| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Vertical motion under gravity |
| Difficulty | Moderate -0.8 This is a straightforward SUVAT question requiring direct application of kinematic equations with constant acceleration. Part (i) uses standard formulas for velocity and displacement; part (ii) requires recognizing the particle reaches maximum height before 0.9s and calculating up-and-down distances separately, but this is a routine textbook exercise with no conceptual challenges beyond basic mechanics. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v = 7 - 0.4 \times 9.8\) | M1 | \(v = 7 +/- 0.4g\) |
| \(v = 3.08 \text{ ms}^{-1}\) | A1 | Exact, or correct to 3sf from \(g=9.81\) (3.076) or 10 (3) |
| \(s = 7 \times 0.4 - 9.8 \times 0.4^2/2\) | M1 | \(s = 7 \times 0.4 +/- g0.4^2/2\) |
| \(s = 2.016 \text{ m}\) | A1 | Exact but accept 2.02. \(g=9.81\) (2.0152) or \(g=10\) (2) |
| OR \(3.08^2 = 7^2 - 2 \times 9.8s\) | M1 | \((\text{cv}(v))^2 = 7^2 +/- 2gs\) |
| \(s = 2.016 \text{ m}\) | A1 | Exact but accept 2.02. \(g=9.81\) (2.0152) or \(g=10\) (2) |
| OR \(v^2 = 7^2 - 2 \times 9.8 \times 2.016\) | M1 | \(v^2 = 7^2 +/- 2g(\text{cv}(s))\) |
| \(v = 3.08 \text{ ms}^{-1}\) | A1 | Exact or correct to 3sf. Accept \(v=3.07\) from \(s=2.02\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(H = \pm 7^2/(2 \times 9.8)\) \((= \pm 2.5)\) | B1 | Greatest Height, \(g=9.81\) (2.497 accept 2.5) \(g=10\) (2.45) |
| \(S = \pm(7 \times 0.9 - \frac{1}{2} \times 9.8 \times 0.9^2)\) \((= \pm 2.331)\) | B1 | Height when \(t = 0.9\), \(g=9.81\) (2.32695) \(g=10\) (2.25) |
| \(D = 2.5 + (2.5 - 2.331)\) | M1 | \(2 \times\) greatest height \(- S(0.9)\) |
| \(D = 2.669 \text{ m}\) | A1 | Exact but accept 2.67, \(g=9.81\) (2.66705) \(g=10\) (2.65) |
| OR using \(t_U = 7/9.8=0.7143\), \(t_D = 0.9-0.7143= 0.1857\) s | [4] | "OR" method uses distance from greatest height |
| \(H = \pm(7\times0.7143 - 9.8\times0.7143^2/2)\) \((= \pm 2.5)\) | B1 | OR \(\pm 9.8\times0.7143^2/2\). Gains B1 for \(H\) as above |
| \(s_D = \pm 9.8\times0.1857^2/2\) \((= \pm0.169)\) | B1 | Equivalent to B1 for \(S\) as above |
| \(D = 2.5 + 0.169\) | M1 | Greatest height + Descent distance \(<< H\) |
| \(D = 2.669 \text{ m}\) | A1 | Exact but accept 2.67, \(g=9.81\) (2.66705) \(g=10\) (2.65) |
# Question 2:
## Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = 7 - 0.4 \times 9.8$ | M1 | $v = 7 +/- 0.4g$ |
| $v = 3.08 \text{ ms}^{-1}$ | A1 | Exact, or correct to 3sf from $g=9.81$ (3.076) or 10 (3) |
| $s = 7 \times 0.4 - 9.8 \times 0.4^2/2$ | M1 | $s = 7 \times 0.4 +/- g0.4^2/2$ |
| $s = 2.016 \text{ m}$ | A1 | Exact but accept 2.02. $g=9.81$ (2.0152) or $g=10$ (2) |
| OR $3.08^2 = 7^2 - 2 \times 9.8s$ | M1 | $(\text{cv}(v))^2 = 7^2 +/- 2gs$ |
| $s = 2.016 \text{ m}$ | A1 | Exact but accept 2.02. $g=9.81$ (2.0152) or $g=10$ (2) |
| OR $v^2 = 7^2 - 2 \times 9.8 \times 2.016$ | M1 | $v^2 = 7^2 +/- 2g(\text{cv}(s))$ |
| $v = 3.08 \text{ ms}^{-1}$ | A1 | Exact or correct to 3sf. Accept $v=3.07$ from $s=2.02$ |
## Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $H = \pm 7^2/(2 \times 9.8)$ $(= \pm 2.5)$ | B1 | Greatest Height, $g=9.81$ (2.497 accept 2.5) $g=10$ (2.45) |
| $S = \pm(7 \times 0.9 - \frac{1}{2} \times 9.8 \times 0.9^2)$ $(= \pm 2.331)$ | B1 | Height when $t = 0.9$, $g=9.81$ (2.32695) $g=10$ (2.25) |
| $D = 2.5 + (2.5 - 2.331)$ | M1 | $2 \times$ greatest height $- S(0.9)$ |
| $D = 2.669 \text{ m}$ | A1 | Exact but accept 2.67, $g=9.81$ (2.66705) $g=10$ (2.65) |
| OR using $t_U = 7/9.8=0.7143$, $t_D = 0.9-0.7143= 0.1857$ s | [4] | "OR" method uses distance from greatest height |
| $H = \pm(7\times0.7143 - 9.8\times0.7143^2/2)$ $(= \pm 2.5)$ | B1 | OR $\pm 9.8\times0.7143^2/2$. Gains B1 for $H$ as above |
| $s_D = \pm 9.8\times0.1857^2/2$ $(= \pm0.169)$ | B1 | Equivalent to B1 for $S$ as above |
| $D = 2.5 + 0.169$ | M1 | Greatest height + Descent distance $<< H$ |
| $D = 2.669 \text{ m}$ | A1 | Exact but accept 2.67, $g=9.81$ (2.66705) $g=10$ (2.65) |
---2 A particle is projected vertically upwards with speed $7 \mathrm {~ms} ^ { - 1 }$ from a point on the ground.\\
(i) Find the speed of the particle and its distance above the ground 0.4 s after projection.\\
(ii) Find the total distance travelled by the particle in the first 0.9 s after projection.
\hfill \mbox{\textit{OCR M1 2012 Q2 [8]}}