| Exam Board | Edexcel |
|---|---|
| Module | FM1 (Further Mechanics 1) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Oblique and successive collisions |
| Type | Oblique collision, direction deflected given angle |
| Difficulty | Challenging +1.2 This is a standard Further Mechanics 1 oblique collision problem requiring resolution of velocities parallel and perpendicular to the line of centres, conservation of momentum along the line of centres, and Newton's experimental law. While it involves multiple steps and careful vector resolution, it follows a well-established procedure taught in FM1 with no novel insights required. The 90° deflection provides a helpful constraint that simplifies the algebra. |
| Spec | 6.03c Momentum in 2D: vector form6.03d Conservation in 2D: vector momentum6.03f Impulse-momentum: relation6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Parallel to line of centres applied correctly | M1 | Use of CLM parallel to line of centres. Need all 4 terms. Dimensionally correct. Condone sign errors and sin/cos confusion. |
| \(9mu\cos30° - 8mu\cos30° = 4mv\cos60° - 3mw\cos\theta\) giving \((u\cos30° = 2v - 3w\cos\theta)\) | A1 | Correct unsimplified equation. Allow \(w_x\) in place of \(w\cos\theta\) and \(v_x\) in place of \(v\cos60°\). Allow division by common factor e.g. \(m\) |
| \(\uparrow A: w\sin\theta = 3u\sin30° \left(= \frac{3u}{2}\right)\) | B1 | No change perpendicular to line of centres for one sphere. Allow \(w_y\) in place of \(w\sin\theta\) |
| \(\uparrow B: v\sin60° = 2u\sin30°(= u)\) giving \(v = \frac{2u}{\sqrt{3}}\) | B1 | No change perpendicular to line of centres for both spheres |
| \((w_x =) w\cos\theta = \frac{1}{3}(2v - u\cos30°) = \frac{5u\sqrt{3}}{18}\), \((w_y =) w\sin\theta = \frac{3u}{2}\), \(\Rightarrow \tan\theta = \frac{9\sqrt{3}}{5}\), \(\theta = 72.2°\) | M1 | Use scalar product or solve simultaneous equations to find \(\theta\) using their \(w_x\). Must reach \(\theta\) = numerical value |
| Direction deflected by \(77.8°\) (\(78°\) or better) | A1 | \(78°\) or better |
| Magnitude of impulse \(= 4m(v\cos60° - (-2u\cos30°))\) | M1 | Use of \(I = mv - mu\) in direction of line of centres. Condone subtraction in either order |
| \(= 4m\left(\frac{1}{2} \times \frac{u}{\sin60°} - \left(-2u\frac{\sqrt{3}}{2}\right)\right)\) | A1 | Correct unsimplified expression. Allow the negative of this |
| \(= \frac{16\sqrt{3}}{3}mu\) | A1 | Any equivalent simplified form. Must be positive. Accept \(9.2(376...)mu\) (2 sf or better) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Using CLM | M1 | Use of CLM. Need all terms. Must be dimensionally correct. Condone sign errors. Accept consistent cancelling of \(m\) |
| \(6mu - 4mu = -3mv + 4mw\) giving \(2u = -3v + 4w\) | A1 | Correct unsimplified equation for CLM. They can have \(v\) in either direction |
| Use of impact law | M1 | Correct use of impact law (used the right way round). Condone sign errors in finding speed of approach and speed of separation |
| \(w + v = e \times 3u\) | A1 | Correct unsimplified equation. Signs consistent with equation for CLM |
| Complete method to find \(w\) | M1 | Complete method e.g. forming simultaneous equations using CLM and Impact Law and solving. Requires both preceding M marks |
| \(\begin{cases} 3w + 3v = 9eu \\ -3v + 4w = 2u \end{cases} \Rightarrow 7w = 9eu + 2u\), \(w = \frac{u}{7}(9e+2)\) | A1* | Obtain given answer from correct working. Accept with \(2 + 9e\) in place of \(9e + 2\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(w' = \frac{1}{2} \times \frac{u}{7}(9e+2) = \frac{u}{14}(9e+2)\) | B1 | Speed of \(Q\) after impact with the wall. Any equivalent form. Correct speed can be implied by a correct negative velocity |
| \(v = \frac{u}{7}(12e-2)\) | B1 | Speed of \(P\) after impact with \(Q\). Accept \(\pm\). Any equivalent form in \(u\) and \(e\) |
| For a second collision: \(w' > v\) | M1 | Form correct inequality using their \(v\) and \(w'\). A correct inequality has \(P\) and \(Q\) both moving away from wall |
| \(9e + 2 > 2(12e-2)\), giving \(0 < e < \frac{2}{5}\) | A1 | Correct interval only. Accept unsimplified fraction. Need both ends of interval. Must be strict inequality at both ends |
