| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2019 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Variable resistance: find k or constants |
| Difficulty | Standard +0.3 This is a straightforward mechanics problem requiring standard application of power = force × velocity and resolving forces on an incline. Part (i) is a 'show that' with clear steps (resolve parallel to slope, use P=Fv). Part (ii) simply substitutes v=5 into R=kv² to find k=60. Part (iii) applies the same principles on level ground. All steps are routine with no novel insight required, making it slightly easier than average. |
| Spec | 6.02l Power and velocity: P = Fv6.02m Variable force power: using scalar product |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(DF = 1500 + 12000 \times g \times 0.08\) \([DF = 11100]\) | M1 | Using \(DF =\) Resistance \(+\) weight component (3 terms) |
| Power \(= DF \times 5\) | M1 | Using \(P = Fv\) (their 2 term \(DF \times 5\)) |
| Power \(= 11100 \times 5 = 55.5\) kW | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(k \times 5^2 = 1500, k = 60\) | B1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(DF = 60v^2\) | B1 | Using \(DF =\) resistance \(= 60v^2\) |
| \(55500 = DF \times v = 60v^2 \times v = 60v^3\) | M1 | \(P = Fv\) used and attempt to solve a 2-term cubic equation for \(v\) |
| \(v = 9.74\) ms\(^{-1}\) | A1 |
## Question 3(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $DF = 1500 + 12000 \times g \times 0.08$ $[DF = 11100]$ | M1 | Using $DF =$ Resistance $+$ weight component (3 terms) |
| Power $= DF \times 5$ | M1 | Using $P = Fv$ (their 2 term $DF \times 5$) |
| Power $= 11100 \times 5 = 55.5$ kW | A1 | AG |
## Question 3(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $k \times 5^2 = 1500, k = 60$ | B1 | AG |
## Question 3(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $DF = 60v^2$ | B1 | Using $DF =$ resistance $= 60v^2$ |
| $55500 = DF \times v = 60v^2 \times v = 60v^3$ | M1 | $P = Fv$ used and attempt to solve a 2-term cubic equation for $v$ |
| $v = 9.74$ ms$^{-1}$ | A1 | |3 A lorry has mass 12000 kg .\\
(i) The lorry moves at a constant speed of $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ up a hill inclined at an angle of $\theta$ to the horizontal, where $\sin \theta = 0.08$. At this speed, the magnitude of the resistance to motion on the lorry is 1500 N . Show that the power of the lorry's engine is 55.5 kW .\\
When the speed of the lorry is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ the magnitude of the resistance to motion is $k v ^ { 2 } \mathrm {~N}$, where $k$ is a constant.\\
(ii) Show that $k = 60$.\\
(iii) The lorry now moves at a constant speed on a straight level road. Given that its engine is still working at 55.5 kW , find the lorry's speed.\\
\hfill \mbox{\textit{CAIE M1 2019 Q3 [7]}}