| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| 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 straightforward SUVAT problem requiring students to set up two equations (one for each stone) with the constraint that they hit the ground simultaneously. It involves standard application of s = ut + ½at² with clear given values, requiring algebraic manipulation but no novel insight—slightly easier than average due to its routine nature. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance Notes |
| M1 | Use \(s = ut + \frac{1}{2}at^2\) or complete suvat route to find \(h\) in terms of \(t\) | |
| \(h = \frac{1}{2}gt^2\) | A1 | Or \(h = \frac{1}{2}g(t+1)^2\). Expression for time used in first equation defines expression expected in second equation |
| \(h = 19.6(t-1) + \frac{1}{2}g(t-1)^2\) | A1 | Or \(h = 19.6(t) + \frac{1}{2}g(t)^2\) or \(h = 4.9 + \left(9.8t + \frac{1}{2}gt^2\right)\) |
| \(\frac{1}{2}gt^2 = 19.6(t-1) + \frac{1}{2}g(t-1)^2\) | M1 | Equate the two expressions for \(h\) |
| DM1 | Solve for \(t\). Dependent on previous M1 | |
| \(t = 1.5\) | A1 | Using the "Or" approach gives \(t = 0.5\) |
| \(h = 11\) m or \(11.0\) m | A1 | Accept 2 or 3 s.f. only |
| Total: 7 |
# Question 4:
| Answer/Working | Marks | Guidance Notes |
|---|---|---|
| | M1 | Use $s = ut + \frac{1}{2}at^2$ or complete suvat route to find $h$ in terms of $t$ |
| $h = \frac{1}{2}gt^2$ | A1 | Or $h = \frac{1}{2}g(t+1)^2$. Expression for time used in first equation defines expression expected in second equation |
| $h = 19.6(t-1) + \frac{1}{2}g(t-1)^2$ | A1 | Or $h = 19.6(t) + \frac{1}{2}g(t)^2$ or $h = 4.9 + \left(9.8t + \frac{1}{2}gt^2\right)$ |
| $\frac{1}{2}gt^2 = 19.6(t-1) + \frac{1}{2}g(t-1)^2$ | M1 | Equate the two expressions for $h$ |
| | DM1 | Solve for $t$. Dependent on previous M1 |
| $t = 1.5$ | A1 | Using the "Or" approach gives $t = 0.5$ |
| $h = 11$ m or $11.0$ m | A1 | Accept 2 or 3 s.f. only |
| **Total: 7** | | |
---\begin{enumerate}
\item A small stone is released from rest from a point $A$ which is at height $h$ metres above horizontal ground. Exactly one second later another small stone is projected with speed $19.6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ vertically downwards from a point $B$, which is also at height $h$ metres above the horizontal ground. The motion of each stone is modelled as that of a particle moving freely under gravity. The two stones hit the ground at the same time.
\end{enumerate}
Find the value of $h$.\\
\hfill \mbox{\textit{Edexcel M1 2015 Q4 [7]}}