Welcome to Impulse and Momentum!
Ever wondered why airbags save lives in car crashes, or why a tennis racket hitting a ball feels so explosive? The answer lies in two fundamental concepts in mechanics: momentum and impulse.
Don't worry if mechanics has felt tricky in the past. We will break everything down into bite-sized steps with clear diagrams in words, easy-to-follow rules, and foolproof methods for tackling exam problems.
1. Linear Momentum
What is Momentum?
Momentum is simply a measure of how difficult it is to stop a moving object. A heavy lorry moving slowly has a lot of momentum, but so does a tiny bullet fired at high speed!
For an object of mass \(m\) moving with velocity \(v\), its linear momentum \(p\) is defined as:
\(p = mv\)
Units: Mass is in kilograms (\(\text{kg}\)) and velocity is in metres per second (\(\text{m s}^{-1}\)), so momentum is measured in \(\text{kg m s}^{-1}\) (or equivalently, Newton-seconds, \(\text{N s}\)).
Crucial Point: Direction Matters (Vectors!)
Momentum is a vector quantity. This means it has both magnitude and direction.
Golden Rule: Always pick a positive direction at the very start of a question (usually to the right).
- If an object travels to the right: \(v = +5\text{ m s}^{-1}\)
- If an object travels to the left: \(v = -5\text{ m s}^{-1}\)
Common Mistake to Avoid: Forgetting to put a minus sign on velocities moving in the opposite direction is the single most common mistake in this topic!
Key Takeaway: Section 1
Momentum is \(\text{mass} \times \text{velocity}\). Always assign a positive direction to keep your signs consistent.
2. Impulse and the Impulse-Momentum Principle
What is Impulse?
When a force acts on an object over a period of time, it changes the object's momentum. This effect is called impulse.
For a constant force \(F\) acting for a time \(t\):
\(\text{Impulse} = Ft\)
Connecting Force and Momentum
Recall Newton's Second Law: \(F = ma\). Since acceleration is \(a = \frac{v - u}{t}\), substituting gives:
\(F = m\left(\frac{v - u}{t}\right)\)
Multiplying both sides by \(t\):
\(Ft = mv - mu\)
This gives us the Impulse-Momentum Principle:
\(\text{Impulse} = \text{Final Momentum} - \text{Initial Momentum} = \Delta p\)
\(\text{Impulse} = mv - mu\)
Did you know? Airbags and crumple zones in cars increase the collision time \(t\). Since the required change in momentum \(\Delta p\) is fixed, increasing \(t\) dramatically reduces the impact force \(F\) on the passengers (\(F = \frac{\Delta p}{t}\)).
Step-by-Step Example: Impulse on a Rebounding Ball
A ball of mass \(0.2\text{ kg}\) hits a vertical wall horizontally at \(14\text{ m s}^{-1}\) and rebounds at \(10\text{ m s}^{-1}\). Find the impulse exerted by the wall on the ball.
Step 1: Choose the positive direction. Let the direction away from the wall (rebound direction) be positive.
Step 2: Identify the given values with correct signs:
Initial velocity, \(u = -14\text{ m s}^{-1}\) (moving towards wall)
Final velocity, \(v = +10\text{ m s}^{-1}\) (rebounding away from wall)
Mass, \(m = 0.2\text{ kg}\)
Step 3: Apply the formula \(\text{Impulse} = mv - mu\):
\(\text{Impulse} = 0.2(10) - 0.2(-14)\)
\(\text{Impulse} = 2 - (-2.8) = 2 + 2.8 = 4.8\text{ N s}\)
Answer: The impulse exerted by the wall is \(4.8\text{ N s}\) away from the wall.
Key Takeaway: Section 2
\(\text{Impulse} = Ft = mv - mu\). When an object reverses direction, the change in velocity involves subtracting a negative number, which means the speeds add together!
3. Principle of Conservation of Linear Momentum (PCLM)
The Principle
When two particles collide or interact, they exert equal and opposite forces on each other (Newton's Third Law). Therefore, if no external forces act on the system:
\(\text{Total momentum before collision} = \text{Total momentum after collision}\)
For two particles \(A\) (mass \(m_1\)) and \(B\) (mass \(m_2\)):
\(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\)
where \(u_1, u_2\) are initial velocities and \(v_1, v_2\) are final velocities.
Special Case: Coalescing Particles
When two objects collide and stick together (coalesce), they move with a single common final velocity \(v\):
\(m_1 u_1 + m_2 u_2 = (m_1 + m_2)v\)
Step-by-Step Example: Coalescing Trolleys
Trolley \(A\) of mass \(3\text{ kg}\) travels right at \(4\text{ m s}^{-1}\) and collides with trolley \(B\) of mass \(2\text{ kg}\) travelling left at \(1\text{ m s}^{-1}\). After the collision, they stick together. Find their common velocity.
Step 1: Take right as positive.
\(m_1 = 3\text{ kg}\), \(u_1 = +4\text{ m s}^{-1}\)
\(m_2 = 2\text{ kg}\), \(u_2 = -1\text{ m s}^{-1}\)
Step 2: Set up PCLM:
\((3)(4) + (2)(-1) = (3 + 2)v\)
\(12 - 2 = 5v\)
\(10 = 5v \implies v = 2\text{ m s}^{-1}\)
Answer: The combined trolleys move to the right at \(2\text{ m s}^{-1}\).
