Momentum

Momentum

Info: Board Coverage AQA Paper 1 | Edexcel CP2 | OCR (A) Paper 1 | CIE P2

Explore the simulation above to develop intuition for this topic.

Intuition

Momentum is “how hard it is to stop something”: A bowling ball moving at 5 m/s is harder to stop than a tennis ball at 5 m/s because it has more momentum. Momentum combines mass and velocity — it’s the product of both. This is why a slow truck can be more dangerous than a fast car — it has more momentum and needs more force to stop.

Why it matters: Momentum conservation explains why cars have crumple zones (extending collision time reduces force), why guns recoil (momentum conservation), and why astronauts float in space (no external forces). It’s the most useful conservation law for solving collision problems.

The key insight: Momentum is always conserved in collisions, but kinetic energy is only conserved in elastic collisions. In inelastic collisions (like car crashes), kinetic energy is converted to heat, sound, and deformation. This is why crumple zones work — they convert kinetic energy to deformation energy, reducing the force on passengers.

1. Linear Momentum

Definition. The linear momentum of a body of mass $m$ moving with velocity $\mathbf{v}$ is:

$\boxed{\mathbf{p} = m\mathbf{v}}$

Momentum is a vector quantity with SI units kg m s$^{-1}$.

2. Conservation of Momentum

Principle of Conservation of Linear Momentum. In the absence of an external net force, the total Momentum of a system is conserved:

$\sum \mathbf{p}_{\mathrm{initial}} = \sum \mathbf{p}_{\mathrm{final}}$

Proof from Newton’s Laws

Consider two bodies $A$ and $B$ that interact with each other but with no external forces.

By Newton’s third law: $\mathbf{F}_{AB} = -\mathbf{F}_{BA}$.

By Newton’s second law: $\mathbf{F}_{AB} = \frac{d\mathbf{p}_A}{dt}$ and $\mathbf{F}_{BA} = \frac{d\mathbf{p}_B}{dt}$.

Therefore:

$\frac{d\mathbf{p}_A}{dt} = -\frac{d\mathbf{p}_B}{dt} \implies \frac{d\mathbf{p}_A}{dt} + \frac{d\mathbf{p}_B}{dt} = 0 \implies \frac{d}{dt}(\mathbf{p}_A + \mathbf{p}_B) = 0$

Hence $\mathbf{p}_A + \mathbf{p}_B = \mathrm{constant}$. $\square$

Intuition. Momentum conservation is more fundamental than energy conservation in collisions Because it holds for all types of collisions — elastic, inelastic, and explosive. Kinetic energy Is only conserved in elastic collisions.

3. Impulse

Definition. The impulse $J$ delivered by a force $\mathbf{F}$ acting over a time interval $\Delta t$ is:

$\boxed{\mathbf{J} = \mathbf{F}\Delta t = \Delta\mathbf{p}}$

For a variable force:

$\mathbf{J} = \int_{t_1}^{t_2} \mathbf{F}(t)\,dt$

Derivation from Newton’s second law.

$\mathbf{F} = \frac{d\mathbf{p}}{dt} \implies \mathbf{F}\,dt = d\mathbf{p} \implies \int \mathbf{F}\,dt = \int d\mathbf{p} = \Delta\mathbf{p}$

$\square$

Units. Impulse has units N s, which are equivalent to kg m s$^{-1}$ (the same as momentum).

Intuition. Impulse explains why airbags save lives. The change in momentum $\Delta p$ is fixed (car stops, so $p$ goes from $mv$ to 0). The airbag increases $\Delta t$So the force $F = \Delta p / \Delta t$ is reduced.

4. Collisions

Coefficient of Restitution

Definition. The coefficient of restitution $e$ measures the elasticity of a collision:

$\boxed{e = -\frac{v_1 - v_2}{u_1 - u_2}}$

Where $u_1, u_2$ are the velocities before collision and $v_1, v_2$ are the velocities after Collision, measured along the line of impact.

Derivation. For a collision between two bodies, Newton’s experimental law of restitution states That the relative speed of separation equals $e$ times the relative speed of approach. Since the Bodies separate after collision ($v_2 > v_1$ if body 2 was struck), and approached before ($u_1 > u_2$), the minus sign ensures $e > 0$.

