Impulse, momentum and energy methods; impact
Impulse–momentum, work–energy, power and angular momentum, with direct central impact and the coefficient of restitution, worked for braking, a rear-end collision and a flywheel.
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Why it matters
Crash pulses, brake stopping distances, flywheel sizing, the energy a bumper must absorb, the bounce of a valve on its seat: these are all problems where integrating F = m·a step by step is slow, but impulse–momentum or work–energy gives the answer in two lines. Choosing the right method is half the skill. Momentum methods handle forces given against time and short, violent impacts; energy methods handle forces given against displacement.
Key ideas
Linear momentum and impulse. Momentum p = m·v is a vector. Integrating Newton's second law over time gives the impulse–momentum principle: the impulse of the resultant external force equals the change in momentum. For a constant force the impulse is F·Δt; for a varying force it is the area under the force–time curve. The average force over a crash or impact is Δp/Δt, which is why crumple zones and airbags reduce injury: they stretch Δt for the same Δp.
Conservation of momentum. If the net external impulse on a system is zero (or negligible, as during a very short impact where only large internal contact forces matter), the total momentum of the system is conserved. Internal forces between the colliding bodies cancel in pairs. Momentum can be conserved in one direction even when it is not in another.
Work and kinetic energy. Integrating along the path gives the work–energy principle: the work done by all forces equals the change in kinetic energy. Work is a scalar: U = ∫F·ds. Forces perpendicular to the motion (normal reactions on a level road) do no work. Kinetic energy is ½·m·v² for a particle; for a rigid body in plane motion it is ½·m·v_G² + ½·I_G·ω² (translation of the centre of mass plus rotation about it).
Conservative forces and potential energy. Gravity and linear springs are conservative, so their work can be written as a loss in potential energy: V_g = m·g·h and V_s = ½·k·x². If only conservative forces do work, T + V stays constant. Friction is non-conservative; its work, −F·s, is lost as heat.
Power and efficiency. Power P = F·v = T·ω is the rate of doing work. Efficiency = output power ÷ input power.
Angular momentum. For a rigid body rotating about a fixed axis, H = I·ω, and the angular impulse ∫M·dt equals the change in H. With no external moment about an axis, H about that axis is conserved, which is the basis of clutch-engagement problems (two shafts reaching a common speed).
Impact (direct central). When two bodies collide along the line joining their centres:
- momentum along the line of impact is conserved;
- the coefficient of restitution e = (relative velocity of separation) ÷ (relative velocity of approach) describes how "bouncy" the contact is: e = 1 perfectly elastic (no kinetic energy loss), e = 0 perfectly plastic (bodies move together), 0 < e < 1 real impacts;
- kinetic energy is lost unless e = 1. The loss goes into permanent deformation, heat, sound and vibration. For oblique impact of smooth bodies, apply both conditions to the velocity components along the line of impact; the tangential components are unchanged. The value of e depends on materials, impact speed and geometry, so take it from the problem or test data.
Formulas
∫F dt = m·v₂ − m·v₁ (for constant F: F·Δt = m·Δv)
- F: resultant external force (N); t: time (s); m: mass (kg); v₁, v₂: velocities (m/s). Impulse in N·s equals momentum in kg·m/s.
m₁·u₁ + m₂·u₂ = m₁·v₁ + m₂·v₂
- u: velocities before impact, v: after (m/s), signs by a chosen positive direction. Valid when external impulse is negligible.
e = (v₂ − v₁) / (u₁ − u₂)
- e: coefficient of restitution (dimensionless), 0 ≤ e ≤ 1. Direct central impact.
ΔKE = [m₁·m₂ / (2·(m₁ + m₂))]·(1 − e²)·(u₁ − u₂)²
- Kinetic energy lost in a direct central impact (J).
U₁₋₂ = T₂ − T₁, with T = ½·m·v² + ½·I_G·ω²
- U: work of all forces (J); I_G: mass moment of inertia about the centre of mass (kg·m²); ω: angular velocity (rad/s).
T₁ + V₁ = T₂ + V₂, with V = m·g·h + ½·k·x²
- k: spring stiffness (N/m); x: spring deflection from its free length (m). Only conservative forces doing work.
P = F·v = T·ω
- P: power (W); T: torque (N·m).
