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.
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
\[
\Delta\mathbf{p}= \int{ti}^{t_f}\mathbf{F}\,dt = \mathbf{J}.
\]
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).
Statement (syllabus 3.2.1): In an isolated system – i.e. a system on which the vector sum of external forces is zero throughout the interaction – the total linear momentum remains constant.
\[
\sum\mathbf{p}{\text{initial}} = \sum\mathbf{p}{\text{final}} .
\]
Reasoning:
If a system contains several particles, the total momentum is the vector sum of the individual momenta:
\[
\mathbf{P}{\text{system}}=\sumi mi\mathbf{v}i .
\]
The motion of the centre of mass (CM) obeys
\[
\mathbf{P}{\text{system}} = M{\text{total}}\mathbf{V}_{\text{CM}},
\]
where \(M{\text{total}}=\sumi mi\) and \(\mathbf{V}{\text{CM}}\) is the velocity of the CM.
In the absence of external forces the CM moves with constant velocity.
| 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.
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 \(v_{\text{rel}}\):
\[
\Delta K = K{\text{initial}}-K{\text{final}} = \frac12\,\mu\,v_{\text{rel}}^{2},
\qquad
\mu=\frac{m1m2}{m1+m2},
\qquad
v{\text{rel}} = |v{1i}-v_{2i}|.
\]
For a perfectly inelastic collision the above expression gives the *maximum* possible loss.
Two carts, \(m1=2.0\;\text{kg}\) at \(v{1i}=3.0\;\text{m s}^{-1}\) and \(m_2=3.0\;\text{kg}\) at rest, stick together.
\[
\mu=\frac{2.0\times3.0}{5.0}=1.2\;\text{kg},\qquad
v_{\text{rel}}=3.0\;\text{m s}^{-1}.
\]
\[
\Delta K = \tfrac12(1.2)(3.0)^2 = 5.4\;\text{J}.
\]
For two masses \(m1,m2\) with initial speeds \(u1,u2\) and final speeds \(v1,v2\):
\[
\begin{aligned}
\text{Momentum:}&\quad m1u1+m2u2 = m1v1+m2v2 \quad (1)\\[4pt]
\text{Kinetic energy:}&\quad \tfrac12m1u1^{2}+\tfrac12m2u2^{2}
= \tfrac12m1v1^{2}+\tfrac12m2v2^{2}\quad (2)
\end{aligned}
\]
Subtracting \((1)\times (v1+v2)\) from \((2)\) eliminates the squared terms and yields
\[
(m1+m2)(v1-u1) = (m2-m1)(v2-u2).
\]
Solving the simultaneous equations (1) and (2) gives the standard results
\[
\boxed{\,v1 = \frac{m1-m2}{m1+m2}\,u1 + \frac{2m2}{m1+m2}\,u2\,},
\qquad
\boxed{\,v2 = \frac{2m1}{m1+m2}\,u1 + \frac{m2-m1}{m1+m2}\,u2\,}.
\]
\[
v_1 = \frac{1.5-2.0}{3.5}(4.0)+\frac{2\times2.0}{3.5}(-2.0)= -0.57\;\text{m s}^{-1},
\]
\[
v_2 = \frac{2\times1.5}{3.5}(4.0)+\frac{2.0-1.5}{3.5}(-2.0)= 3.14\;\text{m s}^{-1}.
\]
Momentum and kinetic energy are both unchanged.
When the bodies stick together, \(v1=v2=v_f\). Conservation of momentum gives
\[
vf = \frac{m1u1+m2u2}{m1+m_2}.
\]
Using the same masses as above but with \(u_2=0\):
\[
v_f = \frac{(2.0)(3.0)}{5.0}=1.2\;\text{m s}^{-1}.
\]
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 \(v1\) making angle \(\theta1\) with the x‑axis, and the target ball moves at speed \(v2\) making angle \(\theta2\).
Conservation of momentum components:
\[
\begin{aligned}
\text{x‑component:}&\quad mu = mv1\cos\theta1 + mv2\cos\theta2 \quad (3)\\
\text{y‑component:}&\quad 0 = mv1\sin\theta1 + mv2\sin\theta2 \quad (4)
\end{aligned}
\]
Conservation of kinetic energy:
\[
\tfrac12mu^{2}= \tfrac12mv1^{2}+ \tfrac12mv2^{2}\quad (5)
\]
From (4) we have \(v1\sin\theta1 = -v2\sin\theta2\). Squaring (3) and (4) and adding eliminates the cross terms, giving
\[
u^{2}=v1^{2}+v2^{2}.
\]
Comparing this with (5) shows that the vector triangle formed by the three velocities is a right‑angled triangle, therefore
\[
\boxed{\theta1+\theta2 = 90^{\circ}}.
\]
For equal masses the speed‑angle relations simplify to
\[
v1 = u\cos\theta2,\qquad v2 = u\cos\theta1.
\]
When kinetic energy is not conserved, only the momentum components are required. The steps are:
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 \(v2\) and direction \(\theta2\) of the second puck.
\[
0.20(5.0)=0.20(3.0\cos30^{\circ})+0.30\,v2\cos\theta2 .
\]
\[
0=0.20(3.0\sin30^{\circ})-0.30\,v2\sin\theta2 .
\]
\[
v2 = 4.2\;\text{m s}^{-1},\qquad \theta2 = -22^{\circ}\;(\text{below the x‑axis}).
\]
\(\displaystyle \Delta K = \tfrac12\mu v{\text{rel}}^{2}\) with \(\mu = \frac{m1m2}{m1+m_2}\).
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