Impulse is a concept that represents a large force acting over a short time. It is very useful for analyzing collisions and explosions. This section defines impulse and shows how to use it to solve elastic and inelastic collision problems.
Definition of impulse
Impulse is the force exerted on an object multiplied by the time duration of that force:
\[ \mathbf{J} = \int_{t_1}^{t_2} \mathbf{F} \, dt \]
For a constant force:
\[ \mathbf{J} = \mathbf{F} \Delta t \]
Unit: newton-second (N·s) or kilogram-meter-per-second (kg·m/s).
Relationship between impulse and change in momentum
From Newton's second law:
\[ \mathbf{F} = \frac{d\mathbf{p}}{dt} \]
Integrating:
\[ \int_{t_1}^{t_2} \mathbf{F} \, dt = \int_{t_1}^{t_2} \frac{d\mathbf{p}}{dt} \, dt = \mathbf{p}_{t_2} - \mathbf{p}_{t_1} \]
\[ \mathbf{J} = \Delta \mathbf{p} \]
Impulse equals the change in momentum of the object.
Example 1: Baseball and bat
A baseball flies toward a bat:
- Ball mass: \( m = 0.145 \) kg
- Velocity before impact: \( v_i = 40 \) m/s
- Velocity after impact: \( v_f = -60 \) m/s (reversed)
- Contact time: \( \Delta t = 0.001 \) s
Question: What is the average force exerted by the bat?
Solution:
Change in momentum: \[ \Delta \mathbf{p} = m(v_f - v_i) = 0.145(-60 - 40) = 0.145(-100) = -14.5 \text{ kg·m/s} \]
Impulse: \[ \mathbf{J} = \Delta \mathbf{p} = -14.5 \text{ N·s} \]
Average force: \[ \mathbf{F}_{\text{avg}} = \frac{\mathbf{J}}{\Delta t} = \frac{-14.5}{0.001} = -14500 \text{ N} \]
The bat exerts a force of 14,500 newtons (roughly 15 tons)!
Collisions — two types
Collisions fall into two categories:
Elastic collision
In an elastic collision:
- Momentum is conserved: \( m_1 v_{1,i} + m_2 v_{2,i} = m_1 v_{1,f} + m_2 v_{2,f} \)
- Kinetic energy is conserved: \( \frac{1}{2}m_1 v_{1,i}^2 + \frac{1}{2}m_2 v_{2,i}^2 = \frac{1}{2}m_1 v_{1,f}^2 + \frac{1}{2}m_2 v_{2,f}^2 \)
Example: billiard balls, atomic collisions.
Inelastic collision
In an inelastic collision:
- Momentum is conserved: \( m_1 v_{1,i} + m_2 v_{2,i} = m_1 v_{1,f} + m_2 v_{2,f} \)
- Kinetic energy is NOT conserved: energy becomes heat and deformation.
Special case — perfectly inelastic: Two objects stick together after collision and move with the same velocity.
Example 2: Elastic 1D collision
Two objects on a line:
- Object 1: mass \( m_1 = 2 \) kg, initial velocity \( v_{1,i} = 5 \) m/s
- Object 2: mass \( m_2 = 3 \) kg, initial velocity \( v_{2,i} = 0 \) m/s
Elastic collision. Final velocities?
Solution:
For elastic collision, we have two equations:
Momentum conservation: \[ 2(5) + 3(0) = 2 v_{1,f} + 3 v_{2,f} \] \[ 10 = 2 v_{1,f} + 3 v_{2,f} \quad \text{...(1)} \]
Energy conservation: \[ \frac{1}{2}(2)(5)^2 + \frac{1}{2}(3)(0)^2 = \frac{1}{2}(2) v_{1,f}^2 + \frac{1}{2}(3) v_{2,f}^2 \] \[ 25 = v_{1,f}^2 + 1.5 v_{2,f}^2 \quad \text{...(2)} \]
From equation (1): \( v_{1,f} = 5 - 1.5 v_{2,f} \)
Substituting into equation (2): \[ 25 = (5 - 1.5 v_{2,f})^2 + 1.5 v_{2,f}^2 \] \[ 25 = 25 - 15 v_{2,f} + 2.25 v_{2,f}^2 + 1.5 v_{2,f}^2 \] \[ 0 = -15 v_{2,f} + 3.75 v_{2,f}^2 \] \[ v_{2,f}(3.75 v_{2,f} - 15) = 0 \]
Two solutions:
- \( v_{2,f} = 0 \) (initial state — no collision)
- \( v_{2,f} = 4 \) m/s (final state — actual collision)
Therefore \( v_{2,f} = 4 \) m/s and: \[ v_{1,f} = 5 - 1.5(4) = 5 - 6 = -1 \text{ m/s} \]
Object 1 reverses and moves backward at 1 m/s.
Check:
- Momentum: \( 2(-1) + 3(4) = -2 + 12 = 10 \) ✓
- Energy: \( (-1)^2 + 1.5(4)^2 = 1 + 24 = 25 \) ✓
Example 3: Inelastic collision — cars stick together
Two cars:
- Car A: mass \( m_A = 1000 \) kg, velocity \( v_{A,i} = 20 \) m/s
- Car B: mass \( m_B = 1500 \) kg, velocity \( v_{B,i} = 0 \) m/s (at rest)
They collide and stick together.
Question: Final velocity?
Solution:
Only momentum conservation (energy is lost):
\[ m_A v_{A,i} + m_B v_{B,i} = (m_A + m_B) v_f \]
\[ 1000(20) + 1500(0) = (1000 + 1500) v_f \]
\[ 20000 = 2500 v_f \]
\[ v_f = 8 \text{ m/s} \]
Energy analysis:
Initial kinetic energy: \[ KE_i = \frac{1}{2}(1000)(20)^2 = 200000 \text{ J} \]
Final kinetic energy: \[ KE_f = \frac{1}{2}(2500)(8)^2 = 80000 \text{ J} \]
Energy lost: \[ \Delta E = 200000 - 80000 = 120000 \text{ J} \]
60 percent of energy became heat and deformation!
Quick formulas for elastic collision
For 1D elastic collision between two objects:
\[ v_{1,f} = \frac{(m_1 - m_2)v_{1,i} + 2m_2 v_{2,i}}{m_1 + m_2} \]
\[ v_{2,f} = \frac{(m_2 - m_1)v_{2,i} + 2m_1 v_{1,i}}{m_1 + m_2} \]
What you should be able to do
- Define impulse: \( \mathbf{J} = \mathbf{F} \Delta t \)
- Connect impulse to momentum change: \( \mathbf{J} = \Delta \mathbf{p} \)
- Distinguish elastic from inelastic collisions
- Apply momentum conservation to all collisions
- Recognize that only elastic collisions conserve energy
Preview of §9.5
Until now we have studied linear motion (translation). But objects like spinning wheels, spinning balls, and neutron stars rotate! The next section introduces angular momentum.
📚 See also: Halliday Vol 1, Ch 9, §9.4.
Have a question? 🤔
If something isn't clear or you have a question, ask it here. The answer will be published on this page.