| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Difficulty | Standard +0.3 Part (i) is a straightforward work-energy calculation with constant speed (forces balanced). Part (ii) requires connecting power, force, and velocity relationships to find the final speed, then applying work-energy theorem—this involves more steps and algebraic manipulation than typical M1 questions but uses standard techniques without requiring novel insight. |
| Spec | 6.02b Calculate work: constant force, resolved component6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| PE gain \(= 1250 \times 10 \times 400 \times 0.125\) | B1 | |
| WD against resistance is \(800 \times 400\) J | B1 | |
| M1 | For using WD by car's engine = Gain in PE + WD against resistance | |
| WD by car's engine is 945 000 J (945 kJ) | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([v_2/6 = 5 \times (1/3)]\) | M1 | For using \(P = Fv \Rightarrow \dfrac{v_2}{v_1} = \dfrac{P_2}{P_1} \times \dfrac{F_1}{F_2}\) |
| \(v_2 = 10\) | A1 | |
| KE gain \(= \frac{1}{2} \times 1250(10^2 - 6^2)\) | B1ft | |
| \([\text{WD by car's engine} = 945000 + 40000]\) | M1 | For using WD by car's engine = (Gain in PE + WD against resistance) + KE gain |
| WD by car's engine is 985 000 J (985 kJ) | A1ft | [5] ft incorrect ans(i) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| M1 | For using Newton's second law with \(a = 0\) | |
| \(DF = 1250g \times 0.125 + 800\) | A1 | |
| M1 | For using \(WD = DF \times 400\) | |
| WD by car's engine is 945 00 J (945 kJ) | A1 | [4] |
# Question 6:
**(i)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| PE gain $= 1250 \times 10 \times 400 \times 0.125$ | B1 | |
| WD against resistance is $800 \times 400$ J | B1 | |
| | M1 | For using WD by car's engine = Gain in PE + WD against resistance |
| WD by car's engine is 945 000 J (945 kJ) | A1 | [4] |
**(ii)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[v_2/6 = 5 \times (1/3)]$ | M1 | For using $P = Fv \Rightarrow \dfrac{v_2}{v_1} = \dfrac{P_2}{P_1} \times \dfrac{F_1}{F_2}$ |
| $v_2 = 10$ | A1 | |
| KE gain $= \frac{1}{2} \times 1250(10^2 - 6^2)$ | B1ft | |
| $[\text{WD by car's engine} = 945000 + 40000]$ | M1 | For using WD by car's engine = (Gain in PE + WD against resistance) + KE gain |
| WD by car's engine is 985 000 J (985 kJ) | A1ft | [5] ft incorrect ans(i) |
**Alternative scheme for part (i):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| | M1 | For using Newton's second law with $a = 0$ |
| $DF = 1250g \times 0.125 + 800$ | A1 | |
| | M1 | For using $WD = DF \times 400$ |
| WD by car's engine is 945 00 J (945 kJ) | A1 | [4] |6 A car of mass 1250 kg travels from the bottom to the top of a straight hill which has length 400 m and is inclined to the horizontal at an angle of $\alpha$, where $\sin \alpha = 0.125$. The resistance to the car's motion is 800 N . Find the work done by the car's engine in each of the following cases.\\
(i) The car's speed is constant.\\
(ii) The car's initial speed is $6 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, the car's driving force is 3 times greater at the top of the hill than it is at the bottom, and the car's power output is 5 times greater at the top of the hill than it is at the bottom.
\hfill \mbox{\textit{CAIE M1 2012 Q6 [9]}}