| Exam Board | Pre-U |
|---|---|
| Module | Pre-U 9794/3 (Pre-U Mathematics Paper 3) |
| Year | 2017 |
| Session | June |
| Marks | 11 |
| Topic | Forces, equilibrium and resultants |
| Type | Lift with occupant problems |
| Difficulty | Moderate -0.3 This is a standard multi-part mechanics question requiring Newton's second law applied to three different acceleration scenarios, basic kinematics, and velocity-time graph interpretation. While it involves several steps and parts, each component uses routine A-level mechanics techniques with no novel problem-solving required. The calculations are straightforward applications of F=ma and SUVAT equations, making it slightly easier than average. |
| Spec | 3.02b Kinematic graphs: displacement-time and velocity-time3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension3.03f Weight: W=mg |
**Question 6(i)**
- Diagram of crate with weight and tension in the cable shown. **B1**
**Question 6(ii)**
- $220a = 220g - T \therefore T = 220(10 - a)$ **M1** (Correct application of N2 used at least once.)
- $a = 1.5 \quad \therefore T = 1870$ (N) **A1** (cao)
- $a = 0 \quad \therefore T = 2200$ (N) **B1** (cao)
- $a = -0.75 \quad \therefore T = 2365$ (N) **A1** (cao)
**Question 6(iii)**
- Trapezium (middle portion horizontal), one vertex at the origin, fourth vertex on the $t$ axis. **B1**
- Third part less steep than first. Axes labelled $t$ and $v$; horizontal section at $v = 3$. **B1**
**Question 6(iv)**
- Acceleration and deceleration stages take $2 + 4 = 6$ sec. **B1** (For acceleration time.) **B1** (For deceleration time.)
- If $t =$ total time of descent then $s = \dfrac{1}{2} \times 3(t + (t-6)) = 15$ **B1** (Area of trapezium $\ldots$)
- $\ldots$ equated to 15. **M1**
- $\therefore t = 8$ (sec) **A1** (cao)
**Total: 11 marks**6 A crate, which has a mass of 220 kg , is being lowered on the end of a cable onto the back of a lorry.\\
(i) Draw a diagram to show the forces acting on the crate.
The crate is lowered in three stages.\\
Stage 1 It starts from rest and accelerates at $1.5 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ until it reaches a speed of $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
Stage 2 It descends at a constant speed of $3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
Stage 3 It decelerates at $0.75 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ and eventually comes to rest.\\
(ii) Find the tension in the cable in each of the three stages.\\
(iii) Sketch the velocity-time graph for the complete downward motion of the crate.\\
(iv) The crate is lowered 15 m altogether. By considering your velocity-time graph, find the total time taken.
\hfill \mbox{\textit{Pre-U Pre-U 9794/3 2017 Q6 [11]}}