Principle (Cambridge wording):
In a closed system the total linear momentum is constant if the resultant external force is zero.
Mathematically, for a system of \(n\) particles:
\[
\sum{i=1}^{n}\mathbf{p}{i,\text{initial}} \;=\; \sum{i=1}^{n}\mathbf{p}{i,\text{final}}
\]
Because momentum is a vector, the conservation law must be applied separately to each independent direction (normally \(x\) and \(y\)).
| Collision type | Quantities conserved | Typical exam information | Key formula(s) |
|---|---|---|---|
| Elastic | Momentum ✓ Kinetic energy ✓ | Masses, all initial speeds (and sometimes directions) | \(vA = \dfrac{(mA-mB)uA+2mBuB}{mA+mB}\) \(vB = \dfrac{(mB-mA)uB+2mAuA}{mA+mB}\) |
| Perfectly inelastic (objects stick) | Momentum ✓ Kinetic energy ✗ (maximum loss) | Masses, all initial speeds (and directions) | \(v = \dfrac{mAuA+mBuB}{mA+mB}\) |
| General (partially) inelastic (separate after impact) | Momentum ✓ Kinetic energy ✗ | Masses, all initial speeds + one extra datum (e.g. a rebound speed, angle, or coefficient of restitution – not required for this syllabus) | Use the momentum components (Eq 5) together with the supplied extra information. |
\[
mA uA + mB uB = mA vA + mB vB \tag{1}
\]
\[
\frac12 mA uA^{2} + \frac12 mB uB^{2}
= \frac12 mA vA^{2} + \frac12 mB vB^{2} \tag{2}
\]
Solving (1) and (2) gives the standard results (3) shown in the table above.
When the bodies stick together, \(vA=vB=v\). From (1):
\[
v = \frac{mA uA + mB uB}{mA+mB} \tag{4}
\]
Only (1) is guaranteed. An additional piece of information (e.g. a given final speed or direction) is required to close the problem.
Momentum must be conserved in each Cartesian component:
\[
\begin{cases}
mA u{Ax}+mB u{Bx}=mA v{Ax}+mB v{Bx}\\[4pt]
mA u{Ay}+mB u{By}=mA v{Ay}+mB v{By}
\end{cases} \tag{5}
\]
For an elastic collision the kinetic‑energy equation (2) is also applied, giving three independent equations for the three unknown speed components.
Problem: A 0.5 kg ball travelling at \(4\ \text{m s}^{-1}\) collides elastically with a stationary 0.2 kg ball. Find the speeds after the impact.
Solution:
\[
v_A = \frac{(0.5-0.2)4 + 2(0.2)(0)}{0.5+0.2}
= \frac{0.3\times4}{0.7}
\approx 1.71\ \text{m s}^{-1}
\]
\[
v_B = \frac{(0.2-0.5)0 + 2(0.5)(4)}{0.5+0.2}
= \frac{4.0}{0.7}
\approx 5.71\ \text{m s}^{-1}
\]
Problem: A 3.0 kg cart moving at \(2\ \text{m s}^{-1}\) collides head‑on with a 2.0 kg cart moving at \(-1\ \text{m s}^{-1}\). The carts stick together. Find their common speed after the collision.
Solution:
\[
v = \frac{(3.0)(2) + (2.0)(-1)}{3.0+2.0}
= \frac{6 - 2}{5}
= 0.8\ \text{m s}^{-1}
\]
The combined 5 kg mass moves forward (positive direction) at \(0.8\ \text{m s}^{-1}\).
Problem: A 2.0 kg puck moving east at \(3\ \text{m s}^{-1}\) collides with a 1.0 kg puck moving north at \(2\ \text{m s}^{-1}\). They stick together. Determine the speed and direction of the combined puck.
Solution:
\[
p_{x,i}= (2.0)(3) + (1.0)(0)=6\ \text{kg·m s}^{-1}
\]
\[
p_{y,i}= (2.0)(0) + (1.0)(2)=2\ \text{kg·m s}^{-1}
\]
\[
vx = \frac{p{x,i}}{M}= \frac{6}{3}=2.0\ \text{m s}^{-1}
\]
\[
vy = \frac{p{y,i}}{M}= \frac{2}{3}=0.67\ \text{m s}^{-1}
\]
\[
v = \sqrt{vx^{2}+vy^{2}} = \sqrt{2^{2}+0.67^{2}} \approx 2.11\ \text{m s}^{-1}
\]
\[
\theta = \tan^{-1}\!\left(\frac{vy}{vx}\right)
= \tan^{-1}\!\left(\frac{0.67}{2}\right)
\approx 18^{\circ}
\]
The combined puck moves at \(2.1\ \text{m s}^{-1}\) at \(18^{\circ}\) north of east.
Problem: Two identical billiard balls (mass \(m\)) move on a frictionless table. Ball A travels east at \(5\ \text{m s}^{-1}\); Ball B is initially at rest. After a perfectly elastic collision Ball A is observed to move at \(30^{\circ}\) north of east with speed \(3\ \text{m s}^{-1}\). Find the speed and direction of Ball B after the impact.
Solution (using the 2‑D quick‑solve template):
\(mA=mB=m\), \(u{Ax}=5,\ u{Ay}=0\), \(u{Bx}=u{By}=0\).
After collision: \(v{Ax}=3\cos30^{\circ}=2.60\), \(v{Ay}=3\sin30^{\circ}=1.50\).
Unknowns: \(v{Bx},\,v{By}\).
\[
\begin{cases}
m(5)+0 = m(2.60)+m v_{Bx}\\[4pt]
0 = m(1.50)+m v_{By}
\end{cases}
\]
Simplify (mass cancels):
\[
\begin{cases}
5 = 2.60 + v_{Bx}\\[4pt]
0 = 1.50 + v_{By}
\end{cases}
\]
Hence \(v{Bx}=2.40\ \text{m s}^{-1}\) (east) and \(v{By}=-1.50\ \text{m s}^{-1}\) (south).
\[
v_B = \sqrt{(2.40)^2+(-1.50)^2}\approx 2.84\ \text{m s}^{-1}
\]
\[
\phi = \tan^{-1}\!\left(\frac{|v{By}|}{v{Bx}}\right)
= \tan^{-1}\!\left(\frac{1.50}{2.40}\right)
\approx 32^{\circ}
\]
Direction: \(32^{\circ}\) south of east.
| Collision type | Conserved quantities | Typical data supplied in exam | Key formula(s) or approach |
|---|---|---|---|
| Elastic (1‑D or 2‑D) | Linear momentum (vector) ✓ Kinetic energy ✓ | Masses, all initial speeds (and directions for 2‑D) | Use momentum components (Eq 1 or 5) + kinetic‑energy equation (2). For 1‑D you may use the compact results (3). |
| Perfectly inelastic (stick) | Linear momentum ✓ Kinetic energy ✗ (maximum loss) | Masses, all initial speeds (directions if 2‑D) | Momentum components only. Final common velocity given by Eq 4 (1‑D) or by component form of Eq 5 (2‑D). |
| General inelastic (separate) | Linear momentum ✓ Kinetic energy ✗ | Masses, all initial speeds + one extra datum (e.g. a rebound speed, angle, or a numerical coefficient of restitution – not required here) | Momentum components (Eq 5) together with the extra information to close the system. |
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