Impact

Dynamics · FE Reference Handbook section

Dynamics

15 formulas

10 exam-style examples

~60 min

All Dynamics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Impact within Dynamics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what impact describes physically and when it applies.
State every one of the 15 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: g = 32.2 ft/s² or 9.81 m/s²; never mix mass and weight.

Lecture

Why this section exists. Impact is the part of Dynamics that lets you connect a particle or rigid body moving under known forces to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as kinematics first, then a work-energy or impulse-momentum shortcut. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. g = 32.2 ft/s² or 9.81 m/s²; never mix mass and weight. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Crane lowering a steel plate girder onto bridge bearings while ironworkers guide it. — Photo 1. Where this shows up in practice: impact.
Capstone Studio instructional photograph

Dynamics — Impact: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes a particle or rigid body moving under known forces. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 15 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

m1v1 + m2v2	Quantity produced by "m1v1 + m2v2 = m1vl1 + m2vl2" — read its definition and unit from the handbook line directly above the equation.
m1, m2	Quantity produced by "m1, m2 = masses of the two bodies" — read its definition and unit from the handbook line directly above the equation.
v1, v2	Quantity produced by "v1, v2 = velocities of the bodies just before impact" — read its definition and unit from the handbook line directly above the equation.
v l1, v l2	Quantity produced by "v l1, v l2 = velocities of the bodies just after impact" — read its definition and unit from the handbook line directly above the equation.
e	Quantity produced by "e = coefficient of restitution" — read its definition and unit from the handbook line directly above the equation.
(vi)n	Quantity produced by "(vi)n = velocity normal to the plane of impact just before impact" — read its definition and unit from the handbook line directly above the equation.
_ v li in	Quantity produced by "_ v li in = velocity normal to the plane of impact just after impact" — read its definition and unit from the handbook line directly above the equation.
^v l1hn	Quantity produced by "^v l1hn = m1 + m2" — read its definition and unit from the handbook line directly above the equation.
^v l2hn	Quantity produced by "^v l2hn = m1 + m2" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

During an impact, momentum is conserved while energy may or may not be conserved. For direct central impact with no
external forces
where
For impacts, the relative velocity expression is
_vl2 in − _vl1 in
_v1 in − _v2 in
where
The value of e is such that
Knowing the value of e, the velocities after the impact are given as

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Perfectly plastic impact — Impact

A 7 kg mass moving at 12 m/s strikes a stationary 15 kg mass and they move together. Find the common velocity and the energy lost.

Given

m₁ = 7 kg, v₁ = 12 m/s
m₂ = 15 kg, v₂ = 0

Find

v' and ΔKE

Start with the thinking

Momentum is conserved in every impact; energy is not.
Plastic impact means one common final velocity.

Step-by-step solution

Momentum
$m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v^{'}$
Substituting
$7 (12) + 15 (0) = (22) v^{'}$
Solve
$v^{'} = 84/22 = 3.818 m / s$
Initial KE
$½ (7) (12)^{2} = 504.0 J$
Final KE
$½ (22) (3.818)^{2} = 160.4 J$
Energy lost — ΔKE = 343.6 J

Answer: v′ ≈ 3.82 m/s; ΔKE ≈ 343.6 J

Why the other options are there

6.00 m/s (masses assumed equal)
ΔKE = 0 (energy assumed conserved)

Reference: FE Reference Handbook — Dynamics → Impact

Example 2

Perfectly plastic impact — Impact (2)

A 11 kg mass moving at 24 m/s strikes a stationary 7 kg mass and they move together. Find the common velocity and the energy lost.

Given

m₁ = 11 kg, v₁ = 24 m/s
m₂ = 7 kg, v₂ = 0

Find

v' and ΔKE

Start with the thinking

Momentum is conserved in every impact; energy is not.
Plastic impact means one common final velocity.

Step-by-step solution

Momentum
$m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v^{'}$
Substituting
$11 (24) + 7 (0) = (18) v^{'}$
Solve
$v^{'} = 264/18 = 14.667 m / s$
Initial KE
$½ (11) (24)^{2} = 3, 168 J$
Final KE
$½ (18) (14.667)^{2} = 1, 936 J$
Energy lost — ΔKE = 1,232 J

Answer: v′ ≈ 14.67 m/s; ΔKE ≈ 1,232 J

Why the other options are there

12.00 m/s (masses assumed equal)
ΔKE = 0 (energy assumed conserved)

Reference: FE Reference Handbook — Dynamics → Impact

Example 3

Perfectly plastic impact — Impact (3)

A 6 kg mass moving at 13 m/s strikes a stationary 14 kg mass and they move together. Find the common velocity and the energy lost.

Given

m₁ = 6 kg, v₁ = 13 m/s
m₂ = 14 kg, v₂ = 0

Find

v' and ΔKE

Start with the thinking

Momentum is conserved in every impact; energy is not.
Plastic impact means one common final velocity.

