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Qwestrum Engineering360 · Mechanical Engineering · Engineering Mechanics

Impulse and Momentum

Linear impulse equals change of momentum: J = ∫F dt = Δ(mv). In collisions momentum is conserved and the coefficient of restitution e = (v₂ − v₁)/(u₁ − u₂) relates approach and separation speeds, as in RK Bansal.
Exam tip: draw a complete FBD first; resolve forces along chosen axes and check moment equilibrium about a convenient point.

Key formulas & points

Skim these first — then read the full notes below.

  • Conservationofmomentum:Σpbefore=ΣpafterConservation of momentum: Σp_{before} = Σp_{after} (no external impulse)
  • Coefficientofrestitutione=(v2−v1)(u1−u2)forimpactCoefficient of restitution e = \frac{(v_{2} - v_{1})}{(u_{1} - u_{2})} for impact
  • Eccentric impact causes impulse at contact plus rotation

Topic details

Introduction

Impulse-momentum is the time-integral counterpart of work-energy and appears in Indian dynamics papers as collision, jet-impact, and variable-mass (rocket) problems. It is preferred when forces act over short times and are hard to detail.

Scope in B.Tech and GATE syllabus

RK Bansal frames the linear form J = FΔt = Δp and the angular form M·Δt = Δ(Iω). Conservation of momentum applies whenever the net external impulse is zero — the key to solving impacts without knowing the internal contact force.

Why this topic matters in practice

The coefficient of restitution characterises the collision: e = 1 is perfectly elastic (kinetic energy conserved), e = 0 is perfectly plastic (bodies stick). Combining momentum conservation with the restitution equation gives two equations for the two unknown final velocities — the standard exam method.

Key relations & formulas

p=m⋅vp = m\cdot v
(linear momentum)
J=∫Fdt=ΔpJ = \int F dt = \Delta p
(impulse-momentum theorem)
H=I⋅ωH = I\cdot \omega
(angular momentum)
ΣM=I⋅αΣM = I\cdot \alpha
(rotational Newton II)

Notation and sign conventions

Relation 1 —
p=m⋅vp = m\cdot v
p=m⋅vp = m\cdot v
(linear momentum)
Write this relation with symbols exactly as in Strength of Materials — RK Bansal before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 2 —
J=∫Fdt=ΔpJ = \int F dt = \Delta p
J=∫Fdt=ΔpJ = \int F dt = \Delta p
(impulse-momentum theorem)
Write this relation with symbols exactly as in Strength of Materials — RK Bansal before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 3 —
H=I⋅ωH = I\cdot \omega
H=I⋅ωH = I\cdot \omega
(angular momentum)
Write this relation with symbols exactly as in Strength of Materials — RK Bansal before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.
Relation 4 —
ΣM=I⋅αΣM = I\cdot \alpha
ΣM=I⋅αΣM = I\cdot \alpha
(rotational Newton II)
Write this relation with symbols exactly as in Strength of Materials — RK Bansal before substituting numbers. Examiners award partial marks for a correct setup even when arithmetic slips.

Fundamentals and definitions

Newton's second law integrated over time gives the impulse-momentum theorem: ∫F dt = Δ(mv). A large force over a short time (a bat hitting a ball) produces a finite impulse and a measurable velocity change.

Governing relations in practice

When no external impulse acts on a system, total momentum is conserved: Σm·u = Σm·v. For a two-body collision this single vector equation is combined with the restitution relation e = (separation velocity)/(approach velocity) to find both final velocities.

Design and analysis considerations

Perfectly elastic impact (e = 1) also conserves kinetic energy; perfectly plastic impact (e = 0) has the bodies move together, dissipating maximum energy. Real impacts lie between, so energy is generally not conserved even though momentum is.

Advanced theory and extensions

The angular analogue is impulse of moment equal to change of angular momentum, M·Δt = Δ(Iω), used for suddenly applied torques and eccentric impacts where a body both translates and rotates. Choosing momentum over energy methods is the strategic insight for these problems.

Assumptions and validity limits

State assumptions explicitly before using any relation for impulse and momentum — steady state, uniform properties, linear elastic material, ideal gas, incompressible flow, etc., as applicable.
Wrong assumptions invalidate the entire solution even when the formula is correct. In Engineering Mechanics viva and GATE descriptive questions, listing valid assumptions often earns separate marks.

Step-by-step problem approach

1. Read the question and list given data with SI units (common in Engineering Mechanics papers).
2. Draw a neat labelled diagram where applicable — examiners in Indian universities award diagram marks even when arithmetic slips.
3. Identify which relation from this topic applies to impulse and momentum.
4. Use equation 1:
p=m⋅vp = m\cdot v
.
5. Use equation 2:
J=∫Fdt=ΔpJ = \int F dt = \Delta p
.
6. Substitute values, compute, and verify units and sign (direction).
7. State conclusion in one line — e.g. safe/unsafe, stable/unstable, feasible/infeasible.

Applications & exam relevance

Impulse and Momentum appears in trusses, frames, and friction problems. In Indian mechanical curricula this topic is tested because it connects theory to force equilibrium and motion of rigid bodies.
GATE and semester exams often combine impulse and momentum with earlier units — revise prerequisites before attempting mixed problems.
Industry interview panels sometimes ask: "Where did you use impulse and momentum?" — answer with a lab, mini-project, or plant visit example if possible.

