OCR MEI AS Paper 1 2023 June — Question 12 8 marks

Exam BoardOCR MEI
ModuleAS Paper 1 (AS Paper 1)
Year2023
SessionJune
Marks8
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicConstant acceleration (SUVAT)
TypeFind acceleration from distances/times
DifficultyStandard +0.3 This is a two-stage kinematics problem with Newton's second law requiring students to find acceleration in each phase, use SUVAT to find velocity at B, then calculate distances and sum them. While it involves multiple steps and two different force scenarios, the approach is methodical and follows standard mechanics procedures without requiring novel insight.
Spec3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension

12 Points A, B and C lie in a straight line in that order on horizontal ground. A box of mass 5 kg is pushed from A to C by a horizontal force of magnitude 8 N . The box is at rest at A and takes 3 seconds to reach B . The ground is smooth between A and B . Between B and C the ground is rough and the resistance to motion is 28 N . The box comes to rest at C . Determine the distance AC.

Question 12:
AnswerMarks Guidance
Answer/WorkingMark Guidance
For AB, Newton's second law \(8 = 5a\); acceleration is \(1.6 \text{ ms}^{-2}\)B1 3.1b — Uses Newton's second law to calculate acceleration
For AB: \(u=0\), \(a=1.6\), \(t=3\); \(s = \dfrac{1}{2}at^2 = \dfrac{1}{2}\times 1.6 \times 3^2 = 7.2\) mM1 1.1a — Uses suvat equation(s) and their \(a\) leading to a value for \(s\)
Velocity at B: \(v = at = 1.6 \times 3 = 4.8 \text{ ms}^{-1}\)M1 3.1b — Uses suvat equation(s) and their \(a\) leading to a value for velocity at B
For BC Newton's second law \(8 - 28 = 5a\); acceleration is \(-4 \text{ ms}^{-2}\)M1 3.1b — Uses Newton's second law to calculate acceleration. Condone missing 8N force. Allow sign errors
A11.1b — soi
For BC: \(u=4.8\), \(v=0\), \(a=-4\); \(0^2 = 4.8^2 - 2\times 4s\)M1 1.1a — Uses suvat equation(s) and their \(a\) leading to a value for \(s\)
\(s = 2.88\) mA1 1.1b — FT their negative \(a\) and their positive velocity at B
Distance AC is \(7.2 + 2.88 = 10.08\) mA1 1.1b — Allow 10 m
Total: [8]
## Question 12:

| Answer/Working | Mark | Guidance |
|---|---|---|
| For AB, Newton's second law $8 = 5a$; acceleration is $1.6 \text{ ms}^{-2}$ | B1 | 3.1b — Uses Newton's second law to calculate acceleration |
| For AB: $u=0$, $a=1.6$, $t=3$; $s = \dfrac{1}{2}at^2 = \dfrac{1}{2}\times 1.6 \times 3^2 = 7.2$ m | M1 | 1.1a — Uses suvat equation(s) and their $a$ leading to a value for $s$ |
| Velocity at B: $v = at = 1.6 \times 3 = 4.8 \text{ ms}^{-1}$ | M1 | 3.1b — Uses suvat equation(s) and their $a$ leading to a value for velocity at B |
| For BC Newton's second law $8 - 28 = 5a$; acceleration is $-4 \text{ ms}^{-2}$ | M1 | 3.1b — Uses Newton's second law to calculate acceleration. Condone missing 8N force. Allow sign errors |
| | A1 | 1.1b — soi |
| For BC: $u=4.8$, $v=0$, $a=-4$; $0^2 = 4.8^2 - 2\times 4s$ | M1 | 1.1a — Uses suvat equation(s) and their $a$ leading to a value for $s$ |
| $s = 2.88$ m | A1 | 1.1b — FT their negative $a$ and their positive velocity at B |
| Distance AC is $7.2 + 2.88 = 10.08$ m | A1 | 1.1b — Allow 10 m |

**Total: [8]**
12 Points A, B and C lie in a straight line in that order on horizontal ground. A box of mass 5 kg is pushed from A to C by a horizontal force of magnitude 8 N . The box is at rest at A and takes 3 seconds to reach B . The ground is smooth between A and B . Between B and C the ground is rough and the resistance to motion is 28 N . The box comes to rest at C .

Determine the distance AC.

\hfill \mbox{\textit{OCR MEI AS Paper 1 2023 Q12 [8]}}