Step-by-step solution

Momentum
$m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v^{'}$
Substituting
$6 (13) + 14 (0) = (20) v^{'}$
Solve
$v^{'} = 78/20 = 3.900 m / s$
Initial KE
$½ (6) (13)^{2} = 507.0 J$
Final KE
$½ (20) (3.900)^{2} = 152.1 J$
Energy lost — ΔKE = 354.9 J

Answer: v′ ≈ 3.90 m/s; ΔKE ≈ 354.9 J

Why the other options are there

6.50 m/s (masses assumed equal)
ΔKE = 0 (energy assumed conserved)

Reference: FE Reference Handbook — Dynamics → Impact

Example 4

Perfectly plastic impact — Impact (4)

A 2 kg mass moving at 17 m/s strikes a stationary 10 kg mass and they move together. Find the common velocity and the energy lost.

Given

m₁ = 2 kg, v₁ = 17 m/s
m₂ = 10 kg, v₂ = 0

Find

v' and ΔKE

Start with the thinking

Momentum is conserved in every impact; energy is not.
Plastic impact means one common final velocity.

Step-by-step solution

Momentum
$m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v^{'}$
Substituting
$2 (17) + 10 (0) = (12) v^{'}$
Solve
$v^{'} = 34/12 = 2.833 m / s$
Initial KE
$½ (2) (17)^{2} = 289.0 J$
Final KE
$½ (12) (2.833)^{2} = 48.2 J$
Energy lost — ΔKE = 240.8 J

Answer: v′ ≈ 2.83 m/s; ΔKE ≈ 240.8 J

Why the other options are there

8.50 m/s (masses assumed equal)
ΔKE = 0 (energy assumed conserved)

Reference: FE Reference Handbook — Dynamics → Impact

Example 5

Perfectly plastic impact — Impact (5)

A 11 kg mass moving at 9 m/s strikes a stationary 4 kg mass and they move together. Find the common velocity and the energy lost.

Given

m₁ = 11 kg, v₁ = 9 m/s
m₂ = 4 kg, v₂ = 0

Find

v' and ΔKE

Start with the thinking

Momentum is conserved in every impact; energy is not.
Plastic impact means one common final velocity.

Step-by-step solution

Momentum
$m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v^{'}$
Substituting
$11 (9) + 4 (0) = (15) v^{'}$
Solve
$v^{'} = 99/15 = 6.600 m / s$
Initial KE
$½ (11) (9)^{2} = 445.5 J$
Final KE
$½ (15) (6.600)^{2} = 326.7 J$
Energy lost — ΔKE = 118.8 J

Answer: v′ ≈ 6.60 m/s; ΔKE ≈ 118.8 J

Why the other options are there

4.50 m/s (masses assumed equal)
ΔKE = 0 (energy assumed conserved)

Reference: FE Reference Handbook — Dynamics → Impact

Example 6

Perfectly plastic impact — Impact (6)

A 4 kg mass moving at 11 m/s strikes a stationary 13 kg mass and they move together. Find the common velocity and the energy lost.

Given

m₁ = 4 kg, v₁ = 11 m/s
m₂ = 13 kg, v₂ = 0

Find

v' and ΔKE

Start with the thinking

Momentum is conserved in every impact; energy is not.
Plastic impact means one common final velocity.

Step-by-step solution

Momentum
$m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v^{'}$
Substituting
$4 (11) + 13 (0) = (17) v^{'}$
Solve
$v^{'} = 44/17 = 2.588 m / s$
Initial KE
$½ (4) (11)^{2} = 242.0 J$
Final KE
$½ (17) (2.588)^{2} = 56.9 J$
Energy lost — ΔKE = 185.1 J

Answer: v′ ≈ 2.59 m/s; ΔKE ≈ 185.1 J

Why the other options are there

5.50 m/s (masses assumed equal)
ΔKE = 0 (energy assumed conserved)

Reference: FE Reference Handbook — Dynamics → Impact

Example 7

Perfectly plastic impact — Impact (7)

A 11 kg mass moving at 23 m/s strikes a stationary 12 kg mass and they move together. Find the common velocity and the energy lost.

Given

m₁ = 11 kg, v₁ = 23 m/s
m₂ = 12 kg, v₂ = 0

Find

v' and ΔKE

Start with the thinking

Momentum is conserved in every impact; energy is not.
Plastic impact means one common final velocity.

Step-by-step solution

Momentum
$m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v^{'}$
Substituting
$11 (23) + 12 (0) = (23) v^{'}$
Solve
$v^{'} = 253/23 = 11.000 m / s$
Initial KE
$½ (11) (23)^{2} = 2, 910 J$
Final KE
$½ (23) (11.000)^{2} = 1, 392 J$
Energy lost — ΔKE = 1,518 J

Answer: v′ ≈ 11.00 m/s; ΔKE ≈ 1,518 J

Why the other options are there

11.50 m/s (masses assumed equal)
ΔKE = 0 (energy assumed conserved)

Reference: FE Reference Handbook — Dynamics → Impact

Example 8

Perfectly plastic impact — Impact (8)

A 4 kg mass moving at 25 m/s strikes a stationary 4 kg mass and they move together. Find the common velocity and the energy lost.

