IB DP Physics Impulse and force-time graphs Study Notes - 2025 Syllabus
IB DP Physics Impulse and force-time graphs Study Notes
IB DP Physics Impulse and force-time graphs Study Notes at IITian Academy focus on specific topic and type of questions asked in actual exam. Study Notes focus on IB Physics syllabus with guiding questions of
- that a resultant external force applied to a system constitutes an impulse J as given by J = FΔt where F is the average resultant force and Δt is the time of contact
- that the applied external impulse equals the change in momentum of the system
Standard level and higher level: 10 hours
Additional higher level: There is no additional higher level content
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 1
- IB DP Physics 2025 SL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
- IB DP Physics 2025 HL- IB Style Practice Questions with Answer-Topic Wise-Paper 2
Impulse and Change in Momentum
A resultant external force applied to a system constitutes an impulse \( J \) as given by:
\( J = F \Delta t \)
- Impulse is the product of the average resultant external force \( F \) and the time duration \( \Delta t \) the force acts.
- It is a vector quantity and has the same direction as the force applied.
- The SI unit of impulse is newton-second (N·s) or equivalently kg·m/s.
- Impulse is not a “thing” but an action that alters momentum – it’s how a force causes a momentum change over time.
Example:
A tennis ball of mass \( 0.06 \, \text{kg} \) is initially at rest. It is struck by a racket with an average force of \( 40 \, \text{N} \) for a contact time of \( 0.02 \, \text{s} \). What is the impulse imparted and the final speed of the ball?
▶️Answer/Explanation
Use impulse formula
\( J = F \Delta t = 40 \times 0.02 = \boxed{0.8 \, \text{N·s}} \)
Use impulse-momentum theorem
\( J = \Delta p = m v – m u \Rightarrow v = \frac{J}{m} = \frac{0.8}{0.06} = \boxed{13.33 \, \text{m/s}} \)
The applied external impulse equals the change in momentum of the system:
\( J = \Delta p = m v – m u \)
- This is known as the impulse-momentum theorem.
- It says that impulse causes a change in momentum.
- If no external force acts (\( F = 0 \)), then \( \Delta p = 0 \) ⇒ momentum is conserved.
- Applicable to any motion: linear, recoil, stopping a vehicle, kicking a ball, rocket propulsion, etc.
Example:
A truck of mass \( 2000 \, \text{kg} \) is moving at \( 15 \, \text{m/s} \) and brakes to a stop in \( 5 \, \text{s} \). Find the impulse and average braking force.
▶️Answer/Explanation
Change in momentum
\( \Delta p = m (v – u) = 2000(0 – 15) = -30,000 \, \text{kg·m/s} \)
Impulse equals change in momentum
\( J = \Delta p = -30,000 \, \text{N·s} \)
Use \( J = F \Delta t \) to find force
\( F = \frac{J}{\Delta t} = \frac{-30,000}{5} = \boxed{-6000 \, \text{N}} \) (Negative sign: force opposes motion)
Example:
A force is applied to a cart for 4 seconds, and the graph of the force versus time is shown below.
If the maximum force is \( F_{\text{max}} = 20 \, \text{N} \) and it increases linearly from 0 to 20 N in 4 s, find the impulse delivered to the cart.
▶️Answer/Explanation
This is a triangular force–time graph, so the impulse is the area under the curve.
Use area of triangle:
\( \text{Impulse} = \text{Area} = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 4 \, \text{s} \times 20 \, \text{N} \)
\( \Rightarrow J = \boxed{40 \, \text{N·s}} \)
Interpretation:
- The cart gained \( 40 \, \text{kg·m/s} \) of momentum.
- If its mass was known, you could calculate its final speed from this impulse.