∫M dt = I·ω₂ − I·ω₁
- M: external moment about the fixed axis (N·m); I: moment of inertia about that axis (kg·m²).
Worked examples
Example 1 (standard): average braking force and distance. Given: a 1200 kg car at 15 m/s brakes to rest in 2.5 s. Find the average braking force and, assuming the force is constant, the stopping distance.
- Impulse–momentum:
F·Δt = m·(v₂ − v₁), so F = 1200 × (0 − 15) / 2.5 = −7200 N. - Work–energy:
−F·s = 0 − ½·m·v₁², so s = ½ × 1200 × 15² / 7200 = 135 000 / 7200 = 18.75 m. - Check with kinematics: a = 15 / 2.5 = 6 m/s² and s = v² / (2a) = 225 / 12 = 18.75 m. Answer: average braking force 7.2 kN; stopping distance 18.75 m.
Example 2 (GATE level): rear-end collision. Given: car A (1500 kg) at 20 m/s strikes car B (1000 kg) moving at 10 m/s in the same direction. Coefficient of restitution e = 0.4. Find the velocities after impact and the kinetic energy lost.
- Momentum: 1500 × 20 + 1000 × 10 = 1500·v_A + 1000·v_B, so 1500·v_A + 1000·v_B = 40 000.
- Restitution:
v_B − v_A = e·(u_A − u_B)= 0.4 × (20 − 10) = 4 m/s. - Substitute v_B = v_A + 4: 2500·v_A + 4000 = 40 000, so v_A = 14.4 m/s and v_B = 18.4 m/s.
- KE before: ½ × 1500 × 20² + ½ × 1000 × 10² = 300 000 + 50 000 = 350 000 J.
- KE after: ½ × 1500 × 14.4² + ½ × 1000 × 18.4² = 155 520 + 169 280 = 324 800 J.
- Loss = 25 200 J. Check by formula: [1500 × 1000 / (2 × 2500)] × (1 − 0.16) × 10² = 300 × 0.84 × 100 = 25 200 J. Answer: v_A = 14.4 m/s, v_B = 18.4 m/s, energy lost 25.2 kJ.
Example 3: flywheel energy. Given: a flywheel with I = 2 kg·m² slows from 3000 to 2900 rev/min during a power stroke demand. Find the energy it supplies.
- ω₁ = 2π × 3000 / 60 = 314.16 rad/s; ω₂ = 2π × 2900 / 60 = 303.69 rad/s.
ΔT = ½·I·(ω₁² − ω₂²)= ½ × 2 × (98 696 − 92 226) = 6470 J. Answer: about 6.47 kJ.
Common mistakes
- Dropping the sign of velocity: in head-on impacts one velocity is negative.
- Assuming kinetic energy is conserved in a collision. Only momentum is, unless e = 1.
- Writing e with the subtraction the wrong way round, which gives a negative e.
- Using conservation of momentum when an external impulse is not negligible, such as a body hitting a fixed wall or a ground reaction during a long contact.
- Forgetting the rotational kinetic energy ½·I·ω² of wheels and rotating parts.
- Mixing up impulse (N·s) and work (N·m = J).
- Applying restitution to the tangential components in oblique impact of smooth bodies.
For GATE ME
Expect direct central impact with a given e (velocities and energy loss), ball dropped on a floor (rebound heights scale with e²), block-and-spring problems with work–energy, momentum conservation with a gun and shell or a bullet and block, and angular momentum in clutch or flywheel problems. Practise setting a sign convention first and checking with the energy-loss formula.
Quick check
- A ball dropped from 2 m rebounds to 0.5 m. What is e?
- Two equal masses collide head-on with equal speeds and e = 0. What is their common velocity?
- A 10 N force acts on a body for 4 s. What impulse does it give?
- In an elastic collision of a moving mass with an equal stationary mass, what happens?
Answers: 1. e = √(0.5/2) = 0.5. 2. Zero. 3. 40 N·s. 4. The moving mass stops and the stationary one moves off with the original velocity.
Interview questions
All Engineering Mechanics and Strength of Materials interview questionsTry answering each one aloud before you open it.
