apply the principle of conservation of momentum to solve simple problems, including elastic and inelastic interactions between objects in both one and two dimensions (knowledge of the concept of coefficient of restitution is not required)

Linear Momentum and its Conservation

Linear momentum is a vector quantity defined for a particle of mass \$m\$ moving with velocity \$\mathbf{v}\$ as

\$\mathbf{p}=m\mathbf{v}\$

Its direction is the same as the velocity vector. The principle of conservation of linear momentum states that, in the absence of external forces, the total momentum of a closed system remains constant.

Key Concepts

• Momentum (\$\mathbf{p}\$): \$m\mathbf{v}\$, SI unit kg·m·s⁻¹.
• Impulse (\$\mathbf{J}\$): Change in momentum, \$\mathbf{J}=\Delta\mathbf{p}=\int\mathbf{F}\,dt\$.
• Conservation Law: For a system of \$n\$ particles,

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

  provided the net external force is zero.

• Elastic Collision: Both momentum and kinetic energy are conserved.
• Inelastic Collision: Momentum is conserved but kinetic energy is not; in a perfectly inelastic collision the objects stick together after impact.

One‑Dimensional Collisions

Consider two objects, \$A\$ and \$B\$, moving along a straight line. Let \$mA\$, \$uA\$ and \$mB\$, \$uB\$ be their masses and initial velocities, and \$vA\$, \$vB\$ their velocities after the collision.

General Momentum Equation

\$mA uA + mB uB = mA vA + mB vB\$

Elastic Collision

In addition to momentum conservation, kinetic energy is conserved:

\$\$\frac{1}{2}mA uA^{2} + \frac{1}{2}mB uB^{2}

= \frac{1}{2}mA vA^{2} + \frac{1}{2}mB vB^{2}\$\$

Solving the two equations simultaneously gives:

\$vA = \frac{(mA - mB)uA + 2mB uB}{mA + mB}\$

\$vB = \frac{(mB - mA)uB + 2mA uA}{mA + mB}\$

Perfectly Inelastic Collision

Objects stick together after impact, so \$vA = vB = v\$.

\$v = \frac{mA uA + mB uB}{mA + mB}\$

Two‑Dimensional Collisions

Momentum must be conserved in each independent direction (usually \$x\$ and \$y\$). For objects \$A\$ and \$B\$:

\$\$\begin{aligned}

mA \mathbf{u}A + mB \mathbf{u}B &= mA \mathbf{v}A + mB \mathbf{v}B \\

\text{or}\quad

\begin{cases}

mA u{Ax} + mB u{Bx} = mA v{Ax} + mB v{Bx} \\

mA u{Ay} + mB u{By} = mA v{Ay} + mB v{By}

\end{cases}

\end{aligned}\$\$

When the collision is elastic, kinetic energy is also conserved in the same way as for 1‑D collisions.

Problem‑Solving Steps

1. Draw a clear diagram and define a coordinate system.
2. List known quantities (masses, initial speeds, directions).
3. Write the momentum conservation equations for each independent direction.
4. If the collision is elastic, write the kinetic‑energy conservation equation.
5. Solve the simultaneous equations for the unknown final velocities.
6. Check units and physical plausibility (e.g., speeds should be non‑negative).

Worked Example – One‑Dimensional Elastic Collision

Problem: A 0.5 kg ball moving at \$4\ \text{m s}^{-1}\$ collides elastically with a stationary 0.2 kg ball. Find the speeds of both balls after the collision.

Solution:

• Take the positive \$x\$‑direction as the initial direction of the 0.5 kg ball.
• Initial data: \$mA=0.5\ \text{kg},\ uA=4\ \text{m s}^{-1}\$; \$mB=0.2\ \text{kg},\ uB=0\$.
• Apply the elastic‑collision formulas:

  \$v_A = \frac{(0.5-0.2)4 + 2(0.2)(0)}{0.5+0.2}= \frac{0.3\times4}{0.7}= \frac{1.2}{0.7}\approx1.71\ \text{m s}^{-1}\$

  \$v_B = \frac{(0.2-0.5)0 + 2(0.5)(4)}{0.5+0.2}= \frac{4.0}{0.7}\approx5.71\ \text{m s}^{-1}\$

• Result: After the collision the 0.5 kg ball moves at \$\approx1.7\ \text{m s}^{-1}\$, and the 0.2 kg ball moves at \$\approx5.7\ \text{m s}^{-1}\$ in the original direction.

Worked Example – Two‑Dimensional Inelastic Collision

Problem: A 2.0 kg puck moving at \$3\ \text{m s}^{-1}\$ east collides with a 1.0 kg puck moving at \$2\ \text{m s}^{-1}\$ north. The pucks stick together. Find the speed and direction of the combined mass after the collision.

Solution:

• Choose \$x\$‑axis east, \$y\$‑axis north.
• Initial momenta:

  \$p_{x,\text{initial}} = (2.0)(3) + (1.0)(0) = 6\ \text{kg·m s}^{-1}\$

  \$p_{y,\text{initial}} = (2.0)(0) + (1.0)(2) = 2\ \text{kg·m s}^{-1}\$

• Total mass after collision: \$M = 2.0 + 1.0 = 3.0\ \text{kg}\$.
• Final velocity components:

  \$vx = \frac{p{x,\text{initial}}}{M} = \frac{6}{3}=2\ \text{m s}^{-1}\$

  \$vy = \frac{p{y,\text{initial}}}{M} = \frac{2}{3}\approx0.67\ \text{m s}^{-1}\$

• Speed:

  \$v = \sqrt{vx^{2}+vy^{2}} = \sqrt{2^{2}+0.67^{2}}\approx2.11\ \text{m s}^{-1}\$

• Direction (angle north of east):

  \$\theta = \tan^{-1}\!\left(\frac{vy}{vx}\right)=\tan^{-1}\!\left(\frac{0.67}{2}\right)\approx18.4^{\circ}\$

• Result: The combined puck moves at \$2.1\ \text{m s}^{-1}\$ at \$18^{\circ}\$ north of east.

Summary Table