Momentum
Definition: Momentum (p) is defined as the product of an object's mass and its velocity.
Equation:
p = mv
Where:
- p = momentum, measured in kilogram metres per second (kg \cdot m/s) or Newton seconds (N \cdot s)
- m = mass, measured in kilograms (kg)
- v = velocity, measured in metres per second (m/s)
Step 1: Convert units.
Mass must be in kg.
m = 58 \text{ g} = 0.058 \text{ kg}
Step 2: Apply the formula.
p = mv
p = 0.058 \times 30
p = 1.74 \text{ kg m/s}
Note: The direction of the momentum is the same as the direction of the velocity (away from the racket).
The Error: Students often treat momentum as a scalar (just a number) and ignore direction. For example, if a ball hits a wall and bounces back, students might add the initial and final speeds instead of subtracting them.
The Correct Understanding: Momentum is a vector. You must define a positive direction (e.g., right is +). If the ball moves left after bouncing, its velocity is negative.
- Initial momentum: +mv
- Final momentum: -mv (if speed is same)
- Change in momentum: p_{final} - p_{initial} = (-mv) - (+mv) = -2mv. The magnitude of the change is 2mv, not zero.
Why examiners accept this: The definition must link force and time explicitly. Simply saying 'change in momentum' is often accepted as an alternative, but the primary definition is force-based.
Correct Phrase: "Impulse is the force multiplied by the time for which the force acts."
Example Usage:
Q: Define impulse.
A: Impulse is the product of the resultant force and the time interval during which it acts (I = F\Delta t). This directly addresses the syllabus requirement to link force, time, and momentum change.
Impulse is the measure of the change in momentum produced by a force acting over a period of time. It explains why airbags work: they increase the time (\Delta t) of impact, which decreases the force (F) required to produce the same change in momentum.
Equation:
\text{Impulse} = F \Delta t = \Delta p = m(v - u)
Where:
- F = resultant force (N)
- \Delta t = time interval (s)
- \Delta p = change in momentum (kg \cdot m/s or N \cdot s)
- m = mass (kg)
- v = final velocity (m/s)
- u = initial velocity (m/s)
Equation:
F = \frac{\Delta p}{\Delta t}
Where:
- F = resultant force (N)
- \Delta p = change in momentum (kg \cdot m/s)
- \Delta t = time taken for the change (s)
Derivation from Newton's Second Law:
We know F = ma.
Acceleration a = \frac{v - u}{t} = \frac{\Delta v}{\Delta t}.
So, F = m \left( \frac{\Delta v}{\Delta t} \right) = \frac{m(v - u)}{\Delta t} = \frac{mv - mu}{\Delta t} = \frac{\Delta p}{\Delta t}.
This shows that force is directly proportional to how quickly momentum changes.
Step 1: Calculate Change in Momentum (\Delta p).
\Delta p = m(v - u)
\Delta p = 0.046 \times (41 - 0) = 1.886 \text{ kg m/s}
Step 2: Calculate Force (F).
F = \frac{\Delta p}{\Delta t}
F = \frac{1.886}{0.0005} = 3772 \text{ N}
Rounding: The answer is approximately 3800 N (to 2 significant figures).
The Correct Understanding: Impulse is Force × Time, so the unit is Newton-seconds (N \cdot s). Alternatively, since Impulse = Change in Momentum, the unit is kilogram metres per second (kg \cdot m/s). These two units are equivalent:
1 \text{ N} = 1 \text{ kg m/s}^2 \implies 1 \text{ N s} = 1 \text{ kg m/s}
Why examiners accept this: Examiners look for the specific condition that allows conservation laws to apply. Momentum is only conserved if no external resultant forces act on the system.
Correct Phrase: "Momentum is conserved because there are no external resultant forces acting on the system (or friction/air resistance are negligible)."
Example Usage:
Q: Explain why momentum is conserved in this collision.
A: Momentum is conserved because the trolleys form an isolated system where the net external force is zero. The internal forces between them cancel out.
Principle: In a closed system (no external forces), the total momentum before an event is equal to the total momentum after the event.
Key Condition: This applies to collisions and explosions. It holds true regardless of whether kinetic energy is conserved (i.e., it works for both elastic and inelastic collisions).
Equation:
\sum p_{\text{before}} = \sum p_{\text{after}}
For two objects (A and B):
m_A u_A + m_B u_B = m_A v_A + m_B v_B
Where:
- u = initial velocity (before collision/explosion)
- v = final velocity (after collision/explosion)
Step 1: Define Direction.
Let the initial direction be positive (+).
Step 2: Calculate Total Momentum Before.
p_{\text{before}} = (m_A u_A) + (m_B u_B)
p_{\text{before}} = (1.8 \times 1.5) + (0.19 \times 0) = 2.7 \text{ kg m/s}
Step 3: Set Up Conservation Equation.
p_{\text{after}} = p_{\text{before}} = 2.7 \text{ kg m/s}
p_{\text{after}} = (m_A v_A) + (m_B v_B)
2.7 = (1.8 \times v_A) + (0.19 \times 6.9)
Step 4: Solve for v_A.
2.7 = 1.8 v_A + 1.311
1.8 v_A = 2.7 - 1.311
1.8 v_A = 1.389
v_A = \frac{1.389}{1.8} \approx 0.77 \text{ m/s}
Answer: The object continues in the original direction at 0.77 m/s.
The Error: In explosion problems or bouncing collisions, students forget that objects move in opposite directions. They add the momenta instead of subtracting them.
The Correct Understanding: Always assign a positive sign to one direction (e.g., right/up) and a negative sign to the opposite direction (left/down).
- If Object A moves right (+v) and Object B moves left (-v), their total momentum is m_A v - m_B v.
- In an explosion from rest, total initial momentum is 0. Therefore, m_A v_A + m_B v_B = 0, meaning m_A v_A = -m_B v_B. The momenta are equal in magnitude but opposite in direction.
Why examiners accept this: It simplifies the equation and shows you understand that zero velocity means zero momentum contribution.
Correct Phrase: "The initial momentum of [Object] is zero because its velocity is zero."
Example Usage:
Q: Write down the total momentum before the collision.
A: Total momentum before is 0 because both objects are at rest (or one is at rest and we define the system appropriately). In an explosion, p_{\text{initial}} = 0, so p_{\text{final}} must also sum to 0.
I = F \Delta t
I = 300 \times 0.02
I = 6 \text{ N s} (or 6 \text{ kg m/s})
Step 2: Momentum before.
p_A = 2 \times 4 = +8 \text{ kg m/s}
p_B = 3 \times (-2) = -6 \text{ kg m/s}
p_{\text{total}} = 8 + (-6) = +2 \text{ kg m/s}
Step 3: Momentum after.
Combined mass = 2 + 3 = 5 \text{ kg}.
p_{\text{after}} = 5v
Step 4: Conservation.
5v = 2
v = 0.4 \text{ m/s}
Answer: 0.4 \text{ m/s} to the right.