Common mistakes in exams

• Assuming kinetic energy is conserved in an inelastic (e < 1) collision
• Getting the sign convention wrong in e = (v₂ − v₁)/(u₁ − u₂)
• Applying momentum conservation when a significant external impulse (e.g. friction over time) acts
• Forgetting that impulse and momentum are vectors — treating components incorrectly

Quick revision checklist

Before attempting impulse and momentum problems, confirm you can:
1.
Conservationofmomentum:Σpbefore=ΣpafterConservation of momentum: Σp_{before} = Σp_{after}
(no external impulse)
2.
Coefficientofrestitutione=(v2−v1)(u1−u2)forimpactCoefficient of restitution e = \frac{(v_{2} - v_{1})}{(u_{1} - u_{2})} for impact

3. Eccentric impact causes impulse at contact plus rotation
Revise the solved examples in Strength of Materials — RK Bansal and one previous-year GATE or university paper for this unit.

Worked examples

Try the problem first — open the solution when you are ready to check.

Impulse on a mass

Problem

A constant force F = 50 N acts on a 5 kg mass, initially at rest, for t = 4 s. Find the final velocity using impulse-momentum.

Solution

J = F·t = 50 × 4 = 200 N·s = Δ(mv) = m·v; v = 200/5 = 40 m/s.

Conceptual check — Impulse and Momentum

Problem

In a Engineering Mechanics semester or GATE paper you are asked: "State the main assumption, the governing relation, and one practical consequence of impulse and momentum." What should a complete answer include?

Practice questions

Most-asked interview and GATE questions for this topic — expand any item for a model answer.

  1. 1
    What is Impulse and Momentum, and why does it appear in B.Tech / GATE syllabi?

    Model answer

    Linear impulse equals change of momentum: J = ∫F dt = Δ(mv). In collisions momentum is conserved and the coefficient of restitution e = (v₂ − v₁)/(u₁ − u₂) relates approach and separation speeds, as in RK Bansal.
  2. 2
    State the relation p = m·v and name each symbol.

    Model answer

    The governing relation is p=m⋅vp = m\cdot v. Write every symbol with SI units before substituting numbers.
  3. 3
    State the relation J = ∫F dt = Δp and name each symbol.

    Model answer

    The governing relation is J=∫Fdt=ΔpJ = \int F dt = \Delta p. Write every symbol with SI units before substituting numbers.
  4. 4
    State the relation H = I·ω and name each symbol.

    Model answer

    The governing relation is H=I⋅ωH = I\cdot \omega. Write every symbol with SI units before substituting numbers.
  5. 5
    State the relation ΣM = I·α and name each symbol.

    Model answer

    The governing relation is ΣM=I⋅αΣM = I\cdot \alpha. Write every symbol with SI units before substituting numbers.
  6. 6
    Explain: Conservation of momentum: Σp_before = Σp_after (no external impulse)

    Model answer

    Conservationofmomentum:Σpbefore=ΣpafterConservation of momentum: Σp_{before} = Σp_{after} (no external impulse) — state the assumption range and one exam trap linked to this point.
  7. 7
    Explain: Coefficient of restitution e = (v₂ − v₁)/(u₁ − u₂) for impact

    Model answer

    Coefficientofrestitutione=(v2−v1)(u1−u2)forimpactCoefficient of restitution e = \frac{(v_{2} - v_{1})}{(u_{1} - u_{2})} for impact — state the assumption range and one exam trap linked to this point.
  8. 8
    Explain: Eccentric impact causes impulse at contact plus rotation

    Model answer

    Eccentric impact causes impulse at contact plus rotation — state the assumption range and one exam trap linked to this point.
  9. 9
    How would you correct this error in a viva: Assuming kinetic energy is conserved in an inelastic (e < 1) collision?

    Model answer

    Identify the wrong assumption or unit mix-up, rewrite the correct relation, and recompute with a one-line sanity check.
  10. 10
    How would you correct this error in a viva: Getting the sign convention wrong in e = (v₂ − v₁)/(u₁ − u₂)?

    Model answer

    Identify the wrong assumption or unit mix-up, rewrite the correct relation, and recompute with a one-line sanity check.
  11. 11
    How would you correct this error in a viva: Applying momentum conservation when a significant external impulse (e.g. friction over time) acts?

    Model answer

    Identify the wrong assumption or unit mix-up, rewrite the correct relation, and recompute with a one-line sanity check.
  12. 12
    How would you correct this error in a viva: Forgetting that impulse and momentum are vectors — treating components incorrectly?

    Model answer

    Identify the wrong assumption or unit mix-up, rewrite the correct relation, and recompute with a one-line sanity check.

Exams & GATE

  • 1
    Draw momentum diagrams for impact problems — RK Bansal Ch. 8.
  • 2
    Avoid: Assuming kinetic energy is conserved in an inelastic (e < 1) collision
  • 3
    Avoid: Getting the sign convention wrong in e = (v₂ − v₁)/(u₁ − u₂)
  • 4
    Avoid: Applying momentum conservation when a significant external impulse (e.g. friction over time) acts

📖 Standard books (India)

  • Strength of Materials — RK Bansal

    Read: Syllabus unit

    SOM — beams, torsion, columns, and deflection

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