state the principle of conservation of momentum

Linear Momentum and Its Conservation

Learning Objective

State the principle of conservation of momentum and apply it to one‑dimensional (1‑D) and two‑dimensional (2‑D) collisions – elastic, inelastic and perfectly inelastic – as required by Cambridge IGCSE 9702.

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1. Linear Momentum

• For a particle of mass m moving with velocity v⃗, the linear momentum is the vector
  
  \(\displaystyle \mathbf{p}=m\mathbf{v}\)
• Direction of p⃗ is the same as the direction of v⃗.
• SI unit: kilogram metre per second (kg·m·s⁻¹).
• Note (Cambridge 9702): Relativistic mass change is not considered – the relation \(\mathbf{p}=m\mathbf{v}\) holds for all syllabus problems.

2. Newton’s Second Law → Impulse–Momentum Theorem

Newton’s second law in vector form is

\[ \sum\mathbf{F}= \frac{d\mathbf{p}}{dt}. \]

Integrating over a time interval \(\Delta t\) gives the impulse–momentum theorem

\[ \Delta\mathbf{p}= \int_{t_i}^{t_f}\mathbf{F}\,dt = \mathbf{J}. \]

For a constant net force F acting for a time \(\Delta t\)

\[ \Delta\mathbf{p}= \mathbf{J}= \mathbf{F}\,\Delta t . \]

This theorem is the formal basis for the conservation of momentum in an isolated system (no external impulse).

3. Principle of Conservation of Linear Momentum

Reasoning:

• From \(\sum\mathbf{F}=d\mathbf{p}/dt\), if \(\sum\mathbf{F}_{\text{ext}}=0\) then \(d\mathbf{p}/dt=0\) ⇒ \(\mathbf{p}\) is constant.
• Internal forces occur in equal‑and‑opposite pairs (Newton’s third law) and cancel in the vector sum.

4. Conditions for Conservation

• The system is closed – no mass enters or leaves during the interaction.
• The vector sum of external forces acting on the system is zero (or negligible) for the whole duration of the interaction.
• All internal forces obey Newton’s third law (equal magnitude, opposite direction).

5. Momentum of a System of Particles & Centre of Mass

If a system contains several particles, the total momentum is the vector sum of the individual momenta:

\[ \mathbf{P}_{\text{system}}=\sum_i m_i\mathbf{v}_i . \]

The motion of the centre of mass (CM) obeys

\[ \mathbf{P}_{\text{system}} = M_{\text{total}}\mathbf{V}_{\text{CM}}, \] where \(M_{\text{total}}=\sum_i m_i\) and \(\mathbf{V}_{\text{CM}}\) is the velocity of the CM. In the absence of external forces the CM moves with constant velocity.

6. Types of Collisions (Cambridge 9702)

Collision Type	Momentum	Kinetic Energy	Typical Example
Elastic	Conserved	Conserved	Billiard balls striking each other
Inelastic	Conserved	Not conserved – some is transformed into deformation, heat, sound, etc.	Car crash where the vehicles deform but do not stick
Perfectly Inelastic	Conserved	Maximum loss of kinetic energy – the bodies stick together	Two clay balls sticking after impact

Syllabus note: The coefficient of restitution is not required for the exam, but remembering that an elastic collision corresponds to a restitution coefficient of 1 and a perfectly inelastic collision to 0 can help you categorise problems.

7. Energy Considerations in Collisions

• Elastic collisions: Both momentum and kinetic energy are conserved.
• Inelastic collisions: Momentum is conserved, but kinetic energy is partially lost.

7.1 Quantitative kinetic‑energy loss

For a 1‑D collision the loss of kinetic energy can be written in terms of the reduced mass \(\mu\) and the relative speed before impact \(v_{\text{rel}}\):

\[ \Delta K = K_{\text{initial}}-K_{\text{final}} = \frac12\,\mu\,v_{\text{rel}}^{2}, \qquad \mu=\frac{m_1m_2}{m_1+m_2}, \qquad v_{\text{rel}} = |v_{1i}-v_{2i}|. \]

For a perfectly inelastic collision the above expression gives the *maximum* possible loss.

Example – kinetic‑energy loss

Two carts, \(m_1=2.0\;\text{kg}\) at \(v_{1i}=3.0\;\text{m s}^{-1}\) and \(m_2=3.0\;\text{kg}\) at rest, stick together.

\[ \mu=\frac{2.0\times3.0}{5.0}=1.2\;\text{kg},\qquad v_{\text{rel}}=3.0\;\text{m s}^{-1}. \] \[ \Delta K = \tfrac12(1.2)(3.0)^2 = 5.4\;\text{J}. \]

8. One‑Dimensional Collisions

8.1 Elastic Collision – Derivation Sketch

For two masses \(m_1,m_2\) with initial speeds \(u_1,u_2\) and final speeds \(v_1,v_2\):

\[ \begin{aligned} \text{Momentum:}&\quad m_1u_1+m_2u_2 = m_1v_1+m_2v_2 \quad (1)\\[4pt] \text{Kinetic energy:}&\quad \tfrac12m_1u_1^{2}+\tfrac12m_2u_2^{2} = \tfrac12m_1v_1^{2}+\tfrac12m_2v_2^{2}\quad (2) \end{aligned} \]

Subtracting \((1)\times (v_1+v_2)\) from \((2)\) eliminates the squared terms and yields

\[ (m_1+m_2)(v_1-u_1) = (m_2-m_1)(v_2-u_2). \]

