| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2019 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Connected particles on inclined plane |
| Difficulty | Standard +0.3 This is a straightforward mechanics problem requiring standard application of Newton's second law to connected particles on an inclined plane. Part (i) involves resolving forces, finding sin α from tan α, and solving simultaneous equations for acceleration and tension—all routine M1 techniques. Part (ii) is a simple kinematics calculation using suvat equations. The problem requires multiple steps but no novel insight or challenging conceptual understanding. |
| Spec | 3.02d Constant acceleration: SUVAT formulae3.03c Newton's second law: F=ma one dimension3.03k Connected particles: pulleys and equilibrium3.03o Advanced connected particles: and pulleys |
| Answer | Marks | Guidance |
|---|---|---|
| 4(i) | Apply Newton’s second law to either or to the system | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 25 25 | A1 | Any two correct. Allow α=16.3 used. |
| Answer | Marks |
|---|---|
| equations for either a or T | M1 |
| a=1.2ms–2 | A1 |
| T=16N | A1 |
| Answer | Marks |
|---|---|
| 4(ii) | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | M1 | Use constant acceleration equation(s) with u=1 and solve a 3 |
| Answer | Marks |
|---|---|
| t= 0.5s | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2 | M1 | Use relevant constant acceleration equations with u=1 in a |
| Answer | Marks |
|---|---|
| t=0.5s | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 4:
--- 4(i) ---
4(i) | Apply Newton’s second law to either or to the system | M1
Block A: T – 4g × 7 = 4a
25
Block B: 36 – T – 5g × 7 = 5a
25
System: 36 – 5g × 7 – 4g × 7 = 9a
25 25 | A1 | Any two correct. Allow α=16.3 used.
Either solving the system for a or solving a pair of simultaneous
equations for either a or T | M1
a=1.2ms–2 | A1
T=16N | A1
5
--- 4(ii) ---
4(ii) | 1
0.65=1×t+ ×1.2t2
2 | M1 | Use constant acceleration equation(s) with u=1 and solve a 3
term quadratic equation to find t
t= 0.5s | A1
Alternative method for question 4(ii)
v2 = 12 + 2 × 1.2 × 0.65 [v = 1.6] and 0.65= 1( 1+v )×t
2 | M1 | Use relevant constant acceleration equations with u=1 in a
complete method to find t
t=0.5s | A1
2
Question | Answer | Marks | Guidance\includegraphics{figure_4}
Two blocks $A$ and $B$ of masses 4 kg and 5 kg respectively are joined by a light inextensible string. The blocks rest on a smooth plane inclined at an angle $\alpha$ to the horizontal, where tan $\alpha = \frac{7}{24}$. The string is parallel to a line of greatest slope of the plane with $B$ above $A$. A force of magnitude 36 N acts on $B$, parallel to a line of greatest slope of the plane (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Find the acceleration of the blocks and the tension in the string. [5]
\end{enumerate}
\begin{enumerate}[label=(\roman*)]
\setcounter{enumi}{1}
\item At a particular instant, the speed of the blocks is 1 m s$^{-1}$. Find the time, after this instant, that it takes for the blocks to travel 0.65 m. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2019 Q4 [7]}}