| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Constant acceleration (SUVAT) |
| Type | Read and interpret velocity-time graph |
| Difficulty | Standard +0.3 This is a standard M1 mechanics question combining kinematics (reading a velocity-time graph for distance and acceleration) with Newton's second law applied to a two-body system. While it requires multiple steps and careful consideration of forces in different stages, all techniques are routine: area under v-t graph, gradient for acceleration, and F=ma with weight. The two-body aspect (elevator + box) adds slight complexity but follows standard textbook methods. Slightly above average due to the multi-stage nature and need to track forces carefully across different accelerations. |
| Spec | 3.02c Interpret kinematic graphs: gradient and area3.02d Constant acceleration: SUVAT formulae3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([s = \frac{1}{2} \times 5 \times 0.4 + 19 \times 0.4 + \frac{1}{2} \times 4 \times 0.4]\) | M1 | For using the area property for distance |
| Distance \(= 9.4\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Acceleration is \(0.08 \text{ ms}^{-2}\) | B1 | |
| Deceleration is \(0.1 \text{ ms}^{-2}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([T - (800 + 100)g = (800 + 100)a]\) | M1 | For applying Newton's 2nd law to the elevator and box |
| \(T - 900g = 900a\) | A1 | |
| \(T = 9072\) N in 1st stage, \(T = 9000\) N in 2nd stage, \(T = 8910\) N in 3rd stage | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([R - 100g = 100a]\) | M1 | For applying Newton's 2nd law to the box |
| \(R = 1008\) N | A1 | For obtaining the greatest value of the force on the box |
| \(R = 990\) N | A1 | For obtaining the least value of the force on the box |
# Question 7 (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[s = \frac{1}{2} \times 5 \times 0.4 + 19 \times 0.4 + \frac{1}{2} \times 4 \times 0.4]$ | M1 | For using the area property for distance |
| Distance $= 9.4$ | A1 | |
---
# Question 7 (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Acceleration is $0.08 \text{ ms}^{-2}$ | B1 | |
| Deceleration is $0.1 \text{ ms}^{-2}$ | B1 | |
---
# Question 7 (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[T - (800 + 100)g = (800 + 100)a]$ | M1 | For applying Newton's 2nd law to the elevator and box |
| $T - 900g = 900a$ | A1 | |
| $T = 9072$ N in 1st stage, $T = 9000$ N in 2nd stage, $T = 8910$ N in 3rd stage | A1 | |
---
# Question 7 (iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[R - 100g = 100a]$ | M1 | For applying Newton's 2nd law to the box |
| $R = 1008$ N | A1 | For obtaining the greatest value of the force on the box |
| $R = 990$ N | A1 | For obtaining the least value of the force on the box |7\\
\includegraphics[max width=\textwidth, alt={}, center]{79b90ef5-ef3a-4c59-b662-d0fbfba813ca-4_492_1365_255_392}
An elevator is pulled vertically upwards by a cable. The velocity-time graph for the motion is shown above. Find\\
(i) the distance travelled by the elevator,\\
(ii) the acceleration during the first stage and the deceleration during the third stage.
The mass of the elevator is 800 kg and there is a box of mass 100 kg on the floor of the elevator.\\
(iii) Find the tension in the cable in each of the three stages of the motion.\\
(iv) Find the greatest and least values of the magnitude of the force exerted on the box by the floor of the elevator.
\hfill \mbox{\textit{CAIE M1 2013 Q7 [10]}}