| Exam Board | OCR |
|---|---|
| Module | Further Mechanics AS (Further Mechanics AS) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | String becomes taut problem |
| Difficulty | Standard +0.3 This is a standard 'string becomes taut' problem requiring conservation of momentum and impulse-momentum calculations. While it involves multiple parts and Further Mechanics content, the solution follows a well-established template with straightforward application of formulas. The calculations are routine once the method is identified, making it slightly easier than average for A-level standard. |
| Spec | 6.03a Linear momentum: p = mv6.03b Conservation of momentum: 1D two particles6.03e Impulse: by a force6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(3.6 \times 7.2 = 3.6v_A + 2.4v_B\) | M1 | Conservation of momentum soi; \((25.92)\) |
| \(v_A = v_B\) | M1 | |
| \(4.32 \text{ ms}^{-1}\) | A1 | May be \(-4.32\) if initial velocity counted as negative |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\pm 3.6 \times 4.32 \mp 3.6 \times 7.2\) | M1 | Using their \(4.32\) from 2(a) provided c.o.m. used; Or \(-(2.4 \times 4.32)\) |
| \(-10.4 \text{ Ns}\) (or \(\text{kg ms}^{-1}\)) | A1 | Or \(10.4\text{ Ns}\) towards \(B\); Must be opposite sign to initial velocity; Deduct final mark if correct direction not soi |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\pm\left(\frac{1}{2} \times 3.6 \times 7.2^2 - \frac{1}{2} \times (3.6+2.4) \times 4.32^2\right)\) | M1 | Using their \(4.32\) from 2(a); Allow one slip in substitution other than sign error; must have 3 terms; \(93.31\ldots-(33.59\ldots+22.39\ldots)\) |
| \(37.3 \text{ J}\) | A1 | Allow \(-37.3\text{ J}\) |
## Question 2:
### Part (a)
| Answer | Mark | Guidance |
|--------|------|----------|
| $3.6 \times 7.2 = 3.6v_A + 2.4v_B$ | M1 | Conservation of momentum soi; $(25.92)$ |
| $v_A = v_B$ | M1 | |
| $4.32 \text{ ms}^{-1}$ | A1 | May be $-4.32$ if initial velocity counted as negative |
### Part (b)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\pm 3.6 \times 4.32 \mp 3.6 \times 7.2$ | M1 | Using their $4.32$ from 2(a) provided c.o.m. used; Or $-(2.4 \times 4.32)$ |
| $-10.4 \text{ Ns}$ (or $\text{kg ms}^{-1}$) | A1 | Or $10.4\text{ Ns}$ towards $B$; Must be opposite sign to initial velocity; Deduct final mark if correct direction not soi |
### Part (c)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\pm\left(\frac{1}{2} \times 3.6 \times 7.2^2 - \frac{1}{2} \times (3.6+2.4) \times 4.32^2\right)$ | M1 | Using their $4.32$ from 2(a); Allow one slip in substitution other than sign error; must have 3 terms; $93.31\ldots-(33.59\ldots+22.39\ldots)$ |
| $37.3 \text{ J}$ | A1 | Allow $-37.3\text{ J}$ |
---2 A particle $A$ of mass 3.6 kg is attached by a light inextensible string to a particle $B$ of mass 2.4 kg .\\
$A$ and $B$ are initially at rest, with the string slack, on a smooth horizontal surface. $A$ is projected directly away from $B$ with a speed of $7.2 \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Calculate the speed of $A$ after the string becomes taut.
\item Find the impulse exerted on $A$ at the instant that the string becomes taut.
\item Find the loss in kinetic energy as a result of the string becoming taut.
\end{enumerate}
\hfill \mbox{\textit{OCR Further Mechanics AS 2019 Q2 [7]}}