IB MYP 4-5 Physics- Momentum its conservation and impulse- Study Notes - New Syllabus
IB MYP 4-5 Physics-Momentum its conservation and impulse- Study Notes
Key Concepts
- Momentum its conservation and impulse
Momentum its conservation and impulse
Momentum
Momentum is the measure of the quantity of motion an object has. It depends on both the object’s mass and its velocity.
Formula: \( p = mv \)
Where: \( p \) = momentum (kg·m/s) \( m \) = mass (kg) \( v \) = velocity (m/s)
- Momentum is a vector quantity it has both magnitude and direction.
- Conservation of Momentum: In a closed system with no external forces, total momentum remains constant.
Example:
A 1,200 kg car moves at 20 m/s. Find its momentum.
▶️ Answer/Explanation
Given: \( m = 1200\ \text{kg},\ v = 20\ \text{m/s} \)
\( p = mv = (1200)(20) = 24000\ \text{kg·m/s} \)
\(\boxed{p = 24000\ \text{kg·m/s}}\)
Example:
A 5 kg ball is moving north at 4 m/s. What is its momentum vector?
▶️ Answer/Explanation
Given: \( m = 5\ \text{kg},\ v = 4\ \text{m/s north} \)
\( p = mv = (5)(4) = 20\ \text{kg·m/s north} \)
\(\boxed{p = 20\ \text{kg·m/s north}}\)
Law of Conservation of Momentum
The Law of Conservation of Momentum states that in an isolated system (no external forces), the total momentum before an interaction is equal to the total momentum after the interaction.
\(\text{Total momentum before} = \text{Total momentum after}\)
\(m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2\)
- \(m_1, m_2\) = masses of the two objects
- \(u_1, u_2\) = initial velocities before collision
- \(v_1, v_2\) = final velocities after collision
Key Points:
- Applies to all types of collisions and explosions.
- Momentum is a vector quantity (direction matters).
- Only valid in absence of external forces.
Example:
A 2 kg trolley moving at \(4 \ \text{m/s}\) collides with a stationary 1 kg trolley. They stick together after the collision. Find their common velocity.
▶️ Answer/Explanation
Step 1: Initial momentum:
\(p_{\text{initial}} = (2)(4) + (1)(0) = 8 \ \text{kg·m/s}\)
Step 2: After collision, total mass = \(2 + 1 = 3 \ \text{kg}\)
Step 3: Using conservation of momentum:
\(8 = (3)(v) \implies v = \dfrac{8}{3} \ \text{m/s}\)
Final Answer: \(\boxed{2.67 \ \text{m/s}}\)
Example:
A 1000 kg car moving at \(20 \ \text{m/s}\) collides head-on with a 1500 kg truck moving in the opposite direction at \(15 \ \text{m/s}\). After the collision, the car moves backward at \(5 \ \text{m/s}\). Find the truck’s velocity after the collision.
▶️ Answer/Explanation
Step 1: Assign forward direction as positive.
\(p_{\text{initial}} = (1000)(20) + (1500)(-15) = 20000 – 22500 = -2500 \ \text{kg·m/s}\)
Step 2: Final momentum:
\(p_{\text{final}} = (1000)(-5) + (1500)(v_{\text{truck}}) = -5000 + 1500v_{\text{truck}}\)
Step 3: Equate initial and final momentum:
\(-2500 = -5000 + 1500v_{\text{truck}}\)
\(1500v_{\text{truck}} = 2500 \implies v_{\text{truck}} = \dfrac{2500}{1500} = 1.67 \ \text{m/s}\)
Final Answer: \(\boxed{1.67 \ \text{m/s} \ \text{forward}}\)
Impulse
Impulse is the change in momentum of an object when a force is applied over a certain time period.
Formula: \( J = F\Delta t = \Delta p \)
Where: \( J \) = impulse (N·s) \( F \) = force (N) \( \Delta t \) = time (s) \( \Delta p \) = change in momentum (kg·m/s)
- Impulse is a vector quantity.
- It can be calculated from the area under a force–time graph.
Example:
A 0.15 kg baseball is hit with a force of 600 N for 0.02 s. Find the impulse.
▶️ Answer/Explanation
Given: \( F = 600\ \text{N},\ \Delta t = 0.02\ \text{s} \)
\( J = F\Delta t = (600)(0.02) = 12\ \text{N·s} \)
\(\boxed{J = 12\ \text{N·s}}\)
Example:
A 50 kg cart is initially at rest. A force acts for 5 s, giving it a momentum of 250 kg·m/s. What is the impulse?
▶️ Answer/Explanation
Impulse equals the change in momentum: \( J = \Delta p = 250 – 0 = 250\ \text{N·s} \)
\(\boxed{J = 250\ \text{N·s}}\)