Momentum and Impulse — A-Level Mathematics Mechanics Study Guide

For: A-Level Mathematics candidates sitting Paper 4 (Mechanics).

Covers: Linear momentum, conservation of momentum, impulse, and direct 1D collisions and explosions as specified in the latest A-Level Mathematics Mechanics syllabus.

You should already know: A-Level Mathematics Pure 1 calculus, basic vectors.

A note on the practice questions: All worked questions in the "Practice Questions" section below are original problems written by us in the A-Level Mathematics style for educational use. They are not reproductions of past Cambridge International examination papers and may differ in wording, numerical values, or context. Use them to practise the technique; cross-check with official Cambridge mark schemes for grading conventions.

1. What Is Momentum and Impulse?

Momentum is a vector quantity that describes the total motion content of a moving object, while impulse quantifies the change in an object’s momentum caused by a force acting over a finite time interval. Together, these two concepts form the core framework for solving collision and explosion problems, which make up 5–10% of total marks on A-Level Mathematics Paper 4 exams annually. Unlike kinetic energy, momentum is conserved in all closed systems with no external forces, so it remains valid even when energy is lost to heat, sound, or deformation during collisions.

2. Linear momentum — $p=mv$

Linear momentum (denoted $p$) is defined as the product of an object’s mass and its instantaneous velocity. The formula is: $$p=mv$$ Where:

• $p$ = linear momentum, units ${\text{kg m s}}^{-1}$ (equivalent to newton-seconds, $\text{N s}$)
• $m$ = mass of the object, in kilograms ($\text{kg}$)
• $v$ = velocity of the object, in ${\text{m s}}^{-1}$ (a vector, so momentum shares the same direction as velocity)

For 1D problems, you will simplify vector calculations by assigning a positive direction (e.g. right = +, left = -) at the start of your working, then treating all velocities as signed scalars. Examiners regularly penalize missing signs or unit errors, so this step is non-negotiable.

Worked Example

A 2.5 kg football is kicked left at 8 ${\text{m s}}^{-1}$. Take right as the positive direction. Calculate the momentum of the football.

1. Identify values: $m=2.5\text{kg}$, $v=-8{\text{m s}}^{-1}$ (negative because motion is left, opposite to positive direction)
2. Substitute into formula: $p=2.5\times (-8)=-20{\text{kg m s}}^{-1}$
3. Interpret result: The momentum is 20 ${\text{kg m s}}^{-1}$ directed to the left, as expected.

3. Conservation of momentum in collisions

The principle of conservation of momentum is derived from Newton’s third law: when two objects collide, the force each exerts on the other is equal in magnitude, opposite in direction, and acts for the same time interval, so the total change in momentum of the system is zero.

The principle states: For a closed system (no external forces such as friction, air resistance, or applied forces act on the system), the total linear momentum before a collision equals the total linear momentum after the collision.

For two colliding objects, the formula is: $$m_1{u}_1+m_2{u}_2=m_1{v}_1+m_2{v}_2$$ Where:

• $u_1,u_2$ = initial velocities of objects 1 and 2 before collision
• $v_1,v_2$ = final velocities of objects 1 and 2 after collision

This rule applies to all collision types, both elastic (kinetic energy conserved) and inelastic (kinetic energy lost), so it is the first formula you should use for any collision problem.

Worked Example

A 4 kg trolley moving right at 5 ${\text{m s}}^{-1}$ collides with a stationary 1 kg trolley. After collision, the 1 kg trolley moves right at 8 ${\text{m s}}^{-1}$. Take right as positive. Find the velocity of the 4 kg trolley after collision.

1. Total initial momentum: $(4\times 5)+(1\times 0)=20{\text{kg m s}}^{-1}$
2. Total final momentum: $4v_1+(1\times 8)=4v_1+8$
3. Equate initial and final momentum: $20=4v_1+8$
4. Solve: $4v_1=12⟹v_1=3{\text{m s}}^{-1}$, so the 4 kg trolley moves right at 3 ${\text{m s}}^{-1}$.

4. Impulse — $\Delta p=F\Delta t$

Impulse (denoted $I$) is defined as the change in an object’s momentum, or equivalently, the product of the average force applied to an object and the time interval the force acts for. It is derived directly from Newton’s second law: $F=ma=m\frac{\Delta v}{\Delta t}$, rearranged to $F\Delta t=m\Delta v=\Delta p$.

