| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2022 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Oblique collision of spheres |
| Difficulty | Challenging +1.8 This is a challenging oblique collision problem requiring resolution of velocities, application of conservation of momentum (parallel and perpendicular to line of centres), Newton's experimental law, and the perpendicularity condition to find k, followed by energy loss calculation. It requires systematic multi-step reasoning with several equations, but follows standard Further Maths mechanics methodology without requiring novel geometric insight. |
| Spec | 6.03b Conservation of momentum: 1D two particles6.03c Momentum in 2D: vector form6.03d Conservation in 2D: vector momentum6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Let speed of A after collision be \(\rightarrow v_A\), speed of B perpendicular to line of centres be \(\downarrow v\). Along line of centres: \(mu - km\frac{5}{8}u\cos\alpha = mv_A\) | M1 | Momentum |
| NEL: \(0 - v_A = e\left(\frac{5}{8}u\cos\alpha + u\right)\) | M1 | NEL |
| \(u - \frac{5}{8}ku\cos\alpha = -\frac{2}{3}\left(\frac{5}{8}u\cos\alpha + u\right)\) | M1 | Solve |
| Substitute for \(\cos\alpha\) to give \(k = 4\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v_B = \frac{5}{8}u\sin\alpha = \frac{3}{8}u\) | B1 | Velocity perpendicular to line of centres |
| \(v_A = -u\) | B1 FT | |
| KE before \(= \frac{1}{2}mu^2 + \frac{1}{2}km\left(\frac{5}{8}u\right)^2 = \frac{1}{2}mu^2 + \frac{25}{32}mu^2 = \frac{41}{32}mu^2\); KE after \(= \frac{1}{2}mv_A^2 + \frac{1}{2}kmv_B^2 = \frac{1}{2}mu^2 + 2m\frac{9}{64}u^2 = \frac{25}{32}mu^2\) | M1 | KE before and after for \(A\) is unchanged. Both. |
| Loss \(= mu^2\left(\frac{41}{32} - \frac{25}{32}\right) = \frac{1}{2}mu^2\) | A1 |
## Question 6:
### Part 6(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Let speed of A after collision be $\rightarrow v_A$, speed of B perpendicular to line of centres be $\downarrow v$. Along line of centres: $mu - km\frac{5}{8}u\cos\alpha = mv_A$ | M1 | Momentum |
| NEL: $0 - v_A = e\left(\frac{5}{8}u\cos\alpha + u\right)$ | M1 | NEL |
| $u - \frac{5}{8}ku\cos\alpha = -\frac{2}{3}\left(\frac{5}{8}u\cos\alpha + u\right)$ | M1 | Solve |
| Substitute for $\cos\alpha$ to give $k = 4$ | A1 | |
### Part 6(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v_B = \frac{5}{8}u\sin\alpha = \frac{3}{8}u$ | B1 | Velocity perpendicular to line of centres |
| $v_A = -u$ | B1 FT | |
| KE before $= \frac{1}{2}mu^2 + \frac{1}{2}km\left(\frac{5}{8}u\right)^2 = \frac{1}{2}mu^2 + \frac{25}{32}mu^2 = \frac{41}{32}mu^2$; KE after $= \frac{1}{2}mv_A^2 + \frac{1}{2}kmv_B^2 = \frac{1}{2}mu^2 + 2m\frac{9}{64}u^2 = \frac{25}{32}mu^2$ | M1 | KE before and after for $A$ is unchanged. Both. |
| Loss $= mu^2\left(\frac{41}{32} - \frac{25}{32}\right) = \frac{1}{2}mu^2$ | A1 | |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{5e95e0c9-d47d-4f2b-89da-ab949b9661f4-10_426_1191_267_438}
Two uniform smooth spheres $A$ and $B$ of equal radii have masses $m$ and $k m$ respectively. The two spheres are moving on a horizontal surface with speeds $u$ and $\frac { 5 } { 8 } u$ respectively. Immediately before the spheres collide, $A$ is travelling along the line of centres, and $B$ 's direction of motion makes an angle $\alpha$ with the line of centres (see diagram). The coefficient of restitution between the spheres is $\frac { 2 } { 3 }$ and $\tan \alpha = \frac { 3 } { 4 }$.
After the collision, the direction of motion of $B$ is perpendicular to the line of centres.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $k$.
\item Find the loss in the total kinetic energy as a result of the collision.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2022 Q6 [8]}}