Impulse, momentum and energy methods; impact

Impulse-momentum, work-energy, rotational KE and power, and direct central impact with the coefficient of restitution, with a conveyor-loading and a restitution example.

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Why it matters

Many mechatronic loads are short and sharp: a gripper closing on a part, a package dropping onto a moving conveyor, a press ram striking a die, a robot that collides with a fixture. Over such events the force history is unknown, but momentum and energy methods still give the velocities, the average forces and the energy lost as heat or deformation. They also give drive power and the energy a motor must supply, which you need to size it.

Key ideas

Linear impulse and momentum. Newton's second law integrated over time: the impulse of the resultant force, ∫F dt, equals the change in linear momentum m·v. It is a vector equation, so apply it component by component. It is ideal when force is given as a function of time, or when the time of an event is wanted.

Conservation of linear momentum. If the net external impulse on a system is zero in some direction, total momentum in that direction is conserved. During a short impact, the large contact forces are internal to the two-body system and ordinary forces such as gravity give negligible impulse, so momentum along the line of impact is conserved.

Work and kinetic energy. The work done by all forces on a body equals its change in kinetic energy: W = ΔKE. Work is a scalar, so it is ideal when force is given as a function of position and the speed after a distance is wanted. If only conservative forces (gravity, springs) do work, mechanical energy KE + PE is conserved. Friction converts mechanical energy to heat.

Rigid bodies. A body in plane motion has KE = ½·m·v_G² + ½·I_G·ω². The angular impulse about a fixed axis equals the change in angular momentum I·ω, and angular momentum is conserved when the net external moment about that axis is zero (for example, a clutch engaging two free shafts).

Power. P = F·v for a force, P = T·ω for a torque. Power sets motor ratings; energy sets battery size and heating.

Direct central impact. Two bodies collide along the line joining their centres of mass. Two equations give the two unknown final velocities:

  • conservation of momentum along the line of impact;
  • Newton's law of restitution: separation speed = e × approach speed. The coefficient of restitution e depends on both materials, impact speed and geometry, so take it from test data. e = 1 is perfectly elastic (no KE loss); e = 0 is perfectly plastic (the bodies move together after impact and the KE loss is greatest). Kinetic energy is not conserved in a real impact; it goes into vibration, heat and permanent deformation. For a ball dropped from height h₁ on a rigid floor, the rebound height is h₂ = e²·h₁.

Choosing the method. Force versus time: impulse-momentum. Force versus position: work-energy. Collision: momentum plus restitution, then energy to find the loss.

Formulas

∫F dt = m·v₂ − m·v₁

  • F: resultant force (N); t: time (s); m: mass (kg); v: velocity (m/s). With constant F, the impulse is F·Δt (N·s).

m₁·u₁ + m₂·u₂ = m₁·v₁ + m₂·v₂

  • u: velocities before impact, v: after (m/s), signed along the line of impact. No net external impulse.

e = (v₂ − v₁) / (u₁ − u₂)

  • e: coefficient of restitution, 0 ≤ e ≤ 1 (dimensionless).

v₁ = [(m₁ − e·m₂)·u₁ + (1 + e)·m₂·u₂] / (m₁ + m₂), v₂ = [(m₂ − e·m₁)·u₂ + (1 + e)·m₁·u₁] / (m₁ + m₂)

  • Direct central impact, from the two equations above.

ΔKE_loss = ½·[m₁·m₂ / (m₁ + m₂)]·(1 − e²)·(u₁ − u₂)²

  • Energy lost in direct central impact (J).

W = ΔKE, KE = ½·m·v² + ½·I_G·ω²

  • W: total work of all forces (J); I_G: mass moment of inertia about the centre of mass (kg·m²); ω: angular velocity (rad/s).

P = F·v = T·ω

  • P: power (W); T: torque (N·m).

Worked examples

Example 1 (standard): package placed on a moving conveyor. Given: a 4 kg package is placed at rest on a belt moving at a constant 2 m/s; μk = 0.4; g = 9.81 m/s². Find the time to reach belt speed, the slip distance, the heat generated and the energy the drive supplies.

  1. Friction on the package: F = μk·m·g = 0.4 × 4 × 9.81 = 15.70 N, forward.
  2. Impulse-momentum: F·t = m·v, so t = 4 × 2/15.70 = 0.510 s.
  3. Package travel: s_p = v²/(2μk·g) = 4/7.848 = 0.510 m. Belt travel in the same time: 2 × 0.510 = 1.019 m.
  4. Slip distance: 1.019 − 0.510 = 0.510 m.
  5. Heat: F × slip = 15.70 × 0.510 = 8.0 J. Kinetic energy gained: ½ × 4 × 2² = 8.0 J.
  6. Drive energy: F × belt travel = 15.70 × 1.019 = 16.0 J. Answer: t ≈ 0.51 s, slip ≈ 0.51 m, heat = 8.0 J, drive energy = 16.0 J. Half the drive energy is always lost as heat when a belt accelerates a load from rest by sliding friction, whatever μk is.

Example 2 (GATE level): impact with restitution. Given: a 3 kg slider moving at 4 m/s strikes a 2 kg block at rest on a smooth guide, e = 0.6. Find the velocities after impact and the energy lost.