The impulse formulas are: $$I=\Delta p=F\Delta t={\int }_{t_1}^{t_2}F(t)dt$$ Where:

• $I$ = impulse, units $\text{N s}$ (equivalent to ${\text{kg m s}}^{-1}$)
• $F$ = average constant force, in newtons ($\text{N}$)
• $\Delta t$ = time interval of force application, in seconds ($\text{s}$)
• The integral form is used for non-constant force, and equals the area under a force-time graph (uses your Pure 1 calculus skills for polynomial force functions).

Impulse is a vector, so its direction matches the direction of the applied force, and you must use sign conventions for 1D problems.

Worked Example

A 0.15 kg cricket ball hits a bat moving right at 30 ${\text{m s}}^{-1}$, then bounces back left at 25 ${\text{m s}}^{-1}$. The collision lasts 0.003 s. Take right as positive. Calculate (a) impulse exerted on the ball by the bat, (b) average force exerted by the bat on the ball.

1. (a) Calculate change in momentum: $\Delta p=m(v-u)=0.15\times (-25-30)=0.15\times (-55)=-8.25\text{N s}$. The negative sign means impulse is directed left.
2. (b) Rearrange impulse formula for force: $F=\frac{I}{\Delta t}=\frac{-8.25}{0.003}=-2750\text{N}$, so the average force is 2750 N directed left.

5. Direct (1D) collisions and explosions

A direct (1D) collision occurs when two objects move along the same straight line before and after impact, so all velocity vectors lie on that line, simplifying calculations to signed scalars. There are two core collision types:

• Perfectly elastic collision: Total kinetic energy is conserved (only applies to ideal rigid objects, e.g. snooker balls, and will be explicitly stated in exam questions)
• Inelastic collision: Total kinetic energy is lost to heat, sound, or deformation; completely inelastic collisions are a special case where objects stick together and move with a common final velocity.

Explosions are the reverse of collisions: a single stationary or moving object breaks into two or more fragments. For stationary objects, total initial momentum is zero, so total final momentum of all fragments must also sum to zero, meaning fragments move in opposite directions to cancel out momentum.

Worked Example 1 (Completely Inelastic Collision)

A 3 kg block moving right at 7 ${\text{m s}}^{-1}$ collides with a 2 kg block moving left at 3 ${\text{m s}}^{-1}$. The blocks stick together after collision. Take right as positive. Find their common final velocity, and the kinetic energy lost in the collision.

1. Total initial momentum: $(3\times 7)+(2\times -3)=21-6=15{\text{kg m s}}^{-1}$
2. Total final momentum: $(3+2)v=5v$
3. Equate: $5v=15⟹v=3{\text{m s}}^{-1}$ (directed right)
4. Initial KE: $\frac{1}{2}\times 3\times 7^2+\frac{1}{2}\times 2\times (-3{)}^2=73.5+9=82.5\text{J}$
5. Final KE: $\frac{1}{2}\times 5\times 3^2=22.5\text{J}$
6. KE lost: $82.5-22.5=60\text{J}$

Worked Example 2 (Explosion)

A stationary 6 kg shell explodes into two fragments: a 2 kg fragment moving left at 12 ${\text{m s}}^{-1}$, and a 4 kg fragment. Find the velocity of the 4 kg fragment.

1. Total initial momentum = 0
2. Total final momentum: $(2\times -12)+4v=-24+4v$
3. Equate to 0: $4v=24⟹v=6{\text{m s}}^{-1}$ (directed right)

6. Common Pitfalls (and how to avoid them)

• Wrong move: Forgetting to assign a positive direction and using speed instead of signed velocity in calculations. Why students do it: They confuse scalar speed with vector velocity. Correct move: Write "Take [direction] as positive" at the start of every working, and use negative values for velocities in the opposite direction.
• Wrong move: Applying conservation of momentum when external forces are active over a long time interval. Why students do it: They assume momentum is always conserved regardless of context. Correct move: Only use conservation of momentum for the short time interval of the collision/explosion, when internal collision forces are far larger than external forces like gravity or friction.
• Wrong move: Calculating impulse as $m(v+u)$ instead of $m(v-u)$. Why students do it: They mix up change in momentum with total momentum. **Correct move: Always write $\Delta p=p_{final}-p_{initial}$ explicitly before substituting values.
• Wrong move: Assuming kinetic energy is conserved in all collisions. Why students do it: They mix up conservation of momentum and conservation of energy. Correct move: Only use KE conservation if the question explicitly states the collision is perfectly elastic.
• Wrong move: Using mass in grams instead of kilograms. Why students do it: Examiners often give masses in grams to test unit conversion. Correct move: Convert all masses to kg before calculation, as the SI unit for momentum is ${\text{kg m s}}^{-1}$.

