| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 8 |
| 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 systematic application of kinematic equations. Part (i) is routine (v²=u²+2as). Part (ii) requires setting up equations for both particles and solving simultaneously, which is a common textbook exercise type. The 'show that' format provides the answer, reducing problem-solving demand. Slightly above average due to the two-particle coordination, but well within standard M1 scope. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.02h Motion under gravity: vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Height reached by first particle: \(0 = 21^2 - 2 \times 9.8 \times s\) | M1 | Other methods must be complete. Allow \(g = \pm 9.8, \pm 10\) |
| So \(s = 22.5\), so 22.5 m | A1 | Accept with consistent signs |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t\) seconds after second particle projected, height is \(15t - 4.9t^2\) | M1 | Allow \(g = \pm 9.8, \pm 10\) |
| A1 | ||
| First particle has height \(22.5 - 4.9t^2\) (or \(21t - 4.9t^2\)) | M1 | Allow \(g = \pm 9.8, \pm 10\) |
| A1 | Award only if used correctly | |
| Either: Sub \(t = 1.5\) to show both have same value; state height as 11.475 m | E1, A1 | (or sub \(t = 3.64\) into \(21t - 4.9t^2\) for 1st & \(t = 1.5\) for 2nd); cao; accept any reasonable accuracy; don't award if only one correctly used equation obtained |
| Or: \(15t - 4.9t^2 = 22.5 - 4.9t^2\) giving \(t = 1.5\) and height as 11.475 m | M1, A1 | Both \(t\) shown; ht cao (to any reasonable accuracy) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(t\) seconds after second particle projected, height is \(15t - 4.9t^2\) | M1 | Allow \(g = \pm 9.8, \pm 10\) |
| A1 | ||
| First particle has fallen \(4.9t^2\) | B1 | |
| Collide when \(15T - 4.9T^2 + 4.9T^2 = 22.5\), so \(T = 1.5\) | M1, E1 | Or other correct method |
| \(H = 22.5 - 4.9 \times 1.5^2 = 11.475\) m | A1 | cao; accept any reasonable accuracy; don't award if only one correctly used equation obtained |
# Question 2:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Height reached by first particle: $0 = 21^2 - 2 \times 9.8 \times s$ | M1 | Other methods must be complete. Allow $g = \pm 9.8, \pm 10$ |
| So $s = 22.5$, so 22.5 m | A1 | Accept with consistent signs |
## Part (ii) — Solution 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t$ seconds after second particle projected, height is $15t - 4.9t^2$ | M1 | Allow $g = \pm 9.8, \pm 10$ |
| | A1 | |
| First particle has height $22.5 - 4.9t^2$ (or $21t - 4.9t^2$) | M1 | Allow $g = \pm 9.8, \pm 10$ |
| | A1 | Award only if used correctly |
| **Either:** Sub $t = 1.5$ to show both have same value; state height as 11.475 m | E1, A1 | (or sub $t = 3.64$ into $21t - 4.9t^2$ for 1st & $t = 1.5$ for 2nd); cao; accept any reasonable accuracy; don't award if only one correctly used equation obtained |
| **Or:** $15t - 4.9t^2 = 22.5 - 4.9t^2$ giving $t = 1.5$ and height as 11.475 m | M1, A1 | Both $t$ shown; ht cao (to any reasonable accuracy) |
## Part (ii) — Solution 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $t$ seconds after second particle projected, height is $15t - 4.9t^2$ | M1 | Allow $g = \pm 9.8, \pm 10$ |
| | A1 | |
| First particle has fallen $4.9t^2$ | B1 | |
| Collide when $15T - 4.9T^2 + 4.9T^2 = 22.5$, so $T = 1.5$ | M1, E1 | Or other correct method |
| $H = 22.5 - 4.9 \times 1.5^2 = 11.475$ m | A1 | cao; accept any reasonable accuracy; don't award if only one correctly used equation obtained |
---2 A particle is projected vertically upwards from a point O at $21 \mathrm {~ms} ^ { - 1 }$.\\
(i) Calculate the greatest height reached by the particle.
When this particle is at its highest point, a second particle is projected vertically upwards from $O$ at $15 \mathrm {~ms} ^ { - 1 }$.\\
(ii) Show that the particles collide 1.5 seconds later and determine the height above O at which the collision takes place.
\hfill \mbox{\textit{OCR MEI M1 Q2 [8]}}