  1. Momentum: 3 × 4 + 0 = 3·v₁ + 2·v₂, so 3·v₁ + 2·v₂ = 12.
  2. Restitution: v₂ − v₁ = 0.6 × (4 − 0) = 2.4.
  3. Solve: substitute v₂ = v₁ + 2.4 into step 1: 5·v₁ + 4.8 = 12, so v₁ = 1.44 m/s and v₂ = 3.84 m/s.
  4. KE before = ½ × 3 × 4² = 24.0 J. KE after = ½ × 3 × 1.44² + ½ × 2 × 3.84² = 3.11 + 14.75 = 17.86 J.
  5. Loss = 24.0 − 17.86 = 6.14 J. Check by formula: ½ × (3 × 2/5) × (1 − 0.36) × 4² = 6.144 J. Answer: v₁ = 1.44 m/s, v₂ = 3.84 m/s (both forward), energy lost ≈ 6.14 J. With e = 1 the results would be 0.8 m/s and 4.8 m/s with no loss; with e = 0 both would move at 2.4 m/s.

Common mistakes

  • Ignoring signs: velocities in opposite directions must have opposite signs in the momentum equation.
  • Assuming kinetic energy is conserved in an impact unless told it is perfectly elastic.
  • Writing e with approach and separation the wrong way round, which gives e > 1.
  • Using impulse = F·Δt for a varying force; integrate, or use the area under the force-time curve.
  • Leaving out rotational KE (½·I·ω²) of wheels, pulleys and rotors.
  • Treating friction work as recoverable potential energy.

For GATE ME

Expect direct central impact with a given e (final velocities, energy loss, rebound height), bullet-and-block and ballistic-pendulum problems that combine momentum during impact with energy afterwards, work-energy with springs and friction, rolling bodies with rotational KE, and power from torque and speed. Practise splitting a problem into the impact phase (momentum) and the motion phases (energy).

Quick check

  1. A constant force of 10 N acts for 4 s on a body. What is the impulse?
  2. A ball dropped from 2 m rebounds to 0.5 m. What is e?
  3. Two equal masses collide head-on with e = 1, one at rest. What happens?
  4. What is the power delivered by a shaft carrying 50 N·m at 100 rad/s?
  5. In a perfectly plastic impact, is momentum conserved? Is KE?

Answers: 1. 40 N·s. 2. e = √(0.5/2) = 0.5. 3. They exchange velocities: the moving one stops and the other moves off at the original speed. 4. 5 kW. 5. Momentum yes; KE no.

Interview questions

All Engineering Mechanics and Strength of Materials interview questions

Try answering each one aloud before you open it.

  1. 1.What is impulse in the context of mechanics?Concept

    Impulse is the time integral of a force, ∫F dt, a vector with unit N·s; for a constant force it is simply F·Δt. By Newton's second law the impulse of the resultant force on a body equals its change in momentum, m·v₂ − m·v₁. Graphically it is the area under the force-time curve, which is how it is used for impacts where the peak force is large but short.

  2. 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. 3.How is kinetic energy related to momentum?Concept

    For a particle, KE = ½·m·v² and p = m·v, so KE = p²/(2m). Momentum is a vector and is conserved in any collision with no external impulse; kinetic energy is a scalar and is conserved only in a perfectly elastic collision. That difference is why collision problems use momentum conservation plus the coefficient of restitution, and energy only afterwards to find the loss.

  4. 4.Why is the impulse-momentum theorem useful in analyzing collisions?Application

    The impulse-momentum theorem is useful because it directly relates the force applied over a time interval to the change in momentum of an object. This is particularly helpful in analyzing collisions where forces are often not constant and can be difficult to measure directly.

  5. 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 before, although momentum is conserved.

  6. 6.How are momentum and energy methods combined to solve problems involving impact?Application

    Across the impact itself, kinetic energy is generally not conserved, so you use conservation of momentum along the line of impact plus the coefficient of restitution to get the velocities just after impact. Before and after the impact, when forces are ordinary and act over a distance, you use work-energy or conservation of mechanical energy. A ballistic pendulum is the classic case: momentum for the bullet entering the block, then energy for the swing up to find the height.

  7. 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 after collision divided by the relative velocity before collision. A value of e = 1 indicates a perfectly elastic collision, while e = 0 indicates a perfectly inelastic collision. It affects how much kinetic energy is conserved in the collision.

  8. 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.

  9. 9.A 2 kg ball moving at 3 m/s collides with a stationary 3 kg ball. If the collision is perfectly elastic, what are their velocities after the collision?Numerical

    Momentum: 2 × 3 = 2·v₁ + 3·v₂. Perfectly elastic means e = 1, so v₂ − v₁ = 3. Solving: v₁ = (m₁ − m₂)·u₁/(m₁ + m₂) = (2 − 3) × 3/5 = −0.6 m/s, and v₂ = 2m₁·u₁/(m₁ + m₂) = 4 × 3/5 = 2.4 m/s. So the 2 kg ball rebounds at 0.6 m/s and the 3 kg ball moves forward at 2.4 m/s; check: KE before 9 J, after 0.36 + 8.64 = 9 J.

  10. 10.When would you use energy methods rather than momentum methods in a dynamics problem?Application

    Use work-energy when forces are known as functions of position and you want a speed after some distance, because work is a scalar and you avoid finding accelerations and times. Use impulse-momentum when forces are known as functions of time, when the duration of an event is wanted, or across an impact, where momentum is conserved but energy is not. Many problems need both: momentum through the collision, energy before and after it.

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