| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2022 |
| 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 advanced 2D collision problem requiring resolution of velocities, application of conservation of momentum and Newton's law of restitution along the line of centres, followed by a geometric constraint. The constraint that B's final direction equals A's initial direction adds significant complexity, requiring simultaneous consideration of parallel and perpendicular components. While systematic, it demands careful vector decomposition and algebraic manipulation beyond standard A-level mechanics questions. |
| Spec | 6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks |
|---|---|
| 7(a) | Let v, w be speeds of A and B along line of centres after collision |
| Answer | Marks | Guidance |
|---|---|---|
| 2 2 | M1 | Momentum: masses correct, opposite signs on RHS. |
| w−v=e(2ucos+ucos) | M1 | NEL: LHS signs must be consistent with momentum equation, |
| Answer | Marks | Guidance |
|---|---|---|
| Use this fact and solve to find w | M1 | Solve to find an expression of the correct form. |
| Answer | Marks |
|---|---|
| 3 4 8 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 7(b) | Perpendicular to line of centres, speed of B is |
| 2usin=2ucos | B1 |
| Answer | Marks |
|---|---|
| w | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| 3 4 8 | M1* | Obtain homogeneous equation in cos and sin or an equation in tan |
| Answer | Marks | Guidance |
|---|---|---|
| 2(tan)2 +13tan−24=0, (2tan−3)(tan+8)=0 | DM1 | Obtain quadratic and solve to find values of tan |
| Answer | Marks |
|---|---|
| 2 | A1 |
Question 7:
--- 7(a) ---
7(a) | Let v, w be speeds of A and B along line of centres after collision
1 1
mv+ mw=mucos− m.2ucos
2 2 | M1 | Momentum: masses correct, opposite signs on RHS.
w−v=e(2ucos+ucos) | M1 | NEL: LHS signs must be consistent with momentum equation,
same sign for both terms on RHS.
+=90, so cos=sin
Use this fact and solve to find w | M1 | Solve to find an expression of the correct form.
2 1 13
w= u sin+ cos
3 4 8 | A1
4
Question | Answer | Marks | Guidance
--- 7(b) ---
7(b) | Perpendicular to line of centres, speed of B is
2usin=2ucos | B1
After, velocity of B makes angle with line of centres, so
2ucos
tan=
w | B1
sin 2ucos
= giving
cos 2 1 13
u sin+ cos
3 4 8 | M1* | Obtain homogeneous equation in cos and sin or an equation in tan
1 13
3(cos)2 = (sin)2 + sincos
4 8
2(tan)2 +13tan−24=0, (2tan−3)(tan+8)=0 | DM1 | Obtain quadratic and solve to find values of tan
3
tan=
2 | A1
5\includegraphics{figure_7}
Two uniform smooth spheres $A$ and $B$ of equal radii have masses $m$ and $\frac{1}{2}m$ respectively. The two spheres are moving on a horizontal surface when they collide. Immediately before the collision, sphere $A$ is travelling with speed $u$ and its direction of motion makes an angle $\alpha$ with the line of centres. Sphere $B$ is travelling with speed $2u$ and its direction of motion makes an angle $\beta$ with the line of centres (see diagram). The coefficient of restitution between the spheres is $\frac{5}{8}$ and $\alpha + \beta = 90°$.
\begin{enumerate}[label=(\alph*)]
\item Find the component of the velocity of $B$ parallel to the line of centres after the collision, giving your answer in terms of $u$ and $\alpha$. [4]
\end{enumerate}
The direction of motion of $B$ after the collision is parallel to the direction of motion of $A$ before the collision.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the value of $\tan \alpha$. [5]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2022 Q7 [9]}}