• Perfectly elastic collision: $e = 1$ (kinetic energy is conserved)
• Perfectly inelastic collision: $e = 0$ (bodies coalesce — maximum KE loss)
• Inelastic collision: $0 < e < 1$

1D Elastic Collision: Final Velocities

For two bodies of masses $m_1$ and $m_2$ with initial velocities $u_1$ and $u_2$:

Conservation of momentum:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2 \quad \mathrm{...(i)}$

Conservation of kinetic energy (elastic, $e = 1$):

$\frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 = \frac{1}{2}m_1 v_1^2 + \frac{1}{2}m_2 v_2^2 \quad \mathrm{...(ii)}$

From (i): $m_1(u_1 - v_1) = m_2(v_2 - u_2)$ … (iii)

From (ii): $m_1(u_1^2 - v_1^2) = m_2(v_2^2 - u_2^2)$ … (iv)

Dividing (iv) by (iii):

$u_1 + v_1 = u_2 + v_2 \implies v_1 - v_2 = -(u_1 - u_2) \quad \mathrm{(Newton\'s restitution with } e = 1\mathrm{)}$

From $v_1 = u_2 + v_2 - u_1$Substituting into (i):

$m_1 u_1 + m_2 u_2 = m_1(u_2 + v_2 - u_1) + m_2 v_2$

$m_1 u_1 + m_2 u_2 = m_1 u_2 + m_1 v_2 - m_1 u_1 + m_2 v_2$

$2m_1 u_1 + (m_2 - m_1)u_2 = (m_1 + m_2)v_2$

$\boxed{v_2 = \frac{(m_2 - m_1)u_2 + 2m_1 u_1}{m_1 + m_2}}$

By symmetry:

$\boxed{v_1 = \frac{(m_1 - m_2)u_1 + 2m_2 u_2}{m_1 + m_2}}$

Special Cases

Equal masses ($m_1 = m_2$): $v_1 = u_2$ and $v_2 = u_1$. The bodies exchange velocities.

Stationary target ($u_2 = 0$):

$v_1 = \frac{(m_1 - m_2)u_1}{m_1 + m_2}, \qquad v_2 = \frac{2m_1 u_1}{m_1 + m_2}$

Heavy stationary target ($m_2 \gg m_1$): $v_1 \approx -u_1$ (light body rebounds), $v_2 \approx 0$ (heavy body barely moves).

Light stationary target ($m_2 \ll m_1$): $v_1 \approx u_1$ (heavy body continues), $v_2 \approx 2u_1$ (light body moves at twice the speed).

Inelastic Collision

For a perfectly inelastic collision ($e = 0$), the bodies coalesce. Let the common final velocity be $v$:

$m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$

$\boxed{v = \frac{m_1 u_1 + m_2 u_2}{m_1 + m_2}}$

Kinetic energy lost:

$\Delta E_k = \frac{1}{2}m_1 u_1^2 + \frac{1}{2}m_2 u_2^2 - \frac{1}{2}(m_1 + m_2)v^2$

$= \frac{m_1 m_2}{2(m_1 + m_2)}(u_1 - u_2)^2$

Note that the energy loss depends on the relative velocity — a faster approach means more energy is Dissipated.

5. 2D Collisions

In two dimensions, momentum conservation applies separately in each direction:

$\sum p_x^{\mathrm{before}} = \sum p_x^{\mathrm{after}}, \qquad \sum p_y^{\mathrm{before}} = \sum p_y^{\mathrm{after}}$

For an elastic collision, we also conserve kinetic energy. The coefficient of restitution applies Along the line of centres.

6. Force-Time Graphs

The area under a force-time graph equals the impulse, which equals the change in momentum.

For a collision, the force rises rapidly, peaks, and falls. A larger maximum force Corresponds to a shorter collision time (for the same impulse).

Problem Set

7. Impulse in 2D

Definition. In two dimensions, impulse is a vector quantity:

$\boxed{\mathbf{J} = \Delta\mathbf{p} = \mathbf{F}\,\Delta t}$

Resolving into components:

$J_x = \Delta p_x = F_x\,\Delta t, \qquad J_y = \Delta p_y = F_y\,\Delta t$

$|\mathbf{J}| = \sqrt{J_x^2 + J_y^2}$

When a ball strikes a smooth wall, only the component of momentum perpendicular to the wall changes. The parallel component is unchanged because the wall exerts no force parallel to its surface.

8. Coefficient of Restitution — Relative Velocity Form

Definition. The coefficient of restitution can be expressed in terms of relative velocities:

$\boxed{e = \frac{|\mathrm{relative speed of separation}|}{|\mathrm{relative speed of approach}|}}$

Derivation for a 1D Collision

Consider two bodies with masses $m_1$ and $m_2$. Before collision they move with velocities $u_1$ And $u_2$; after collision their velocities are $v_1$ and $v_2$.

Relative speed of approach. The speed at which the bodies approach each other is $|u_1 - u_2|$.

Relative speed of separation. The speed at which the bodies move apart after collision is $|v_2 - v_1|$.

Newton’s law of restitution states:

$e = \frac{|v_2 - v_1|}{|u_1 - u_2|}$

If body 1 catches up to body 2 ($u_1 \gt u_2$) and they separate after collision ($v_2 \gt v_1$), Then:

$e = \frac{v_2 - v_1}{u_1 - u_2}$

This is algebraically equivalent to the form $e = -\dfrac{v_1 - v_2}{u_1 - u_2}$ given in Section 4. $\square$

Intuition. The relative velocity form shows that $e$ is a property of the collision itself (the Materials involved), not of the individual bodies. For a perfectly elastic collision ($e = 1$), the Relative speed is unchanged — the bodies bounce off each other just as fast as they approached.

