| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2021 |
| Session | November |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Type | Oblique collision of spheres |
| Difficulty | Challenging +1.8 This is an oblique collision problem requiring resolution of velocities along and perpendicular to the line of centres, application of conservation of momentum and Newton's restitution law in the impulse direction, then calculation of energy loss. While it involves multiple steps and careful vector resolution, it follows a standard Further Mechanics framework with no unusual conceptual leaps. The given coefficient of restitution and specific angle make the algebra manageable, placing it well above average but not at the extreme difficulty level. |
| Spec | 6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks |
|---|---|
| 5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or |
| Answer | Marks |
|---|---|
| 5(a) | Let speeds of A and B along line of centres after collision be v and |
| Answer | Marks | Guidance |
|---|---|---|
| 1 2 2 2 4 | M1 | Momentum with masses correct. |
| Answer | Marks | Guidance |
|---|---|---|
| 2 1 3 | M1 | Restitution, with consistent signs on LHS. |
| Answer | Marks |
|---|---|
| v 1 =− 2 u v 2 = 2 u | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | B1 | |
| Direction of B is now 60° above line of centres. | M1 | |
| Angle of deflection is 60°. | A1 FT | FT (120° – their direction of B angle) |
| Answer | Marks |
|---|---|
| 5(b) | 1 1 3m 5 |
| Answer | Marks |
|---|---|
| 2 2 2 4 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| | B1 FT | FT only their speeds from (a) |
| Answer | Marks |
|---|---|
| 8 | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 5:
5 | Where a candidate has misread a number in the question and used that value consistently throughout, provided that number does not alter the difficulty or
the method required, award all marks earned and deduct just 1 mark for the misread.
--- 5(a) ---
5(a) | Let speeds of A and B along line of centres after collision be v and
1
v
2
3 3 u
mv + mv =− mucos60°+mu =
1 2 2 2 4 | M1 | Momentum with masses correct.
v −v =− 2(−ucos60°−u ) (=u)
2 1 3 | M1 | Restitution, with consistent signs on LHS.
1 1
v 1 =− 2 u v 2 = 2 u | A1
3
Perpendicular to line of centres, speed of B is usin60°= u
2 | B1
Direction of B is now 60° above line of centres. | M1
Angle of deflection is 60°. | A1 FT | FT (120° – their direction of B angle)
6
--- 5(b) ---
5(b) | 1 1 3m 5
KE before = mu2 + . u2 = mu2
2 2 2 4 | B1
1 u 2 1 3m u 2 3u 2 7
KE after = m + . + = mu2
2 2 2 2 2 2 8
| B1 FT | FT only their speeds from (a)
3
Loss in KE = mu2
8 | B1
3
Question | Answer | Marks | Guidance\includegraphics{figure_5}
Two uniform smooth spheres $A$ and $B$ of equal radii have masses $m$ and $\frac{2}{3}m$ respectively. The two spheres are each moving with speed $u$ on a horizontal surface when they collide. Immediately before the collision $A$'s direction of motion is along the line of centres, and $B$'s direction of motion makes an angle of $60°$ with the line of centres (see diagram). The coefficient of restitution between the spheres is $\frac{2}{3}$.
\begin{enumerate}[label=(\alph*)]
\item Find the angle through which the direction of motion of $B$ is deflected by the collision. [6]
\item Find the loss in the total kinetic energy of the system as a result of the collision. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2021 Q5 [9]}}