| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2020 |
| Session | November |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Energy method - driving force on horizontal road |
| Difficulty | Moderate -0.3 This is a straightforward multi-part mechanics question testing standard work-energy principles and impulse. Part (a) is direct application of W=Fd, part (b) uses work-energy theorem with given velocities, part (c) applies power formula, and part (d) uses impulse-momentum. All parts follow routine procedures with no novel problem-solving required, making it slightly easier than average for A-level Further Maths mechanics. |
| Spec | 6.02b Calculate work: constant force, resolved component6.02i Conservation of energy: mechanical energy principle6.02m Variable force power: using scalar product6.03b Conservation of momentum: 1D two particles6.03e Impulse: by a force6.03k Newton's experimental law: direct impact |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(30 \times 10\) | M1 | Using Work done \(= Fd\) |
| \(300\) J | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2} \times 2.4 \times 12^2 + 300 = \frac{1}{2} \times 2.4 \times 18^2 + W\) | M1 | Where \(W\) is energy loss (could be e.g. \(10R\)); allow 1 slip e.g. \(R\) instead of \(W\) |
| \(10R = 84\) | M1 | Use of energy loss \(= 10R\); must be e.g. \(10R\), not \(R\) |
| \(R = 8.4\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a = \frac{18^2 - 12^2}{2 \times 10} = 9\) | M1 | Using \(v^2 = u^2 + 2as\) to find \(a\) |
| \(30 - R = 2.4 \times 9\), \(R = 8.4\) | M1, A1 | Use of \(F = ma\) with their \(a\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(t = \frac{2 \times 10}{18 + 12}\) | M1 | Using \(s = \left(\frac{v+u}{2}\right)t\); or use Average power \(=\) Force \(\times\) average speed, i.e. \(P = F\frac{v+u}{2}\) |
| \(t = \frac{2}{3}\) | A1 | |
| \(\frac{84}{\frac{2}{3}} = 126\) W | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(t = \frac{18-12}{9} = \frac{2}{3}\) | M1, A1 | Using \(v = u + at\) to find \(t\); NB \(a = 9\) from (b) |
| \(\frac{84}{\frac{2}{3}} = 126\) W | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(17.28 = 1.6 \times v_Q\) | M1 | |
| \(v_Q = 10.8\) m s\(^{-1}\) | A1 | Do not allow \(-10.8\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2.4 \times 18 = 2.4 \times v_P + 1.6 \times 10.8\) | M1 | Attempt conservation of momentum |
| So \(v_P = 10.8 = v_Q\), so the particles coalesce and the collision is therefore inelastic. | A1 | Find equal velocities and conclude; do not allow use of KE |
## Question 4:
### Part (a)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $30 \times 10$ | M1 | Using Work done $= Fd$ |
| $300$ J | A1 | |
### Part (b)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2} \times 2.4 \times 12^2 + 300 = \frac{1}{2} \times 2.4 \times 18^2 + W$ | M1 | Where $W$ is energy loss (could be e.g. $10R$); allow 1 slip e.g. $R$ instead of $W$ |
| $10R = 84$ | M1 | Use of energy loss $= 10R$; must be e.g. $10R$, not $R$ |
| $R = 8.4$ | A1 | |
**Alternative method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a = \frac{18^2 - 12^2}{2 \times 10} = 9$ | M1 | Using $v^2 = u^2 + 2as$ to find $a$ |
| $30 - R = 2.4 \times 9$, $R = 8.4$ | M1, A1 | Use of $F = ma$ with their $a$ |
### Part (c)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $t = \frac{2 \times 10}{18 + 12}$ | M1 | Using $s = \left(\frac{v+u}{2}\right)t$; or use Average power $=$ Force $\times$ average speed, i.e. $P = F\frac{v+u}{2}$ |
| $t = \frac{2}{3}$ | A1 | |
| $\frac{84}{\frac{2}{3}} = 126$ W | A1 | |
**Alternative method:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $t = \frac{18-12}{9} = \frac{2}{3}$ | M1, A1 | Using $v = u + at$ to find $t$; NB $a = 9$ from (b) |
| $\frac{84}{\frac{2}{3}} = 126$ W | A1 | |
### Part (d)(i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $17.28 = 1.6 \times v_Q$ | M1 | |
| $v_Q = 10.8$ m s$^{-1}$ | A1 | Do not allow $-10.8$ |
### Part (d)(ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2.4 \times 18 = 2.4 \times v_P + 1.6 \times 10.8$ | M1 | Attempt conservation of momentum |
| So $v_P = 10.8 = v_Q$, so the particles coalesce and the collision is therefore inelastic. | A1 | Find equal velocities and conclude; do not allow use of KE |
---4 A particle $P$ of mass 2.4 kg is moving in a straight line $O A$ on a horizontal plane. $P$ is acted on by a force of magnitude 30 N in the direction of motion. The distance $O A$ is 10 m .
\begin{enumerate}[label=(\alph*)]
\item Find the work done by this force as $P$ moves from $O$ to $A$.
The motion of $P$ is resisted by a constant force of magnitude $R \mathrm {~N}$. The velocity of $P$ increases from $12 \mathrm {~ms} ^ { - 1 }$ at $O$ to $18 \mathrm {~ms} ^ { - 1 }$ at $A$.
\item Find the value of $R$.
\item Find the average power used in overcoming the resistance force on $P$ as it moves from $O$ to $A$.
When $P$ reaches $A$ it collides directly with a particle $Q$ of mass 1.6 kg which was at rest at $A$ before the collision. The impulse exerted on $Q$ by $P$ as a result of the collision is 17.28 Ns .
\item \begin{enumerate}[label=(\roman*)]
\item Find the speed of $Q$ after the collision.
\item Hence show that the collision is inelastic.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics AS 2020 Q4 [12]}}