| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Displacement expressions and comparison |
| Difficulty | Standard +0.3 This is a standard M1 two-particle meeting problem requiring SUVAT equations and careful tracking of different motion phases. While it involves multiple parts and requires systematic working through each stage, the techniques are routine and the problem structure is typical for this topic with no novel insights needed. |
| Spec | 3.02d Constant acceleration: SUVAT formulae6.02a Work done: concept and definition |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(A\) cycles \((= 20 \times 1) = 20\) km | B1 | |
| \(B\) walks \(= 20/4\) h | M1 | |
| Time \(= 5\) hours | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(20 \times 1 + 5(T-1)\) | B1 | Total \(A\) or \(B\) distance correct; Accept cv(5) for time |
| \(= 4 \times 5 + 15(T-5)\) | M1 | Equates total distances for \(A\) and \(B\); Using \(t\) instead of \((T-5)\) and finding \(t=2\) gets B1 M1 |
| \(T = 7\) | A1 | |
| *OR* | ||
| \(5(T-1)\) | B1 | A walking distance; Needs consistency of T (or t) for M1 |
| \(= 15(T-5)\) | M1 | Equates \(A\) walking and \(B\) cycling distances |
| \(T = 7\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Total distance \((A) = 20 \times 1 + 5(7-1)\) | M1 | Or \((B)\ 4 \times 5 + 15 \times (7-5)\); cv(7) and, for B cv(5) |
| \(J = 50\) km | A1 |
## Question 3:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $A$ cycles $(= 20 \times 1) = 20$ km | B1 | |
| $B$ walks $= 20/4$ h | M1 | |
| Time $= 5$ hours | A1 | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $20 \times 1 + 5(T-1)$ | B1 | Total $A$ or $B$ distance correct; Accept cv(5) for time |
| $= 4 \times 5 + 15(T-5)$ | M1 | Equates total distances for $A$ and $B$; Using $t$ instead of $(T-5)$ and finding $t=2$ gets B1 M1 |
| $T = 7$ | A1 | |
| *OR* | | |
| $5(T-1)$ | B1 | A walking distance; Needs consistency of T (or t) for M1 |
| $= 15(T-5)$ | M1 | Equates $A$ walking and $B$ cycling distances |
| $T = 7$ | A1 | |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| Total distance $(A) = 20 \times 1 + 5(7-1)$ | M1 | Or $(B)\ 4 \times 5 + 15 \times (7-5)$; cv(7) and, for B cv(5) |
| $J = 50$ km | A1 | |
---(i) Calculate the distance $A$ cycles, and hence find the period of time for which $B$ walks before finding the bicycle.\\
(ii) Find $T$.\\
(iii) Calculate the distance $A$ and $B$ each travel.
\hfill \mbox{\textit{OCR M1 2015 Q3 [8]}}