| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | October |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Two particles: different start times, same height |
| Difficulty | Standard +0.3 This is a standard two-particle SUVAT problem requiring students to set up equations for two stones with different initial conditions but the same displacement. It involves routine application of kinematic equations (s = ut + ½at²) and simultaneous equations, with convenient numbers (g = 9.8, initial speed 19.6) that simplify calculations. Slightly easier than average due to its straightforward structure and minimal algebraic manipulation required. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| VILU SIHI NI III M I ION OC | VIIV 5141 NI JINAM ION OC | VI4V SIHI NI JIIYM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| EITHER: \(h = -19.6(t+3) + \frac{1}{2}g(t+3)^2\) and \(h = \frac{1}{2}gt^2\) | M1A1A1 | Use of \(s = ut + \frac{1}{2}at^2\), correct no. of terms, condone sign errors |
| OR: \(h = -19.6T + \frac{1}{2}gT^2\) and \(h = \frac{1}{2}g(T-3)^2\) | M1A1A1 | Both A marks if \(s\) used instead of \(h\) consistently |
| \(-19.6T + \frac{1}{2}gT^2 = \frac{1}{2}g(T-3)^2\) OR \(-19.6(t+3) + \frac{1}{2}g(t+3)^2 = \frac{1}{2}gt^2\) | M1 | Eliminate \(h\) |
| \(T = 4.5\) | A1 (i) | |
| \(h = \frac{1}{2} \times 9.8 \times (T-3)^2\) oe | M1 | Use \(T\) or \(t\) value in equation |
| \(h = 11\) or \(11.0\) | A1 (ii)[7] |
# Question 3:
## Parts (i) and (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| **EITHER:** $h = -19.6(t+3) + \frac{1}{2}g(t+3)^2$ and $h = \frac{1}{2}gt^2$ | M1A1A1 | Use of $s = ut + \frac{1}{2}at^2$, correct no. of terms, condone sign errors |
| **OR:** $h = -19.6T + \frac{1}{2}gT^2$ and $h = \frac{1}{2}g(T-3)^2$ | M1A1A1 | Both A marks if $s$ used instead of $h$ consistently |
| $-19.6T + \frac{1}{2}gT^2 = \frac{1}{2}g(T-3)^2$ **OR** $-19.6(t+3) + \frac{1}{2}g(t+3)^2 = \frac{1}{2}gt^2$ | M1 | Eliminate $h$ |
| $T = 4.5$ | A1 (i) | |
| $h = \frac{1}{2} \times 9.8 \times (T-3)^2$ oe | M1 | Use $T$ or $t$ value in equation |
| $h = 11$ or $11.0$ | A1 (ii)[7] | |
---\begin{enumerate}
\item At time $t = 0$, a stone is thrown vertically upwards with speed $19.6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from a point $A$ which is $h$ metres above horizontal ground. At time $t = 3 \mathrm {~s}$, another stone is released from rest from a point $B$ which is also $h$ metres above the same horizontal ground. Both stones hit the ground at time $t = T$ seconds. The motion of each stone is modelled as that of a particle moving freely under gravity.
\end{enumerate}
Find\\
(i) the value of $T$,\\
(ii) the value of $h$.\\
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VILU SIHI NI III M I ION OC & VIIV 5141 NI JINAM ION OC & VI4V SIHI NI JIIYM ION OO \\
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\hfill \mbox{\textit{Edexcel M1 2018 Q3 [7]}}