Always draw a "before / after" diagram with arrows and write your chosen positive direction next to it. Then read every velocity off the diagram with its correct sign.

In a direct (head-on) impact the two spheres move along the same straight line. You have two unknowns after impact, $v_1$ and $v_2$, and two equations:

Equation 1 — conservation of momentum: $m_1u_1 + m_2u_2 = m_1v_1 + m_2v_2$

Equation 2 — Newton's experimental law. The speed of separation is $e$ times the speed of approach. With a consistent positive direction: $v_2 - v_1 = e\,(u_1 - u_2)$

(Here the spheres approach with sphere 1 catching sphere 2, so $u_1 > u_2$ and they separate with $v_2 > v_1$.)

The method:

1. Draw before/after, fix the positive direction, assign signs.
2. Write CLM (Equation 1).
3. Write NEL (Equation 2) — separation over approach.
4. Solve the two linear equations simultaneously for $v_1$ and $v_2$.
5. Interpret signs: a negative $v$ means that sphere moves in the negative direction.

A reliable check. Substitute your $v_1, v_2$ back into CLM and confirm the momentum balances numerically — this catches most arithmetic slips.

Cambridge tip. Keep NEL as separation = e × approach with both differences taken in the same order (e.g. always "sphere 1 minus sphere 2" before, "sphere 2 minus sphere 1" after). A sign slip in NEL is the most common way to lose all the marks in this question.

When two smooth spheres collide obliquely, the collision force acts only along the line of centres (the line joining the two centres at the instant of contact). So resolve every velocity into:

• the component along the line of centres, and
• the component perpendicular to it.

Then apply two principles:

• Perpendicular to the line of centres: each sphere's component is UNCHANGED (the spheres are smooth — no force acts in this direction).
• Along the line of centres: the components behave exactly like a direct impact — they obey CLM and NEL together.

The method:

1. Identify the line of centres; resolve each velocity into "along" and "perpendicular" parts.
2. Perpendicular components: write them down unchanged for each sphere.
3. Along components: treat as a direct impact — write CLM and NEL, solve simultaneously.
4. Recombine each sphere's two final components (Pythagoras + $\tan$) for the final speed and direction.

Cambridge tip. A frequent error is applying NEL or CLM to the perpendicular direction. CLM and NEL operate only along the line of centres; the perpendicular components simply pass through unchanged.

Kinetic energy is generally NOT conserved in a collision. Unless the impact is perfectly elastic ($e = 1$), some KE is converted into heat, sound and deformation:

$\text{KE loss} = \left(\tfrac12 m_1u_1^2 + \tfrac12 m_2u_2^2\right) - \left(\tfrac12 m_1v_1^2 + \tfrac12 m_2v_2^2\right)$

Compute the four velocities first (from CLM + NEL), then plug into this. The result must be $\geq 0$ — a negative "loss" signals an arithmetic error.

Impulse = change in momentum. The impulse $J$ delivered to a sphere during the impact equals its change in momentum:

$J = m v - m u$

The two spheres receive equal and opposite impulses (Newton's third law), which is exactly why total momentum is conserved. The magnitude of the impulse on each sphere is the same; their signs are opposite.

A worked sketch of the chain:

• Find $v_1, v_2$ from CLM + NEL.
• Impulse on sphere 1: $J_1 = m_1 v_1 - m_1 u_1$.
• Impulse on sphere 2: $J_2 = m_2 v_2 - m_2 u_2 = -J_1$.
• KE loss: subtract total final KE from total initial KE.

Perfectly inelastic special case ($e = 0$). The spheres coalesce, so $v_1 = v_2 = v$. CLM alone gives $v = \frac{m_1u_1 + m_2u_2}{m_1 + m_2}$, and the KE loss is then maximal for the given masses and speeds.

Cambridge tip. When asked for the impulse "exerted on" a particular sphere, use that sphere's mass and velocity change, with a clear sign convention. Quoting the impulse on the wrong sphere (or with the wrong sign) is a routine lost mark.