M1.4: Momentum and Impulse (Motion in a Straight Line)
Welcome to one of the most practical and exciting chapters in Mechanics! If you’ve ever wondered what truly happens when two objects collide, or why padding is used in sports equipment, this chapter holds the mathematical answers.
In this section, we will explore the fundamental concepts of momentum and impulse, which govern all interactions between moving bodies. Remember, we are restricting our study to motion that happens only along a single straight line, so direction will be handled using positive and negative signs.
Don't worry if this seems tricky at first. Momentum is just a fancy way of quantifying "how much motion" an object has!
1. The Concept of Momentum
Momentum is a measure of the quantity of motion of a body. It depends on both the object's mass and its velocity.
Momentum is a vector quantity, meaning its direction is crucial. Since we are restricted to a straight line, we use positive values for motion in one direction and negative values for motion in the opposite direction.
Definition and Formula
Momentum (\(p\)) is defined as the product of mass (\(m\)) and velocity (\(v\)).
\[p = mv\]
- Mass (\(m\)): Always measured in kilograms (\(\text{kg}\)).
- Velocity (\(v\)): Measured in metres per second (\(\text{m s}^{-1}\)).
- Momentum (\(p\)): Measured in kilogram metres per second (\(\text{kg m s}^{-1}\)).
Analogy: Imagine trying to stop two things: a tennis ball moving at \(5 \text{ m s}^{-1}\) and a bowling ball moving at \(5 \text{ m s}^{-1}\). The bowling ball is harder to stop because it has much greater mass, and therefore much greater momentum.
Momentum is also harder to stop if the velocity is higher, even if the mass is the same (e.g., a fast pitch vs. a gentle lob).
Quick Review: Key Characteristics
- If an object is stationary (\(v=0\)), its momentum is zero.
- The direction of the momentum is always the same as the direction of the velocity.
2. Impulse and the Change in Momentum
A collision or impact involves forces acting over a short time. The measure of the effect of this force is called Impulse.
Definition and Formula
Impulse (\(I\)) is the product of the force (\(F\)) applied and the time interval (\(t\)) over which it acts.
\[I = F t\]
- Force (\(F\)): Measured in Newtons (\(\text{N}\)).
- Time (\(t\)): Measured in seconds (\(\text{s}\)).
- Impulse (\(I\)): Measured in Newton seconds (\(\text{N s}\)).
Did you know? Since \(1 \text{ N} = 1 \text{ kg m s}^{-2}\), the unit for Impulse (\(\text{N s}\)) is equivalent to the unit for momentum (\(\text{kg m s}^{-1}\)). This makes sense because they are linked by the Impulse-Momentum Principle.
The Impulse-Momentum Principle (The Link)
The most important connection in this chapter is that the impulse applied to an object is exactly equal to the change in its momentum.
\[\text{Impulse} = \text{Change in Momentum}\]
If an object has an initial velocity \(u\) and a final velocity \(v\), the change in momentum is \(m(v - u)\).
\[I = m v - m u\]
\[I = m(v - u)\]
This principle is essentially a restatement of Newton’s Second Law (\(F=ma\)) because acceleration \(a = \frac{v-u}{t}\). Substituting this back into \(F=ma\) gives:
\[F = m \left( \frac{v-u}{t} \right)\]
Multiplying by \(t\) gives:
\[F t = m(v - u)\]
So, the two concepts are inextricably linked.
Analogy: When you catch a ball, you pull your hand back (increasing the time \(t\)). Since the change in momentum \(\Delta p\) is fixed (the ball has to stop), increasing \(t\) means the average force \(F\) felt on your hand decreases (\(F = I/t\)). This prevents stinging!
Common Mistake Alert: Always ensure you subtract the initial momentum from the final momentum to find the change: \(\text{Final} - \text{Initial}\). If the object reverses direction, remember to include the negative sign for the reversing velocity!
3. The Principle of Conservation of Momentum
This principle is the cornerstone of analyzing collisions and explosions in Mechanics.
The Law
For a system of interacting particles (like a collision) where no external forces are acting (e.g., ignoring friction or air resistance), the total momentum of the system remains constant.
In simpler terms:
\[\text{Total Momentum BEFORE Interaction} = \text{Total Momentum AFTER Interaction}\]
Applying the Principle (Two Particles)
Consider two particles, A and B, moving along the same straight line, which collide.