Given

m₁ = 4 kg, v₁ = 25 m/s
m₂ = 4 kg, v₂ = 0

Find

v' and ΔKE

Start with the thinking

Momentum is conserved in every impact; energy is not.
Plastic impact means one common final velocity.

Step-by-step solution

Momentum
$m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v^{'}$
Substituting
$4 (25) + 4 (0) = (8) v^{'}$
Solve
$v^{'} = 100/8 = 12.500 m / s$
Initial KE
$½ (4) (25)^{2} = 1, 250 J$
Final KE
$½ (8) (12.500)^{2} = 625.0 J$
Energy lost — ΔKE = 625.0 J

Answer: v′ ≈ 12.50 m/s; ΔKE ≈ 625.0 J

Why the other options are there

12.50 m/s (masses assumed equal)
ΔKE = 0 (energy assumed conserved)

Reference: FE Reference Handbook — Dynamics → Impact

Example 9

Perfectly plastic impact — Impact (9)

A 11 kg mass moving at 10 m/s strikes a stationary 11 kg mass and they move together. Find the common velocity and the energy lost.

Given

m₁ = 11 kg, v₁ = 10 m/s
m₂ = 11 kg, v₂ = 0

Find

v' and ΔKE

Start with the thinking

Momentum is conserved in every impact; energy is not.
Plastic impact means one common final velocity.

Step-by-step solution

Momentum
$m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v^{'}$
Substituting
$11 (10) + 11 (0) = (22) v^{'}$
Solve
$v^{'} = 110/22 = 5.000 m / s$
Initial KE
$½ (11) (10)^{2} = 550.0 J$
Final KE
$½ (22) (5.000)^{2} = 275.0 J$
Energy lost — ΔKE = 275.0 J

Answer: v′ ≈ 5.00 m/s; ΔKE ≈ 275.0 J

Why the other options are there

5.00 m/s (masses assumed equal)
ΔKE = 0 (energy assumed conserved)

Reference: FE Reference Handbook — Dynamics → Impact

Example 10

Perfectly plastic impact — Impact (10)

A 9 kg mass moving at 11 m/s strikes a stationary 4 kg mass and they move together. Find the common velocity and the energy lost.

Given

m₁ = 9 kg, v₁ = 11 m/s
m₂ = 4 kg, v₂ = 0

Find

v' and ΔKE

Start with the thinking

Momentum is conserved in every impact; energy is not.
Plastic impact means one common final velocity.

Step-by-step solution

Momentum
$m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) v^{'}$
Substituting
$9 (11) + 4 (0) = (13) v^{'}$
Solve
$v^{'} = 99/13 = 7.615 m / s$
Initial KE
$½ (9) (11)^{2} = 544.5 J$
Final KE
$½ (13) (7.615)^{2} = 377.0 J$
Energy lost — ΔKE = 167.5 J

Answer: v′ ≈ 7.62 m/s; ΔKE ≈ 167.5 J

Why the other options are there

5.50 m/s (masses assumed equal)
ΔKE = 0 (energy assumed conserved)

Reference: FE Reference Handbook — Dynamics → Impact

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given a particle or rigid body moving under known forces, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Impact contains 15 relations; you must be able to find this page in under 15 seconds.
Exam style: kinematics first, then a work-energy or impulse-momentum shortcut.
Unit rule: g = 32.2 ft/s² or 9.81 m/s²; never mix mass and weight.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

g = 32.2 ft/s² or 9.81 m/s²; never mix mass and weight
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.

Impact

Learning objectives

Lecture

Theory, developed

Notation used in this section

Handbook notes for this section

Core formulas for this FE topic

m1​v1​+m2​v2​=m1​vl1​+m2​vl2​

m1​,m2​=massesofthetwobodies

v1​,v2​=velocitiesofthebodiesjustbeforeimpact

vl1​,vl2​=velocitiesofthebodiesjustafterimpact

e=Needs technical review

e=coefficientofrestitution

(vi)n=velocitynormaltotheplaneofimpactjustbeforeimpact

v​liin=velocitynormaltotheplaneofimpactjustafterimpact

0≤e≤1,withlimitingvalues

e=1,perfectlyelastic(energyconserved)

e=0,perfectlyplastic(norebound)

m2 ^v2hn ^1 + eh + ^m1 - em2h ^v1hnNeeds technical review

vl1​hn=m1​+m2​

m1 ^v1hn ^1 + eh - ^em1 - m2h ^v2hnNeeds technical review

vl2​hn=m1​+m2​

Worked exam-style examples

Self-check

Chapter summary

Common traps in this section