| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Multi-stage motion with velocity-time graph given |
| Difficulty | Standard +0.3 This is a multi-part kinematics question using velocity-time graphs with constant acceleration. While it requires careful bookkeeping across multiple stages and uses both graphical interpretation and SUVAT equations, all techniques are standard M1 material with straightforward application. The question is slightly above average difficulty due to the multiple stages and need to track cumulative displacement, but involves no novel problem-solving or conceptual challenges. |
| Spec | 3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(\frac{1}{2}25v_m = 8\) or \(\frac{1}{2}Tv_m+\frac{1}{2}(25 - T)v_m =\) | B*1 | Do not accept solution based on isosceles or right angled triangle |
| Greatest speed is 0.64 m s\(^{-1}\) | D*B 2 1 | |
| (ii) \(V = 0.02 \times 40\) \(V = 0.8\) | M1 A1 A1 | For using v = u + at or the idea that gradient represents acceleration |
| (iii) \(\frac{1}{2}(70 + T) \times 0.8 = 40 - 8\) Duration is 10s | M1 A1ft A1 | For using the idea that the area represents displacement. nb trapezium area is 16+8+8 For A = \(\frac{1}{2}(L_1 + L_2)h\) or other appropriate breakdown \(\frac{1}{2}(30 + T) \times 0.8 = 40 - 8- \frac{1}{2} \times 40 \times 0.8\) ft cv(0.8) |
| (iv) \(0=0.8+a(30-10)\) Deceleration is \(0.04ms^{-2}\) Or \(40-8-\frac{1}{2} \times 40 \times 0.8-\) 10x0.8 \(=0.8(30-10)-a(30-10)^2/2\) Deceleration is \(0.04ms^{-2}\) | M1 A1ft A1 | For using v = u + at or the idea that gradient represents acceleration ft cv(10) and cv(0.8) Using the idea that the area represents displacement. ft cv(0.8 and 10) Accept -0.04 from correct work. d=-0.04 A0 |
(i) $\frac{1}{2}25v_m = 8$ or $\frac{1}{2}Tv_m+\frac{1}{2}(25 - T)v_m =$ | B*1 | Do not accept solution based on isosceles or right angled triangle
Greatest speed is 0.64 m s$^{-1}$ | D*B 2 1 |
(ii) $V = 0.02 \times 40$ $V = 0.8$ | M1 A1 A1 | For using v = u + at or the idea that gradient represents acceleration
(iii) $\frac{1}{2}(70 + T) \times 0.8 = 40 - 8$ Duration is 10s | M1 A1ft A1 | For using the idea that the area represents displacement. nb trapezium area is 16+8+8 For A = $\frac{1}{2}(L_1 + L_2)h$ or other appropriate breakdown $\frac{1}{2}(30 + T) \times 0.8 = 40 - 8- \frac{1}{2} \times 40 \times 0.8$ ft cv(0.8)
(iv) $0=0.8+a(30-10)$ Deceleration is $0.04ms^{-2}$ Or $40-8-\frac{1}{2} \times 40 \times 0.8-$ 10x0.8 $=0.8(30-10)-a(30-10)^2/2$ Deceleration is $0.04ms^{-2}$ | M1 A1ft A1 | For using v = u + at or the idea that gradient represents acceleration ft cv(10) and cv(0.8) Using the idea that the area represents displacement. ft cv(0.8 and 10) Accept -0.04 from correct work. d=-0.04 A0
---\includegraphics{figure_6}
The diagram shows the $(t, v)$ graph for the motion of a hoist used to deliver materials to different levels at a building site. The hoist moves vertically. The graph consists of straight line segments. In the first stage the hoist travels upwards from ground level for 25 s, coming to rest 8 m above ground level.
\begin{enumerate}[label=(\roman*)]
\item Find the greatest speed reached by the hoist during this stage. [2]
\end{enumerate}
The second stage consists of a 40 s wait at the level reached during the first stage. In the third stage the hoist continues upwards until it comes to rest 40 m above ground level, arriving 135 s after leaving ground level. The hoist accelerates at $0.02 \text{ m s}^{-2}$ for the first 40 s of the third stage, reaching a speed of $V \text{ m s}^{-1}$. Find
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item the value of $V$, [3]
\item the length of time during the third stage for which the hoist is moving at constant speed, [4]
\item the deceleration of the hoist in the final part of the third stage. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR M1 2007 Q6 [12]}}