7. Practice Questions (Paper 4 Style)

Question 1

A 0.08 kg golf ball is struck by a club. The ball approaches the club horizontally at 10 ${\text{m s}}^{-1}$, and leaves the club in the opposite direction at 50 ${\text{m s}}^{-1}$. The club is in contact with the ball for 0.002 s. (a) Calculate the magnitude of the impulse exerted on the ball by the club. [2 marks] (b) Find the average force exerted by the club on the ball. [2 marks]

Solution

(a) Take direction of ball leaving the club as positive: $u=-10{\text{m s}}^{-1}$, $v=50{\text{m s}}^{-1}$, $m=0.08\text{kg}$ $I=\Delta p=m(v-u)=0.08\times (50-(-10))=0.08\times 60=4.8\text{N s}$ Magnitude = $4.8\text{N s}$ (b) $I=F\Delta t⟹F=\frac{I}{\Delta t}=\frac{4.8}{0.002}=2400\text{N}$

Question 2

Two vans are moving along a straight road. Van X, mass 1500 kg, is moving east at 12 ${\text{m s}}^{-1}$. Van Y, mass 1200 kg, is moving west at 18 ${\text{m s}}^{-1}$. The vans collide head-on, and after collision, Van Y moves east at 2 ${\text{m s}}^{-1}$. Assume no external forces act during the collision. (a) Find the velocity of Van X after the collision, stating its direction. [3 marks] (b) Determine if the collision is elastic, showing your working. [3 marks]

Solution

(a) Take east as positive: Total initial momentum = $(1500\times 12)+(1200\times -18)=18000-21600=-3600{\text{kg m s}}^{-1}$ Total final momentum = $1500v_X+(1200\times 2)=1500v_X+2400$ Equate: $-3600=1500v_X+2400⟹1500v_X=-6000⟹v_X=-4{\text{m s}}^{-1}$ Van X moves west at 4 ${\text{m s}}^{-1}$. (b) Initial KE = $\frac{1}{2}\times 1500\times 12^2+\frac{1}{2}\times 1200\times (-18{)}^2=108000+194400=302400\text{J}$ Final KE = $\frac{1}{2}\times 1500\times (-4{)}^2+\frac{1}{2}\times 1200\times 2^2=12000+2400=14400\text{J}$ Final KE < Initial KE, so the collision is inelastic.

Question 3

A stationary 2 kg firework explodes into three fragments. Fragment 1 has mass 0.7 kg and moves horizontally north at 30 ${\text{m s}}^{-1}$. Fragment 2 has mass 0.8 kg and moves horizontally south at 20 ${\text{m s}}^{-1}$. Fragment 3 has mass 0.5 kg. (a) Calculate the velocity of Fragment 3 immediately after the explosion, stating its direction. [4 marks] (b) State why the velocity you calculated is only valid immediately after the explosion. [1 mark]

Solution

(a) Take north as positive: Total initial momentum = 0 Total final momentum = $(0.7\times 30)+(0.8\times -20)+0.5v_3=21-16+0.5v_3=5+0.5v_3$ Equate to 0: $0.5v_3=-5⟹v_3=-10{\text{m s}}^{-1}$ Fragment 3 moves south at 10 ${\text{m s}}^{-1}$. (b) External forces (gravity and air resistance) act on the fragment after the explosion, so momentum is no longer conserved, and the velocity changes immediately.

8. Quick Reference Cheatsheet

Exam tip: Always show units and sign conventions in working to earn full method marks.

9. What's Next

This topic is foundational for both standard Paper 4 problems and further mechanics content if you are studying A-Level Further Mathematics. On A-Level Mathematics Paper 4, momentum and impulse concepts are often combined with kinematics, friction, and Newton’s laws to create multi-step questions worth 6–8 marks, so mastering these rules will help you score highly on extended response items. For further study, you will encounter 2D oblique collisions, coefficient of restitution, and centre of mass calculations that build directly on the momentum principles covered here.

If you are struggling with sign conventions, calculation steps, or any of the worked examples in this guide, you can ask Ollie, our AI tutor, for personalized explanations and extra practice questions tailored to your weak areas. You can also find more Paper 4 study resources, full past paper walkthroughs, and topic quizzes on the homepage to prepare for your exam.

Aligned with the Cambridge International AS & A Level Mathematics 9709 syllabus. OwlsAi is not affiliated with Cambridge Assessment International Education.