1.What is impulse in the context of mechanics?Concept
Impulse is the time integral of a force, J = ∫F dt; for a constant force it is simply F·Δt, and in general it is the area under the force–time curve. It is a vector and, by the impulse–momentum principle, the impulse of the resultant external force equals the change in momentum, m·v₂ − m·v₁. Its unit is the newton-second, which is the same as kg·m/s. It is the natural quantity for short, large forces such as impacts, where the force history is unknown but the velocity change is.
2.Explain the principle of conservation of momentum.Concept
The principle of conservation of momentum states that in a closed system with no external forces, the total momentum before an event is equal to the total momentum after the event. This principle is fundamental in analyzing collisions and explosions.
3.How is kinetic energy related to momentum?Concept
For a particle, KE = ½·m·v² and p = m·v, so KE = p²/(2·m) and p = √(2·m·KE). Momentum is a vector and is conserved in any collision when external impulse is negligible; kinetic energy is a scalar and is conserved only in a perfectly elastic collision. So two bodies with the same momentum can carry very different energies: the lighter one has more kinetic energy, which is why a light, fast projectile is more damaging than a heavy, slow one with equal momentum.
4.Why is the impulse-momentum method used in crash analysis?Application
In a crash the force history is complicated and unknown, but the velocity change of the vehicle and occupants is known or measurable. Impulse–momentum gives the average force directly, F_avg = m·Δv/Δt, which shows why crumple zones, seat-belt load limiters and airbags work: they lengthen the stopping time for the same Δv and so cut the force and deceleration. Conservation of momentum across the impact also lets investigators reconstruct pre-impact speeds from post-impact motion.
5.What happens to the kinetic energy during a perfectly inelastic collision?Application
In a perfectly inelastic collision, the colliding objects stick together after the collision, and some kinetic energy is converted into other forms of energy, such as heat or sound. As a result, the total kinetic energy after the collision is less than the total kinetic energy before the collision, although momentum is conserved.
6.Explain how energy methods are used to analyze the impact of a falling object.Application
The speed at impact comes from energy conservation, m·g·h = ½·m·v², so v = √(2·g·h) if air resistance is negligible. To find the force or deflection on impact, the work–energy principle is applied through the stopping distance: the energy m·g·(h + δ) must be absorbed by the structure or spring over its deflection δ, for example ½·k·δ² for a spring. That is why a short stopping distance gives a very large impact force, and why a suddenly applied load gives twice the static deflection.
7.How does the coefficient of restitution affect the outcome of a collision?Application
The coefficient of restitution (e) measures the elasticity of a collision, defined as the relative velocity of separation divided by the relative velocity of approach. A value of e = 1 indicates a perfectly elastic collision, where kinetic energy is conserved, while e = 0 indicates a perfectly inelastic collision, where the objects stick together. The value of e affects the post-collision velocities of the objects.
8.Calculate the impulse experienced by a 5 kg object subjected to a force of 10 N for 3 seconds.Numerical
Impulse (J) is calculated as the product of force (F) and time (t): J = F * t. Here, J = 10 N * 3 s = 30 N·s. The object experiences an impulse of 30 Newton-seconds.
9.A 2 kg ball moving at 3 m/s collides with a stationary 3 kg ball. If the collision is perfectly elastic, what is the velocity of the 2 kg ball after the collision?Numerical
For a perfectly elastic collision with the second ball at rest, momentum and kinetic energy conservation give v₁ = (m₁ − m₂)/(m₁ + m₂)·u₁ and v₂ = 2·m₁/(m₁ + m₂)·u₁. Here v₁ = (2 − 3)/5 × 3 = −0.6 m/s and v₂ = 4/5 × 3 = 2.4 m/s. So the 2 kg ball rebounds at 0.6 m/s (opposite to its original direction) and the 3 kg ball moves off at 2.4 m/s. Check: momentum 2 × (−0.6) + 3 × 2.4 = 6 kg·m/s, and kinetic energy 0.36 + 8.64 = 9 J, both unchanged.
10.What is the significance of using energy methods in analyzing vehicle crashes?Application
Energy methods are significant in analyzing vehicle crashes because they allow engineers to understand how energy is distributed and dissipated during a crash. By analyzing the conversion of kinetic energy into deformation energy, heat, and sound, engineers can design vehicles that better absorb impact energy, enhancing passenger safety.
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