9. Bouncing Ball: Proof that $e = \sqrt{h'/h}$

Theorem. For a ball dropped from height $h$ onto a hard floor, rebounding to height $h'$The Coefficient of restitution satisfies $e = \sqrt{h'/h}$.

Proof (energy method). The ball falls from height $h$ under gravity. By conservation of energy:

$mgh = \tfrac{1}{2}mu^2 \implies u = \sqrt{2gh}$

Where $u$ is the speed just before impact (downward).

After impact, the ball rises to height $h'$. By conservation of energy:

$\tfrac{1}{2}mv^2 = mgh' \implies v = \sqrt{2gh'}$

Where $v$ is the speed just after impact (upward).

The floor has effectively infinite mass and does not move, so the relative speed of approach is $u$ And the relative speed of separation is $v$:

$e = \frac{v}{u} = \frac{\sqrt{2gh'}}{\sqrt{2gh}} = \sqrt{\frac{h'}{h}}$

$\square$

10. Explosions and Rocket Propulsion

Explosions

Intuition. In an explosion, a body at rest (or moving) breaks into fragments due to internal Forces. Since internal forces come in equal and opposite pairs (Newton’s third law), the total External force on the system is zero and momentum is conserved.

If a body of mass $M$ at rest explodes into two fragments of masses $m_1$ and $m_2$:

$0 = m_1\mathbf{v}_1 + m_2\mathbf{v}_2$

The fragments move in opposite directions with speeds inversely proportional to their masses:

$\frac{|\mathbf{v}_1|}{|\mathbf{v}_2|} = \frac{m_2}{m_1}$

Kinetic energy in an explosion. KE increases as chemical (or nuclear) energy is converted to KE Of fragments:

$\Delta E_k = \tfrac{1}{2}m_1v_1^2 + \tfrac{1}{2}m_2v_2^2 - 0$

Since $m_1v_1 = m_2v_2 = p$ (momentum conservation):

$\Delta E_k = \frac{p^2}{2m_1} + \frac{p^2}{2m_2} = \frac{M\,p^2}{2m_1 m_2}$

Rocket Propulsion

Intuition. A rocket moves forward by ejecting exhaust gases backward. The rocket does not “push Against” the air or the ground — it works in vacuum by momentum conservation.

Consider a rocket of mass $m$ moving at velocity $v$. In time $dt$It ejects fuel of mass $dm$ at Exhaust velocity $u$ relative to the rocket. The absolute velocity of the exhaust is $v - u$.

Momentum conservation (no external forces):

$m\,v = (m - dm)(v + dv) + dm\,(v - u)$

Expanding and neglecting $dm\,dv$:

$mv = mv + m\,dv - v\,dm - dm\,dv + v\,dm - u\,dm$

$0 = m\,dv - u\,dm$

$\boxed{m\,dv = -u\,dm}$

This is the rocket equation (Tsiolkovsky equation in differential form). Since $dm \lt 0$ (the Rocket loses mass), $dv \gt 0$ (the rocket accelerates).

Integrating from initial mass $m_0$ to final mass $m_f$:

$\int_0^{\Delta v} dv = -u\int_{m_0}^{m_f}\frac{dm}{m}$

$\Delta v = u\ln\left(\frac{m_0}{m_f}\right)$

Intuition. The rocket equation shows that $\Delta v$ depends on exhaust velocity $u$ and mass Ratio $m_0/m_f$. This is why multi-stage rockets are used — dropping empty stages reduces $m_f$ Without reducing $m_0$.

Further Problems

Common Pitfalls

1. Rounding intermediate answers too early, which compounds errors in multi-step calculations.

2. Misidentifying the system boundary when applying conservation laws. Define what is included before writing equations.

3. Confusing scalar and vector quantities. Always check whether direction matters for the quantity in question.

4. Confusing displacement with distance, or velocity with speed, particularly in graphs and calculations.

Summary

The key principles covered in this topic are linked in the sub-pages above. Focus on understanding the definitions, applying the formulas or frameworks, and evaluating strengths and limitations of each approach.

Worked Examples

Worked examples demonstrating the application of key concepts are covered in the detailed sub-pages linked above.

Cross-References

• Dynamics — Momentum conservation is derived from Newton’s second and third laws, which are covered in the dynamics topic.
• Work, Energy and Power — The work-energy theorem connects impulse and change in momentum to kinetic energy.
• Oscillations — Momentum principles apply to oscillating systems where forces vary with position.
• Circular Motion — 2D collision problems often require resolving momentum into components, similar to circular motion vector analysis.