Solving the simultaneous equations (1) and (2) gives the standard results

\[ \boxed{\,v_1 = \frac{m_1-m_2}{m_1+m_2}\,u_1 + \frac{2m_2}{m_1+m_2}\,u_2\,}, \qquad \boxed{\,v_2 = \frac{2m_1}{m_1+m_2}\,u_1 + \frac{m_2-m_1}{m_1+m_2}\,u_2\,}. \]

Numerical Example

• \(m_1=1.5\;\text{kg},\; u_1=+4.0\;\text{m s}^{-1}\) (to the right)
• \(m_2=2.0\;\text{kg},\; u_2=-2.0\;\text{m s}^{-1}\) (to the left)

\[ v_1 = \frac{1.5-2.0}{3.5}(4.0)+\frac{2\times2.0}{3.5}(-2.0)= -0.57\;\text{m s}^{-1}, \] \[ v_2 = \frac{2\times1.5}{3.5}(4.0)+\frac{2.0-1.5}{3.5}(-2.0)= 3.14\;\text{m s}^{-1}. \] Momentum and kinetic energy are both unchanged.

8.2 Perfectly Inelastic Collision

When the bodies stick together, \(v_1=v_2=v_f\). Conservation of momentum gives

\[ v_f = \frac{m_1u_1+m_2u_2}{m_1+m_2}. \]

Using the same masses as above but with \(u_2=0\):

\[ v_f = \frac{(2.0)(3.0)}{5.0}=1.2\;\text{m s}^{-1}. \]

9. Two‑Dimensional Collisions

9.1 Elastic Collision of Equal Masses – Right‑Angle Result

Let a cue ball (mass \(m\)) travel along the +x‑axis with speed \(u\) and strike an identical stationary target ball. After impact the cue ball moves at speed \(v_1\) making angle \(\theta_1\) with the x‑axis, and the target ball moves at speed \(v_2\) making angle \(\theta_2\).

Conservation of momentum components:

\[ \begin{aligned} \text{x‑component:}&\quad mu = mv_1\cos\theta_1 + mv_2\cos\theta_2 \quad (3)\\ \text{y‑component:}&\quad 0 = mv_1\sin\theta_1 + mv_2\sin\theta_2 \quad (4) \end{aligned} \]

Conservation of kinetic energy:

\[ \tfrac12mu^{2}= \tfrac12mv_1^{2}+ \tfrac12mv_2^{2}\quad (5) \]

From (4) we have \(v_1\sin\theta_1 = -v_2\sin\theta_2\). Squaring (3) and (4) and adding eliminates the cross terms, giving

\[ u^{2}=v_1^{2}+v_2^{2}. \]

Comparing this with (5) shows that the vector triangle formed by the three velocities is a right‑angled triangle, therefore

\[ \boxed{\theta_1+\theta_2 = 90^{\circ}}. \]

For equal masses the speed‑angle relations simplify to

\[ v_1 = u\cos\theta_2,\qquad v_2 = u\cos\theta_1. \]

Numerical Illustration

• \(u = 3.0\;\text{m s}^{-1}\), \(\theta_1 = 30^{\circ}\)
• Then \(\theta_2 = 60^{\circ}\) (since the angles are complementary)
• \(v_1 = 3.0\cos60^{\circ}=1.5\;\text{m s}^{-1}\)
• \(v_2 = 3.0\cos30^{\circ}=2.60\;\text{m s}^{-1}\)

9.2 Inelastic 2‑D Collision – General Vector Method

When kinetic energy is not conserved, only the momentum components are required. The steps are:

1. Resolve all velocities into orthogonal components (usually x and y).
2. Apply \(\sum p_x\) and \(\sum p_y\) before and after the impact.
3. Solve the resulting simultaneous equations for the unknown speeds or angles.

Example – Inelastic Puck Collision

A 0.20 kg puck moves at \(5.0\;\text{m s}^{-1}\) along +x and strikes a stationary 0.30 kg puck. After impact the first puck deflects at \(30^{\circ}\) to the original line with speed \(3.0\;\text{m s}^{-1}\). Find the speed \(v_2\) and direction \(\theta_2\) of the second puck.

1. x‑momentum: \[ 0.20(5.0)=0.20(3.0\cos30^{\circ})+0.30\,v_2\cos\theta_2 . \]
2. y‑momentum (initial y‑momentum = 0): \[ 0=0.20(3.0\sin30^{\circ})-0.30\,v_2\sin\theta_2 . \]
3. Solving the two equations gives \[ v_2 = 4.2\;\text{m s}^{-1},\qquad \theta_2 = -22^{\circ}\;(\text{below the x‑axis}). \]

10. Summary

• Linear momentum \(\mathbf{p}=m\mathbf{v}\) is a conserved vector quantity in an isolated system.
• Conservation follows directly from Newton’s second law \(\sum\mathbf{F}=d\mathbf{p}/dt\) and the third law (internal forces cancel).
• The total momentum of a system equals the total mass times the centre‑of‑mass velocity.
• In any collision, total momentum is conserved; kinetic energy is conserved only for elastic collisions.
• For 1‑D problems use the scalar form of the momentum equation; for 2‑D problems resolve vectors into perpendicular components and apply conservation to each component.
• Energy loss in an inelastic collision can be quantified by \(\displaystyle \Delta K = \tfrac12\mu v_{\text{rel}}^{2}\) with \(\mu = \frac{m_1m_2}{m_1+m_2}\).
• Remember: the syllabus does **not** require the coefficient of restitution, but knowing that elastic ⇔ \(e=1\) and perfectly inelastic ⇔ \(e=0\) can help you identify the type of collision.