## Question 4:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Parallel to line of centres applied correctly | M1 | Use of CLM parallel to line of centres. Need all 4 terms. Dimensionally correct. Condone sign errors and sin/cos confusion. |
| $9mu\cos30° - 8mu\cos30° = 4mv\cos60° - 3mw\cos\theta$ giving $(u\cos30° = 2v - 3w\cos\theta)$ | A1 | Correct unsimplified equation. Allow $w_x$ in place of $w\cos\theta$ and $v_x$ in place of $v\cos60°$. Allow division by common factor e.g. $m$ |
| $\uparrow A: w\sin\theta = 3u\sin30° \left(= \frac{3u}{2}\right)$ | B1 | No change perpendicular to line of centres for one sphere. Allow $w_y$ in place of $w\sin\theta$ |
| $\uparrow B: v\sin60° = 2u\sin30°(= u)$ giving $v = \frac{2u}{\sqrt{3}}$ | B1 | No change perpendicular to line of centres for both spheres |
| $(w_x =) w\cos\theta = \frac{1}{3}(2v - u\cos30°) = \frac{5u\sqrt{3}}{18}$, $(w_y =) w\sin\theta = \frac{3u}{2}$, $\Rightarrow \tan\theta = \frac{9\sqrt{3}}{5}$, $\theta = 72.2°$ | M1 | Use scalar product or solve simultaneous equations to find $\theta$ using their $w_x$. Must reach $\theta$ = numerical value |
| Direction deflected by $77.8°$ ($78°$ or better) | A1 | $78°$ or better |
| Magnitude of impulse $= 4m(v\cos60° - (-2u\cos30°))$ | M1 | Use of $I = mv - mu$ in direction of line of centres. Condone subtraction in either order |
| $= 4m\left(\frac{1}{2} \times \frac{u}{\sin60°} - \left(-2u\frac{\sqrt{3}}{2}\right)\right)$ | A1 | Correct unsimplified expression. Allow the negative of this |
| $= \frac{16\sqrt{3}}{3}mu$ | A1 | Any equivalent simplified form. Must be positive. Accept $9.2(376...)mu$ (2 sf or better) |
---
## Question 5a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using CLM | M1 | Use of CLM. Need all terms. Must be dimensionally correct. Condone sign errors. Accept consistent cancelling of $m$ |
| $6mu - 4mu = -3mv + 4mw$ giving $2u = -3v + 4w$ | A1 | Correct unsimplified equation for CLM. They can have $v$ in either direction |
| Use of impact law | M1 | Correct use of impact law (used the right way round). Condone sign errors in finding speed of approach and speed of separation |
| $w + v = e \times 3u$ | A1 | Correct unsimplified equation. Signs consistent with equation for CLM |
| Complete method to find $w$ | M1 | Complete method e.g. forming simultaneous equations using CLM and Impact Law and solving. Requires both preceding M marks |
| $\begin{cases} 3w + 3v = 9eu \\ -3v + 4w = 2u \end{cases} \Rightarrow 7w = 9eu + 2u$, $w = \frac{u}{7}(9e+2)$ | A1* | Obtain given answer from correct working. Accept with $2 + 9e$ in place of $9e + 2$ |
---
## Question 5b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $w' = \frac{1}{2} \times \frac{u}{7}(9e+2) = \frac{u}{14}(9e+2)$ | B1 | Speed of $Q$ after impact with the wall. Any equivalent form. Correct speed can be implied by a correct negative velocity |
| $v = \frac{u}{7}(12e-2)$ | B1 | Speed of $P$ after impact with $Q$. Accept $\pm$. Any equivalent form in $u$ and $e$ |
| For a second collision: $w' > v$ | M1 | Form correct inequality using their $v$ and $w'$. A correct inequality has $P$ and $Q$ both moving away from wall |
| $9e + 2 > 2(12e-2)$, giving $0 < e < \frac{2}{5}$ | A1 | Correct interval only. Accept unsimplified fraction. Need both ends of interval. Must be strict inequality at both ends |4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{86a37170-046f-46e5-9c8c-06d5f98ca4fe-12_387_929_246_568}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Two smooth uniform spheres, $A$ and $B$, have equal radii. The mass of $A$ is $3 m$ and the mass of $B$ is $4 m$. The spheres are moving on a smooth horizontal plane when they collide obliquely. Immediately before they collide, $A$ is moving with speed $3 u$ at $30 ^ { \circ }$ to the line of centres of the spheres and $B$ is moving with speed $2 u$ at $30 ^ { \circ }$ to the line of centres of the spheres. The direction of motion of $B$ is turned through an angle of $90 ^ { \circ }$ by the collision, as shown in Figure 3.\\
(i) Find the size of the angle through which the direction of motion of $A$ is turned as a result of the collision.\\
(ii) Find, in terms of $m$ and $u$, the magnitude of the impulse received by $B$ in the collision.
\hfill \mbox{\textit{Edexcel FM1 2022 Q4 [9]}}