| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Work done and energy |
| Type | Energy method - driving force on horizontal road |
| Difficulty | Standard +0.3 This is a multi-part work-energy question with standard mechanics techniques: calculating work done (resolving forces), applying work-energy theorem, finding power, and motion on an inclined plane. All parts follow routine procedures taught in M1 with no novel problem-solving required, making it slightly easier than average. |
| Spec | 6.02a Work done: concept and definition6.02i Conservation of energy: mechanical energy principle6.02k Power: rate of doing work6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Work done \(= 50\cos10 \times 20\) | M1 | Using \(WD = Fd\cos\theta\) |
| \(= 984.8\text{ J}\) | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([984.8 = \frac{1}{2} \times 25v^2 + 30 \times 20]\) | M1 | Using WD by DF \(=\) KE gain \(+\) WD against Res. |
| \(v = 5.55\text{ ms}^{-1}\) | A1 | [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (Power \(= Fv\)) | M1 | For using Power \(= Fv\) |
| Max power \(= 50\cos10 \times 5.55 = 273\text{ W}\) | A1 | [2] Greatest power is at \(v_{\max}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([50\cos10 - 30 - 25g\sin5 = 25a]\) | M1 | For using Newton's 2nd law up the plane |
| \(a = -0.102\text{ ms}^{-2}\) | A1 | |
| \([0 = 5.55 - 0.102t]\) | M1 | For using \(v = u + at\) |
| Time \(t = 54.4\text{ s}\) | A1 | [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| (WD by DF \(+\) KE loss \(=\) PE gain \(+\) WD against Res. to find distance \(s\) up plane) | M1 | |
| \(50\cos10 \times s + \frac{1}{2} \times 25 \times 5.55^2 = 25g \times s\sin5 + 30 \times s\) | A1 | \(s = 151\text{ m}\) |
| (Using \(s = \frac{1}{2}(u+v)t\)) | M1 | |
| \(t = 302/5.55 = 54.4\text{ s}\) | A1 | [4] |
## Question 6(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Work done $= 50\cos10 \times 20$ | M1 | Using $WD = Fd\cos\theta$ |
| $= 984.8\text{ J}$ | A1 | [2] |
## Question 6(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[984.8 = \frac{1}{2} \times 25v^2 + 30 \times 20]$ | M1 | Using WD by DF $=$ KE gain $+$ WD against Res. |
| $v = 5.55\text{ ms}^{-1}$ | A1 | [2] |
## Question 6(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| (Power $= Fv$) | M1 | For using Power $= Fv$ |
| Max power $= 50\cos10 \times 5.55 = 273\text{ W}$ | A1 | [2] Greatest power is at $v_{\max}$ |
## Question 6(iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[50\cos10 - 30 - 25g\sin5 = 25a]$ | M1 | For using Newton's 2nd law up the plane |
| $a = -0.102\text{ ms}^{-2}$ | A1 | |
| $[0 = 5.55 - 0.102t]$ | M1 | For using $v = u + at$ |
| Time $t = 54.4\text{ s}$ | A1 | [4] |
**Alternative for 6(iv):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| (WD by DF $+$ KE loss $=$ PE gain $+$ WD against Res. to find distance $s$ up plane) | M1 | |
| $50\cos10 \times s + \frac{1}{2} \times 25 \times 5.55^2 = 25g \times s\sin5 + 30 \times s$ | A1 | $s = 151\text{ m}$ |
| (Using $s = \frac{1}{2}(u+v)t$) | M1 | |
| $t = 302/5.55 = 54.4\text{ s}$ | A1 | [4] |
---6 A block of mass 25 kg is pulled along horizontal ground by a force of magnitude 50 N inclined at $10 ^ { \circ }$ above the horizontal. The block starts from rest and travels a distance of 20 m . There is a constant resistance force of magnitude 30 N opposing motion.\\
(i) Find the work done by the pulling force.\\
(ii) Use an energy method to find the speed of the block when it has moved a distance of 20 m .\\
(iii) Find the greatest power exerted by the 50 N force.\\
\includegraphics[max width=\textwidth, alt={}, center]{a92f97e2-343f-4cac-ae38-f18a4ad49055-3_236_1027_2161_566}
After the block has travelled the 20 m , it comes to a plane inclined at $5 ^ { \circ }$ to the horizontal. The force of 50 N is now inclined at an angle of $10 ^ { \circ }$ to the plane and pulls the block directly up the plane (see diagram). The resistance force remains 30 N .\\
(iv) Find the time it takes for the block to come to rest from the instant when it reaches the foot of the inclined plane.\\[0pt]
[4]
\hfill \mbox{\textit{CAIE M1 2016 Q6 [10]}}