Impulse
Newton's Laws & Momentum - OCR A-Level Physics
Key Definition
Impulse
The product of force and the time for which the force acts. Impulse is a vectorA quantity with both magnitude and direction (e.g. force, velocity, displacement). quantity in the direction of the force. Measured in $\text{N s}$ (equivalent to $\text{kg m s}^{-1}$). Impulse is equal to the change in momentum it causes.
The product of force and the time for which the force acts. Impulse is a vectorA quantity with both magnitude and direction (e.g. force, velocity, displacement). quantity in the direction of the force. Measured in $\text{N s}$ (equivalent to $\text{kg m s}^{-1}$). Impulse is equal to the change in momentum it causes.
$$F \, \Delta t = \Delta p = m v - m u$$
- Impulse follows directly from Newton's second law $\Sigma F = \Delta p / \Delta t$ by multiplying both sides by $\Delta t$.
- For a constant force, impulse equals $F \times \Delta t$. For a varying force, impulse equals the AREA under the force-time graph (this is the link to the next note).
- Rebound questions: $u$ and $v$ are vectors with different signs. A 0.20 kg ball hitting a wall at $8.0 \; \text{m s}^{-1}$ and rebounding at $6.0 \; \text{m s}^{-1}$ has $\Delta p = (0.20)(-6.0) - (0.20)(8.0) = -2.8 \; \text{kg m s}^{-1}$. Magnitude of impulse is 2.8 N s.
- Extending the collision time reduces the peak force: same $\Delta p$, but a larger $\Delta t$ means a smaller average $F$. This is the safety-device principle.
- Worked example: a 0.058 kg tennis ball changes from $40 \; \text{m s}^{-1}$ to $-50 \; \text{m s}^{-1}$ during a 5.0 ms contact. $\Delta p = (0.058)(-50) - (0.058)(40) = -5.22 \; \text{kg m s}^{-1}$. Average force $|F| = 5.22 / 0.0050 \approx 1.0 \times 10^{3} \; \text{N}$.
Common Mistake
HIGH
Wrong: Treating impulse as a scalar in a rebound. Writing $\Delta p = (0.20)(6.0) - (0.20)(8.0) = -0.4 \; \text{kg m s}^{-1}$ instead of $-2.8 \; \text{kg m s}^{-1}$.
Right: Velocity is a vector. After a rebound, $v$ has the opposite sign to $u$. Use $\Delta p = m v - m u = (0.20)(-6.0) - (0.20)(8.0) = -2.8 \; \text{kg m s}^{-1}$.
Right: Velocity is a vector. After a rebound, $v$ has the opposite sign to $u$. Use $\Delta p = m v - m u = (0.20)(-6.0) - (0.20)(8.0) = -2.8 \; \text{kg m s}^{-1}$.
Examiner Tips and Tricks
- Always state the direction of impulse, not just its magnitude. "2.8 N s directed away from the wall" gets the vector mark.
- If a question asks for AVERAGE force during a collision, use impulse: $F_{\text{avg}} = \Delta p / \Delta t$. Quote your answer to the same number of significant figures as the data.
- Impulse and change in momentum are NUMERICALLY EQUAL and have the same units. Mentioning both in working scores both marks.