| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Two particles: same start time, different heights |
| Difficulty | Standard +0.3 This is a standard M1 two-particle collision problem requiring SUVAT equations for both stones and solving simultaneous equations. The question helpfully guides students to equate speeds, making it slightly easier than average. The algebra is straightforward once the setup is correct, and all three unknowns can be found systematically. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\uparrow v_A = 29.4 - 9.8T\), \(\downarrow v_B = 9.8T\) | M1 | Either attempted. Allow sign errors and \(g=9.81\) etc |
| Both correct | A1 | |
| For same speed: \(29.4 - 9.8T = 9.8T\) | M1 | Attempt to equate. Accept sign errors and \(T=1.5\) substituted in both |
| \(T = 1.5\) | E1 | If 2 subs there must be a statement about equality |
| \(V = 14.7\) | F1 | FT \(T\) or \(V\), whichever found second |
| \(H = 29.4\times1.5 - 0.5\times9.8\times1.5^2 + 0.5\times9.8\times1.5^2\) | M1 | Sum of distances travelled by each attempted |
| \(= 44.1\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(V^2 = 29.4^2 - 2\times9.8\times x = 2\times9.8\times(H-x)\) | M1 | Attempts at \(V^2\) for each particle equated. Allow sign errors, \(9.81\) etc. Allow \(h_1, h_2\) without \(h_1 = H - h_2\) |
| Both correct, requiring \(h_1 = H - h_2\) but not an equation | B1 | |
| \(29.4^2 = 19.6H\) so \(H = 44.1\) | A1 | cao |
| Relative velocity is \(29.4\) | M1 | Any method leading to \(T\) or \(V\) |
| \(T = \frac{44.1}{29.4}\) | E1 | |
| Using \(v = u + at\), \(V = 0 + 9.8\times1.5 = 14.7\) | M1, F1 | Any method leading to other variable |
| 7 |
# Question 6:
**Method 1:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\uparrow v_A = 29.4 - 9.8T$, $\downarrow v_B = 9.8T$ | M1 | Either attempted. Allow sign errors and $g=9.81$ etc |
| Both correct | A1 | |
| For same speed: $29.4 - 9.8T = 9.8T$ | M1 | Attempt to equate. Accept sign errors and $T=1.5$ substituted in both |
| $T = 1.5$ | E1 | If 2 subs there must be a statement about equality |
| $V = 14.7$ | F1 | FT $T$ or $V$, whichever found second |
| $H = 29.4\times1.5 - 0.5\times9.8\times1.5^2 + 0.5\times9.8\times1.5^2$ | M1 | Sum of distances travelled by each attempted |
| $= 44.1$ | A1 | cao |
**Method 2:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $V^2 = 29.4^2 - 2\times9.8\times x = 2\times9.8\times(H-x)$ | M1 | Attempts at $V^2$ for each particle equated. Allow sign errors, $9.81$ etc. Allow $h_1, h_2$ without $h_1 = H - h_2$ |
| Both correct, requiring $h_1 = H - h_2$ but not an equation | B1 | |
| $29.4^2 = 19.6H$ so $H = 44.1$ | A1 | cao |
| Relative velocity is $29.4$ | M1 | Any method leading to $T$ or $V$ |
| $T = \frac{44.1}{29.4}$ | E1 | |
| Using $v = u + at$, $V = 0 + 9.8\times1.5 = 14.7$ | M1, F1 | Any method leading to other variable |
| | **7** | |
---6 Small stones A and B are initially in the positions shown in Fig. 6 with B a height $H \mathrm {~m}$ directly above A.
\begin{figure}[h]
\begin{center}
\begin{tikzpicture}[>=latex]
% Horizontal dashed lines
\draw[dashed] (-1.2,3) -- (2.5,3);
\draw[dashed] (-1.2,0) -- (2.5,0);
% Vertical line connecting A and B
\draw[dashed] (0,-0.5) -- (0,3.5);
% Point B (top)
\fill (0,3) circle (4pt);
\node[above left] at (0,3) {B};
% Point A (bottom)
\fill (0,0) circle (4pt);
\node[below left] at (0,0) {A};
% Dimension arrow for H m
\draw[<->] (2,3) -- node[right] {$H$\,m} (2,0);
\end{tikzpicture}
\captionsetup{labelformat=empty}
\caption{Fig. 6}
\end{center}
\end{figure}
At the instant when B is released from rest, A is projected vertically upwards with a speed of $29.4 \mathrm {~m} \mathrm {~s} \mathrm {~s} ^ { - 1 }$. Air resistance may be neglected.
The stones collide $T$ seconds after they begin to move. At this instant they have the same speed, $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$, and A is still rising.
By considering when the speed of A upwards is the same as the speed of B downwards, or otherwise, show that $T = 1.5$ and find the values of $V$ and $H$.
Section B (36 marks)\\
\hfill \mbox{\textit{OCR MEI M1 2009 Q6 [7]}}