recall that, for an elastic collision, total kinetic energy is conserved and the relative speed of approach is equal to the relative speed of separation

Linear Momentum and Its Conservation

1. Momentum – definition and units

• Momentum (vector) \$\mathbf p\$: \$\displaystyle \mathbf p = m\mathbf v\$

  • \$m\$ – mass (kg)
  • \$\mathbf v\$ – velocity vector (m s⁻¹)
  • Units: kg·m s⁻¹

• Weight as a special case of force: \$\displaystyle \mathbf W = m\mathbf g\$, with \$\mathbf g\approx9.81\;\text{m s}^{-2}\$ downwards.

2. Newton’s II law in momentum form

Force is the time‑rate of change of momentum

\[

\mathbf F = \frac{d\mathbf p}{dt}.

\]

For a particle of constant mass this reduces to the familiar \$\mathbf F = m\mathbf a\$.

3. Impulse–Momentum Theorem

The impulse \$\mathbf J\$ exerted on a body during the interval \$t1\to t2\$ is

\[

\mathbf J = \int{t1}^{t_2}\mathbf F\,dt = \Delta\mathbf p .

\]

Thus a non‑zero net impulse produces a change in momentum.

4. Conservation of Linear Momentum

• For a closed (isolated) system – i.e. a set of particles on which the resultant external force is zero – the total momentum is constant:

  \[

  \sum{i=1}^{n}\mathbf p{i,\text{initial}}=

  \sum{i=1}^{n}\mathbf p{i,\text{final}} .

  \]

• Choosing the system: Include all bodies that interact during the collision and exclude external agents (e.g. the table on which the balls roll) so that the net external force is negligible.
• In one dimension the vector equation reduces to a scalar form

  \$\displaystyle \sum mi v{i,\text{i}} = \sum mi v{i,\text{f}}\$.

• In two or three dimensions each Cartesian component is conserved separately:

  \[

  \sum mi v{i,x}^{\text{i}} = \sum mi v{i,x}^{\text{f}},\qquad

  \sum mi v{i,y}^{\text{i}} = \sum mi v{i,y}^{\text{f}},\;\text{etc.}

  \]

5. Elastic vs. Inelastic Collisions

6. Derivation – Relative speed of approach = relative speed of separation

Two particles \$A\$ and \$B\$ move along a straight line. Let \$uA,uB\$ be their speeds before the impact and \$vA,vB\$ after the impact (positive direction chosen arbitrarily).

1. Momentum conservation (scalar)

   \[

   mA uA + mB uB = mA vA + mB vB .

   \]

2. Kinetic‑energy conservation

   \[

   \tfrac12 mA uA^{2} + \tfrac12 mB uB^{2}

   = \tfrac12 mA vA^{2} + \tfrac12 mB vB^{2}.

   \]

3. Eliminate the masses by subtracting the momentum equation multiplied by \$(uA+uB)\$ from the energy equation multiplied by \$2\$:

   \[

   mA(uA^{2}-vA^{2}) + mB(uB^{2}-vB^{2})

   =(uA+uB)\bigl[mA(uA-vA)+mB(uB-vB)\bigr].

   \]

4. Factor each difference of squares and cancel the common terms:

   \[

   (uA-uB)^{2} = (vA-vB)^{2}.

   \]

5. Take the positive square‑root (speeds are non‑negative):

   \[

   |uA-uB| = |vA-vB|,

   \]

   i.e. the relative speed of approach equals the relative speed of separation.

7. One‑Dimensional Elastic‑Collision Formulae

For two objects \$1\$ and \$2\$ (masses \$m1,m2\$) with initial speeds \$u1,u2\$:

\[

\boxed{\,v1 = \frac{(m1-m2)u1 + 2m2u2}{m1+m2}\,},\qquad

\boxed{\,v2 = \frac{(m2-m1)u2 + 2m1u1}{m1+m2}\,}.

\]

Worked Example

• \$m1 = 0.5\;\text{kg}\$, \$m2 = 1.0\;\text{kg}\$
• Before collision: \$u1 = +4\;\text{m s}^{-1}\$ (right), \$u2 = -2\;\text{m s}^{-1}\$ (left)

Applying the formulae:

\[

v_1 = \frac{(0.5-1)4 + 2(1)(-2)}{0.5+1}

= -4\;\text{m s}^{-1},

\qquad

v_2 = \frac{(1-0.5)(-2) + 2(0.5)(4)}{1.5}

= +2\;\text{m s}^{-1}.

\]

Check the relative‑speed condition:

\[

|u{\text{rel}}| = |u1-u_2| = |4-(-2)| = 6\;\text{m s}^{-1},

\qquad

|v{\text{rel}}| = |v1-v_2| = |-4-2| = 6\;\text{m s}^{-1}.

\]

Equality confirms an elastic collision.

8. Two‑Dimensional Elastic Collisions (identical spheres)

1. Resolve all velocities into components parallel (\$\parallel\$) and perpendicular (\$\perp\$) to the line of centres at the instant of impact.
2. The relative velocity component parallel to the line of centres reverses sign:

   \[

   (\mathbf v{1}-\mathbf v{2}){\parallel}^{\text{f}} = -(\mathbf v{1}-\mathbf v{2}){\parallel}^{\text{i}} .

   \]

3. The perpendicular component is unchanged:

   \[

   (\mathbf v{1}-\mathbf v{2}){\perp}^{\text{f}} = (\mathbf v{1}-\mathbf v{2}){\perp}^{\text{i}} .

   \]

4. Apply the component‑wise momentum conservation to obtain the final velocities of each sphere.

Example – Billiard balls

A cue ball (mass \$m\$) strikes an identical stationary target ball. After impact the cue ball moves off at \$30^{\circ}\$ to the original line of motion and the target ball at \$60^{\circ}\$. Using the component method (parallel/perpendicular to the line of centres) one finds both momentum components and the total kinetic energy are conserved, confirming the collision is perfectly elastic.

9. Common Pitfalls & Quick‑Check Checklist

• System selection: always include *all* bodies that interact during the collision; otherwise external forces will appear to violate momentum conservation.
• Sign convention: be consistent when assigning positive directions; remember that a negative speed simply means motion opposite to the chosen positive axis.
• Elastic‑collision condition: both momentum *and* kinetic energy must be conserved. Verifying only one is insufficient.
• Relative‑speed rule applies only in one dimension. In two dimensions use the parallel/perpendicular decomposition.
• When masses are equal, the formulas reduce to a simple exchange of velocities along the line of centres.

10. Summary of Key Points

• Momentum \$\mathbf p = m\mathbf v\$ is a vector; \$\mathbf F = d\mathbf p/dt\$ (Newton II).
• Impulse \$\mathbf J = \Delta\mathbf p\$ links forces to changes in momentum.
• For a closed system \$\displaystyle\sum\mathbf p{\text{i}} = \sum\mathbf p{\text{f}}\$.
• Elastic collisions: both momentum and kinetic energy are conserved, leading to \$|u{\text{rel}}| = |v{\text{rel}}|\$ in 1‑D.
• In 2‑D, conserve each component of momentum separately and apply the “parallel component reverses, perpendicular component unchanged” rule for identical spheres.
• Remember to define the system, keep a consistent sign convention, and check both conservation laws.