Newton's Law of Restitution (Edexcel A-Level Further Mathematics): Revision Notes

Example 1: Finding Velocities After a Collision

Problem

Two spheres, $A$ (2 kg) and $B$ (3 kg), collide directly. Initially:

• $A$ is moving at 5 ms⁻¹
• $B$ is stationary ($u_B = 0$) The coefficient of restitution is $e = 0.8$

Find the velocities of $A$ and $B$ after the collision.

Step 1: Apply Conservation of Momentum

Substitute the values:

Simplify:

Step 2: Apply Newton's Law of Restitution

Substitute $e = 0.8, u_A = 5$, and $u_B = 0$

Simplify:

Step 3: Solve Simultaneous Equations

From equations (1) and (2):

1. $10 = 2v_A + 3v_B$
2. $v_B - v_A = 4 \quad \Rightarrow \quad v_B = v_A + 4$ Substitute $v_B = v_A + 4$ into (1):

Simplify:

Solve for $v_A$:

Substitute $v_A = -0.4$ into $v_B = v_A + 4$:

Final Answer:

• Velocity of $A$: -0.4 ms⁻¹ (moves left),
• Velocity of $B$: 3.6 ms⁻¹

Example 2: Loss of Kinetic Energy

Problem

Using the data from Example 1, calculate the loss of kinetic energy during the collision.

Step 1: Initial Kinetic Energy

Substitute $m_A = 2 \, \text{kg}, u_A = 5 \, \text{ms}^{-1}, m_B = 3 \, \text{kg}, u_B = 0$

Simplify:

Step 2: Final Kinetic Energy

Substitute $v_A = -0.4 \, \text{ms}^{-1}, v_B = 3.6 \, \text{ms}^{-1}$

Simplify:

Step 3: Loss of Kinetic Energy

Substitute:

Final Answer:

The loss of kinetic energy is 5.4 J

Common Mistakes

1. Forgetting to conserve momentum: Always use $m_A u_A + m_B u_B = m_A v_A + m_B v_B$
2. Misinterpreting ee: Ensure $e$ is applied correctly as $\frac{\text{speed of separation}}{\text{speed of approach}}$
3. Mixing up initial and final velocities: Clearly distinguish $u$ (initial) from $v$ (final).
4. Ignoring special cases:

• $e = 1$: Perfectly elastic collision (no kinetic energy lost).
• $e = 0$: Perfectly inelastic collision (maximum energy lost).

Key Formulas

1. Newton's Law of Restitution:

2. Conservation of Momentum:

3. Loss of Kinetic Energy: