| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Particle on inclined plane |
| Difficulty | Standard +0.3 This is a standard two-equation SUVAT problem requiring students to set up simultaneous equations from two time intervals, solve for u and a, then use a=g sin α. While it involves multiple steps and algebraic manipulation, the approach is methodical and commonly practiced in M1, making it slightly easier than average. |
| Spec | 3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| For using \(s = ut + \frac{1}{2}at^2\) for the first stage | M1 | |
| \(2 = 0.8u + \frac{1}{2}a(0.8)^2\) | A1 | |
| For obtaining another equation in \(u\) and \(a\) | M1 | With relevant values of velocity, displacement and time |
| \(8 = 2u + \frac{1}{2}a \cdot 2^2\) or \(6 = 1.2(u+0.8a)+\frac{1}{2}a(1.2)^2\) or \(6 = 1.2(2\times2\div0.8-u)+\frac{1}{2}a(1.2)^2\) | A1 | |
| For eliminating \(a\) or \(u\) | M1 | |
| \(u = 1.5\) | A1 | |
| Acceleration is \(2.5\) ms\(^{-2}\) | A1 | Total: 7 |
| Answer | Marks | Guidance |
|---|---|---|
| For using \(s = vt - \frac{1}{2}at^2\) for the first stage | M1 | |
| \(2 = 0.8v - \frac{1}{2}a(0.8)^2\) | A1 | |
| For using \(s = ut + \frac{1}{2}at^2\) for the second stage | M1 | |
| \(6 = 1.2v + \frac{1}{2}a(1.2)^2\) | A1 | |
| For obtaining values of \(a\) and \(v\) and using \(v = u + at\) for first stage to find \(u\) | M1 | |
| Acceleration is \(2.5\) ms\(^{-2}\) \((v=3.5)\) | A1 | |
| \(u = 1.5\) | A1 | Total: 7 |
| Answer | Marks | Guidance |
|---|---|---|
| \(2\div0.8\) ms\(^{-1}\) and \(6\div1.2\) ms\(^{-1}\) | M1 | For finding average speeds in both intervals |
| \(= 2.5\) ms\(^{-1}\) and \(5\) ms\(^{-1}\) | A1 | |
| \(t_1 = 0.4\) and \(t_2 = (0.8+)0.6\) | B1 | For finding mid-interval times |
| \(5 = 2.5 + a(1.4 - 0.4)\) | M1 | For using \(v = u + at\) between the mid-interval times |
| Acceleration is \(2.5\) ms\(^{-2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(2.5 = u + 2.5\times0.4\) or \(5 = u + 2.5\times1.4\) | M1 | For using \(v = u + at\) between \(t=0\) and one of the mid-interval times |
| \(u = 1.5\) | A1 | Total: 7 |
| Answer | Marks | Guidance |
|---|---|---|
| \(2.5 = 9.8\sin\alpha\) | M1 | For using \((m)a = (m)g\sin\alpha\) |
| \(\alpha = 14.8°\) | A1ft | ft value of acceleration |
# Question 4:
## Part (i) α
| For using $s = ut + \frac{1}{2}at^2$ for the first stage | M1 | |
| $2 = 0.8u + \frac{1}{2}a(0.8)^2$ | A1 | |
| For obtaining another equation in $u$ and $a$ | M1 | With relevant values of velocity, displacement and time |
| $8 = 2u + \frac{1}{2}a \cdot 2^2$ or $6 = 1.2(u+0.8a)+\frac{1}{2}a(1.2)^2$ or $6 = 1.2(2\times2\div0.8-u)+\frac{1}{2}a(1.2)^2$ | A1 | |
| For eliminating $a$ or $u$ | M1 | |
| $u = 1.5$ | A1 | |
| Acceleration is $2.5$ ms$^{-2}$ | A1 | **Total: 7** |
## Part (i) β
| For using $s = vt - \frac{1}{2}at^2$ for the first stage | M1 | |
| $2 = 0.8v - \frac{1}{2}a(0.8)^2$ | A1 | |
| For using $s = ut + \frac{1}{2}at^2$ for the second stage | M1 | |
| $6 = 1.2v + \frac{1}{2}a(1.2)^2$ | A1 | |
| For obtaining values of $a$ and $v$ and using $v = u + at$ for first stage to find $u$ | M1 | |
| Acceleration is $2.5$ ms$^{-2}$ $(v=3.5)$ | A1 | |
| $u = 1.5$ | A1 | **Total: 7** |
## Part (i) γ
| $2\div0.8$ ms$^{-1}$ and $6\div1.2$ ms$^{-1}$ | M1 | For finding average speeds in both intervals |
| $= 2.5$ ms$^{-1}$ and $5$ ms$^{-1}$ | A1 | |
| $t_1 = 0.4$ and $t_2 = (0.8+)0.6$ | B1 | For finding mid-interval times |
| $5 = 2.5 + a(1.4 - 0.4)$ | M1 | For using $v = u + at$ between the mid-interval times |
| Acceleration is $2.5$ ms$^{-2}$ | A1 | |
## Part (i) continued (γ extension)
| $2.5 = u + 2.5\times0.4$ or $5 = u + 2.5\times1.4$ | M1 | For using $v = u + at$ between $t=0$ and one of the mid-interval times |
| $u = 1.5$ | A1 | **Total: 7** |
## Part (ii)
| $2.5 = 9.8\sin\alpha$ | M1 | For using $(m)a = (m)g\sin\alpha$ |
| $\alpha = 14.8°$ | A1ft | ft value of acceleration | **Total: 2** |
---4 A particle moves downwards on a smooth plane inclined at an angle $\alpha$ to the horizontal. The particle passes through the point $P$ with speed $u \mathrm {~ms} ^ { - 1 }$. The particle travels 2 m during the first 0.8 s after passing through $P$, then a further 6 m in the next 1.2 s . Find\\
(i) the value of $u$ and the acceleration of the particle,\\
(ii) the value of $\alpha$ in degrees.
\hfill \mbox{\textit{OCR M1 2005 Q4 [9]}}