Apply the principle of the conservation of momentum to solve simple problems in one dimension

Momentum – IGCSE Physics 0625 (Topic 1.6)

Objective

Apply the principle of conservation of momentum to solve simple problems in one dimension.

Key Concepts

• Momentum (\$\mathbf{p}\$) is a vector quantity defined as the product of mass and velocity.
• Formula: \$\mathbf{p}=m\mathbf{v}\$ where \$m\$ is the mass (kg) and \$\mathbf{v}\$ is the velocity (m s\(^{-1}\)).
• SI unit of momentum is kilogram‑metre per second (kg m s\(^{-1}\)).
• Momentum is conserved in an isolated system where no external forces act.
• Conservation law (one‑dimensional): \$m1v{1i}+m2v{2i}=m1v{1f}+m2v{2f}\$
• Types of collisions:

  • Elastic – kinetic energy is conserved as well as momentum.
  • Inelastic – kinetic energy is not conserved; objects may stick together.

Conservation of Momentum – Derivation (One Dimension)

Consider two objects of masses \$m1\$ and \$m2\$ moving along a straight line. The net external force on the system is zero, so the total momentum before interaction equals the total momentum after interaction:

\$m1v{1i}+m2v{2i}=m1v{1f}+m2v{2f}\$

Where the subscript \$i\$ denotes initial (before) and \$f\$ denotes final (after) velocities.

Solving Momentum Problems – Step‑by‑Step Method

1. Read the question carefully and draw a simple diagram of the situation.
2. Identify all objects involved and note their masses and velocities (including direction).
3. Choose a sign convention (e.g., rightward = positive, leftward = negative).
4. Write the conservation equation: \$\sum m v{\text{initial}} = \sum m v{\text{final}}\$.
5. If the problem involves an elastic collision, also write the kinetic‑energy equation: \$\frac12 m1v{1i}^2+\frac12 m2v{2i}^2=\frac12 m1v{1f}^2+\frac12 m2v{2f}^2\$.
6. Solve the simultaneous equations for the unknown velocity (or velocities).
7. Check the sign and magnitude of your answer; ensure it is physically reasonable.

Common Pitfalls

• Forgetting that momentum is a vector – directions must be treated with signs.
• Mixing up initial and final velocities in the conservation equation.
• Assuming kinetic energy is conserved in an inelastic collision.
• Neglecting to include all objects that form the isolated system.

Summary Table – Collision Types

Worked Example

Problem: A 0.5 kg cart moving to the right at 4 m s\(^{-1}\) collides head‑on with a 0.8 kg cart moving to the left at 2 m s\(^{-1}\). After the collision the carts stick together. Find their common velocity after the collision.

1. Choose rightward as positive. Initial velocities: \$v{1i}=+4\$ m s\(^{-1}\), \$v{2i}=-2\$ m s\(^{-1}\).
2. Since the carts stick together, treat them as a single mass after the collision: \$m_{\text{total}}=0.5+0.8=1.3\$ kg.
3. Apply conservation of momentum:

   \$0.5(+4)+0.8(-2)=1.3\,v_f\$

   \$2.0-1.6=1.3\,v_f\$

   \$0.4=1.3\,v_f\$

   \$v_f=\frac{0.4}{1.3}\approx0.31\ \text{m s}^{-1}\$

4. Result: The combined carts move to the right at approximately \$0.31\ \text{m s}^{-1}\$.

Suggested Diagram

Practice Questions

1. A 1.2 kg ball moving at 3 m s\(^{-1}\) collides elastically with a stationary 0.8 kg ball. Find the velocities of both balls after the collision.
2. A 2.0 kg sled moving at 5 m s\(^{-1}\) picks up a 0.5 kg snowball that was initially at rest. What is the speed of the sled‑snowball system after the collision?
3. Two ice skaters, A (mass 50 kg) and B (mass 70 kg), push off each other. If A moves away at 2 m s\(^{-1}\), what is B’s speed?