Let:
- \(m_A\), \(m_B\) be the masses of A and B.
- \(u_A\), \(u_B\) be the initial velocities of A and B.
- \(v_A\), \(v_B\) be the final velocities of A and B.
The formula for the conservation of momentum is:
\[m_A u_A + m_B u_B = m_A v_A + m_B v_B\]
Step-by-Step Problem Solving
Solving collision problems requires careful handling of direction:
- Define the Positive Direction: Choose one direction (e.g., right) as positive.
- Assign Velocities (Initial): If a particle moves left, its initial velocity (\(u\)) must be negative.
- Set up the Equation: Substitute all masses and signed velocities into the conservation equation.
- Solve for the Unknown Velocity: If the unknown velocity (\(v\)) comes out negative, it means the particle moves in the direction you defined as negative (the left direction).
Example Scenario (Collision): Particle A (mass 2 kg, speed \(5 \text{ m s}^{-1}\) right) collides with particle B (mass 3 kg, speed \(1 \text{ m s}^{-1}\) left). If A continues right at \(1 \text{ m s}^{-1}\), find \(v_B\).
Let right be positive:
- \(u_A = +5\), \(u_B = -1\)
- \(v_A = +1\)
Equation:
\[(2)(+5) + (3)(-1) = (2)(+1) + (3)(v_B)\]
\[10 - 3 = 2 + 3v_B\]
\[7 = 2 + 3v_B\]
\[5 = 3v_B\]
\[v_B = \frac{5}{3} \approx 1.67 \text{ m s}^{-1}\]
Since \(v_B\) is positive, Particle B moves to the right after the collision.
Key Takeaway: Momentum is always conserved in closed systems. Always choose a positive direction and stick to it religiously!
4. Direct Impact with a Fixed Surface
The syllabus requires us to analyze situations where a particle strikes a fixed smooth surface perpendicularly (i.e., straight on).
The Scenario
A particle hits a wall (the fixed surface) and bounces straight back. Since the wall is fixed and massive, it is treated as having infinite mass and zero velocity—it does not gain momentum, so the conservation law used above for two particles is not applicable.
Calculating Impulse during Impact
We use the Impulse-Momentum Principle: \(I = m(v - u)\).
The impulse is delivered by the wall onto the particle to change the particle’s momentum.
Step-by-Step Example:
A ball of mass 0.5 kg strikes a wall at \(10 \text{ m s}^{-1}\) and rebounds at \(6 \text{ m s}^{-1}\).
- Define Direction: Let the initial direction (towards the wall) be positive.
- Assign Velocities:
- Initial velocity \(u = +10 \text{ m s}^{-1}\)
- Final velocity \(v = -6 \text{ m s}^{-1}\) (It rebounded, so it's moving in the negative direction)
- Calculate Impulse (\(I\)):
\[I = m(v - u)\]
\[I = 0.5 ((-6) - (+10))\]
\[I = 0.5 (-16)\]
\[I = -8 \text{ N s}\]
The impulse is \(-8 \text{ N s}\). The negative sign tells us the impulse acts in the negative direction, which is away from the wall. This makes perfect physical sense—the wall pushes the ball away from it.
If the question asks for the magnitude of the impulse, the answer is \(8 \text{ N s}\).
Crucial Point: When a particle hits a fixed surface and rebounds, the impulse is often much larger than a non-rebounding impact because the momentum change is twice as large (it goes from positive momentum to negative momentum).
Summary: Key Takeaways and Formulas
Keep these core formulas and concepts in your memory box!
Formulas to Remember:
Momentum (\(p\)):
\[p = m v\]
Impulse (\(I\)):
\[I = F t\]
Impulse-Momentum Principle:
\[I = m v - m u\]
Conservation of Momentum (Two Particles A and B):
\[m_A u_A + m_B u_B = m_A v_A + m_B v_B\]
Memory Aids:
- The letter \(p\) is often used for momentum, just think of \(p\) as standing for "power of movement."
- Impulse (\(I\)) is the result of Force (\(F\)) over time (\(t\)). \(I = Ft\).
- When using the conservation law, if you calculate a negative velocity, it simply means the object travels in the direction opposite to the one you chose as positive.
You’ve mastered the essential theory behind impacts! Now, practice applying the sign convention correctly, and you'll be set for any straight-line momentum problem.