Key Takeaway: Section 3
In any closed collision, momentum is conserved: \(\sum m u = \sum m v\). Always check your positive and negative direction signs before calculating.
4. Newton's Experimental Law of Restitution
What is Restitution?
When two objects collide and separate, how "bouncy" the collision is depends on the materials. Isaac Newton discovered an experimental rule connecting the speeds before and after impact.
Newton's Law of Restitution:
\(\text{Speed of separation} = e \times \text{Speed of approach}\)
Here, \(e\) is the coefficient of restitution, where \(0 \le e \le 1\).
Understanding the Values of \(e\)
- \(e = 1\) (Perfectly Elastic): No kinetic energy is lost; objects bounce off each other with full elasticity (e.g., idealised snooker balls).
- \(0 < e < 1\) (Inelastic): Most real collisions; some energy is lost as heat and sound.
- \(e = 0\) (Completely Inelastic / Coalescing): Objects do not separate after impact; they stick together (\(\text{speed of separation} = 0\)).
Direct Impact between Two Particles
Let particle \(A\) move with initial velocity \(u_1\) and particle \(B\) with \(u_2\). After collision, their velocities are \(v_1\) and \(v_2\) respectively (all measured in the same positive direction):
\(v_2 - v_1 = -e(u_2 - u_1)\)
or written in terms of approach and separation speeds:
\(\frac{v_2 - v_1}{u_1 - u_2} = e\) (where \(u_1 > u_2\))
Impact with a Fixed Smooth Surface (e.g., Wall or Floor)
When a particle hits a fixed plane normally (at \(90^\circ\)) with speed \(u\) and rebounds with speed \(v\):
\(v = eu\)
The direction of motion reverses, so velocity after impact is \(-eu\).
Step-by-Step Example: Standard Direct Collision Problem
Sphere \(A\) of mass \(0.5\text{ kg}\) moves at \(6\text{ m s}^{-1}\) towards sphere \(B\) of mass \(1\text{ kg}\) which is at rest on a smooth horizontal table. The coefficient of restitution between the spheres is \(e = 0.5\). Find the velocity of each sphere after the collision.
Step 1: Set up variables (taking direction of \(A\)'s motion as positive)
\(m_A = 0.5\text{ kg}\), \(u_A = 6\text{ m s}^{-1}\)
\(m_B = 1.0\text{ kg}\), \(u_B = 0\text{ m s}^{-1}\)
Let \(v_A\) and \(v_B\) be the velocities after impact.
Step 2: Apply Principle of Conservation of Linear Momentum (PCLM)
\(m_A u_A + m_B u_B = m_A v_A + m_B v_B\)
\(0.5(6) + 1.0(0) = 0.5 v_A + 1.0 v_B\)
\(3 = 0.5 v_A + v_B\)
Multiply by 2 for simpler numbers:
\(v_A + 2v_B = 6\) --- (Equation 1)
Step 3: Apply Newton's Law of Restitution (NLR)
\(\text{Speed of separation} = e \times \text{Speed of approach}\)
\(v_B - v_A = e(u_A - u_B)\)
\(v_B - v_A = 0.5(6 - 0)\)
\(v_B - v_A = 3\) --- (Equation 2)
Step 4: Solve the simultaneous equations
From Equation 2: \(v_B = v_A + 3\)
Substitute into Equation 1:
\(v_A + 2(v_A + 3) = 6\)
\(v_A + 2v_A + 6 = 6\)
\(3v_A = 0 \implies v_A = 0\text{ m s}^{-1}\)
Now find \(v_B\):
\(v_B = 0 + 3 = 3\text{ m s}^{-1}\)
Answer: Sphere \(A\) comes to rest (\(v_A = 0\text{ m s}^{-1}\)) and Sphere \(B\) moves forward at \(3\text{ m s}^{-1}\).
Key Takeaway: Section 4
Most two-body collision problems are solved using a standard 2-step recipe: write the PCLM equation, write the Restitution equation, and solve them simultaneously!
5. Exam Strategy & Common Traps Summary
Quick Review Checklist
- Momentum: \(p = mv\) (Vector: always define a positive direction!).
- Impulse: \(I = Ft = mv - mu = \Delta p\).
- Conservation of Momentum: \(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\).
- Restitution: \(\text{separation speed} = e \times \text{approach speed}\).
- Wall Impact: Rebound speed \(v = eu\).
Top 3 Exam Tips
1. Draw a clear "Before" and "After" diagram: Draw two small sketches showing masses and arrowed velocities for every collision question.
2. Watch your signs: If a particle bounces backwards, its velocity changes sign. If \(u = +8\) and it rebounds at \(4\), then \(v = -4\).
3. Loss of Kinetic Energy: If an exam asks for "loss of kinetic energy", calculate \(\text{Initial } KE - \text{Final } KE\), where \(KE = \frac{1}{2}mv^2\). The result should always be positive (or zero for \(e=1\)).