Edexcel A Level (IAL) Physics-4.3 Applying Conservation of Linear Momentum- Study Notes- New Syllabus
Edexcel A Level (IAL) Physics -4.3 Applying Conservation of Linear Momentum- Study Notes- New syllabus
Edexcel A Level (IAL) Physics -4.3 Applying Conservation of Linear Momentum- Study Notes -Edexcel A level Physics – per latest Syllabus.
Key Concepts:
- understand how to apply conservation of linear momentum to problems in two dimensions
Conservation of Linear Momentum in Two Dimensions
In many real situations, objects move and collide in more than one direction. The principle of conservation of linear momentum still applies, but momentum must be treated as a vector quantity and resolved into perpendicular components.
Principle of Conservation of Linear Momentum
If no external resultant force acts on a system:
Total momentum before interaction = total momentum after interaction
This applies independently in each perpendicular direction.
Why Momentum Must Be Treated as a Vector
- Momentum has both magnitude and direction.
- In two dimensions, motion usually occurs along two perpendicular axes (e.g. x and y).
- Momentum must be resolved into components before applying conservation.
Resolving Momentum into Components
Momentum is given by:
\( p = mv \)
If an object moves at an angle \( \theta \) to the horizontal:
\( p_x = mv\cos\theta \)
\( p_y = mv\sin\theta \)
These components are treated separately.
Applying Conservation in Two Dimensions
For an isolated system:
- Total momentum in the x-direction is conserved.
- Total momentum in the y-direction is conserved.
\( \sum p_{x,\text{before}} = \sum p_{x,\text{after}} \)
\( \sum p_{y,\text{before}} = \sum p_{y,\text{after}} \)
General Method for Solving Problems
- Choose perpendicular axes (usually horizontal and vertical).
- Resolve all velocities into x and y components.
- Write momentum conservation equations separately for x and y directions.
- Solve the simultaneous equations.
- Recombine components if magnitude or direction is required.
Key Assumptions
- No external forces act on the system (or they are negligible during interaction).
- Interaction time is short (e.g. collisions, explosions).
- For explosions, initial total momentum may be zero.
Example (Easy)
An object at rest explodes into two fragments of equal mass. Explain why the fragments move in opposite directions.
▶️ Answer / Explanation
- Initial total momentum is zero.
- Momentum must be conserved.
- The two fragments must have equal and opposite momenta.
- This ensures total momentum remains zero.
Example (Medium)
A ball of mass \( 0.20\ \mathrm{kg} \) moves at \( 10\ \mathrm{m\,s^{-1}} \) along the x-axis. After collision, it moves at \( 6\ \mathrm{m\,s^{-1}} \) at \( 30^\circ \) above the x-axis. Find the change in momentum in the y-direction.
▶️ Answer / Explanation
Initial y-momentum:
\( p_{y,i} = 0 \)
Final y-momentum:
\( p_{y,f} = 0.20 \times 6 \sin 30^\circ = 0.20 \times 6 \times 0.5 = 0.60\ \mathrm{kg\,m\,s^{-1}} \)
Change in y-momentum:
\( \Delta p_y = 0.60\ \mathrm{kg\,m\,s^{-1}} \)
Example (Hard)
A particle of mass \( 2.0\ \mathrm{kg} \) moving east at \( 5.0\ \mathrm{m\,s^{-1}} \) collides with a stationary particle of mass \( 3.0\ \mathrm{kg} \). After the collision, the first particle moves north at \( 3.0\ \mathrm{m\,s^{-1}} \). Find the velocity of the second particle.
▶️ Answer / Explanation
X-direction momentum:
Initial: \( 2.0 \times 5.0 = 10\ \mathrm{kg\,m\,s^{-1}} \)
Final: \( 0 + 3.0 v_{2x} \)
\( 3.0 v_{2x} = 10 \Rightarrow v_{2x} = 3.33\ \mathrm{m\,s^{-1}} \)
Y-direction momentum:
Initial: \( 0 \)
Final: \( 2.0 \times 3.0 + 3.0 v_{2y} \)
\( 6 + 3.0 v_{2y} = 0 \Rightarrow v_{2y} = -2.0\ \mathrm{m\,s^{-1}} \)
Velocity of second particle:
\( v = \sqrt{(3.33)^2 + (2.0)^2} \approx 3.9\ \mathrm{m\,s^{-1}} \)
Direction: south of east.