| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Newton's laws and connected particles |
| Type | Particle on incline, hanging counterpart |
| Difficulty | Standard +0.3 This is a standard connected particles problem requiring Newton's second law applied to two bodies with resolution of forces on an incline. Part (i) is routine mechanics (find acceleration and tension), while part (ii) adds friction and requires calculating distance traveled in two phases (before and after B hits the barrier). The multi-step nature and friction calculation elevate it slightly above average, but it follows a well-practiced template for M1 students. |
| Spec | 3.03n Equilibrium in 2D: particle under forces3.03o Advanced connected particles: and pulleys3.03v Motion on rough surface: including inclined planes |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \([T = 1.6a,\ 2.4g\sin 30 - T = 2.4a]\) System is \(2.4g\sin 30 = 4a\) | M1 | Attempt Newton's 2nd law for \(A\) or \(B\) or for the system |
| Two correct equations | A1 | Two correct equations |
| M1 | Solve for \(a\) or \(T\) | |
| \(a = 3\) | A1 | |
| \(T = 4.8\) | A1 | |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Friction force on \(A\) is \(F = 0.2 \times 1.6g\ [= 3.2]\) | B1 | From \(F = \mu R\) |
| \(T - F = 1.6a\) ; \(2.4g\sin 30 - T = 2.4a\) ; System is \(2.4g\sin 30 - F = 4a\) | M1 | Attempt Newton's 2nd law for both particles or for the system |
| A1 | Correct equations for \(A\) and \(B\) or correct system equation | |
| M1 | Attempt to solve for \(a\) | |
| \(a = 2.2\) | A1 | |
| \(v^2 = 2 \times 2.2 \times 1\) | M1 | Attempt to find \(v\) or \(v^2\) when \(B\) reaches the barrier |
| Subsequent acceleration of \(A\) is \(-2\) | B1 | |
| \(4.4 = 2 \times 2 \times s\) | M1 | Attempt to find distance \(A\) travels while decelerating to \(v = 0\) |
| Total distance travelled is \(2.1\) m | A1 | |
| Total: 9 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(F = 0.2 \times 1.6g\ [= 3.2]\) | B1 | From \(F = \mu R = 0.2 \times 1.6g = 3.2\) |
| M1 | Attempt PE loss as \(B\) reaches the barrier | |
| PE loss \(= 2.4g\sin 30\ [= 12]\) | A1 | |
| M1 | Attempt KE gain for both \(A\) and \(B\) | |
| KE gain \(= \frac{1}{2}(1.6 + 2.4)v^2\ [= 2v^2]\) | A1 | |
| \([2.4g\sin 30 = \frac{1}{2} \times 4 \times v^2 + 3.2 \times 1]\); \([v^2 = 4.4]\) | M1 | Apply work-energy equation for motion until \(B\) reaches barrier (three relevant terms) |
| KE loss \(= \frac{1}{2} \times 1.6 \times 4.4\) | B1 | Find KE loss as \(A\) comes to rest after \(B\) has stopped |
| \([\frac{1}{2} \times 1.6 \times 4.4 = 3.2d]\); \([d = 1.1]\) | M1 | Apply work-energy equation where \(d\) is the extra distance travelled by \(A\) leading to a positive value for \(d\) |
| Total distance \(= 2.1\) m | A1 | Distance \(= d + 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Friction \(= 0.2 \times 1.6g\ [= 3.2]\) | B1 | |
| \(2.4g\sin 30 - T = 2.4a\); \(T - F = 1.6a\) | M1 | Apply Newton's 2nd law to \(A\) and \(B\) and solve for \(T\) |
| \(T = 6.72\) | A1 | |
| \([\frac{1}{2} \times 1.6 \times v^2]\) | M1 | Attempt KE for \(A\) only |
| A1 | Correct KE for \(A\) | |
| \([6.72 \times 1 = \frac{1}{2} \times 1.6 \times v^2 + 3.2 \times 1]\) | M1 | Use work-energy equation for \(A\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Friction \(= 0.2 \times 1.6g\ [= 3.2]\) | B1 | |
| \(2.4g\sin 30 - T = 2.4a\); \(T - F = 1.6a\) | M1 | Apply Newton's 2nd law to \(A\) and \(B\) and solve for \(T\) |
| \(T = 6.72\) | A1 | |
| M1 | Find energy loss/gain for \(B\); allow either term | |
| \(\pm(\frac{1}{2} \times 2.4 \times v^2 - 2.4g\sin 30)\) | A1 | |
| \(2.4g\sin 30 = \frac{1}{2} \times 2.4 \times v^2 + 6.72 \times 1\) | M1 | Use work-energy equation for \(B\) |
## Question 7(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $[T = 1.6a,\ 2.4g\sin 30 - T = 2.4a]$ System is $2.4g\sin 30 = 4a$ | M1 | Attempt Newton's 2nd law for $A$ or $B$ or for the system |
| Two correct equations | A1 | Two correct equations |
| | M1 | Solve for $a$ or $T$ |
| $a = 3$ | A1 | |
| $T = 4.8$ | A1 | |
| **Total: 5** | | |
---
## Question 7(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Friction force on $A$ is $F = 0.2 \times 1.6g\ [= 3.2]$ | B1 | From $F = \mu R$ |
| $T - F = 1.6a$ ; $2.4g\sin 30 - T = 2.4a$ ; System is $2.4g\sin 30 - F = 4a$ | M1 | Attempt Newton's 2nd law for both particles or for the system |
| | A1 | Correct equations for $A$ and $B$ or correct system equation |
| | M1 | Attempt to solve for $a$ |
| $a = 2.2$ | A1 | |
| $v^2 = 2 \times 2.2 \times 1$ | M1 | Attempt to find $v$ or $v^2$ when $B$ reaches the barrier |
| Subsequent acceleration of $A$ is $-2$ | B1 | |
| $4.4 = 2 \times 2 \times s$ | M1 | Attempt to find distance $A$ travels while decelerating to $v = 0$ |
| Total distance travelled is $2.1$ m | A1 | |
| **Total: 9** | | |
---
## Question 7(ii) Alternative [Work-Energy applied to $A$ and $B$]:
| Answer | Mark | Guidance |
|--------|------|----------|
| $F = 0.2 \times 1.6g\ [= 3.2]$ | B1 | From $F = \mu R = 0.2 \times 1.6g = 3.2$ |
| | M1 | Attempt PE loss as $B$ reaches the barrier |
| PE loss $= 2.4g\sin 30\ [= 12]$ | A1 | |
| | M1 | Attempt KE gain for both $A$ and $B$ |
| KE gain $= \frac{1}{2}(1.6 + 2.4)v^2\ [= 2v^2]$ | A1 | |
| $[2.4g\sin 30 = \frac{1}{2} \times 4 \times v^2 + 3.2 \times 1]$; $[v^2 = 4.4]$ | M1 | Apply work-energy equation for motion until $B$ reaches barrier (three relevant terms) |
| KE loss $= \frac{1}{2} \times 1.6 \times 4.4$ | B1 | Find KE loss as $A$ comes to rest after $B$ has stopped |
| $[\frac{1}{2} \times 1.6 \times 4.4 = 3.2d]$; $[d = 1.1]$ | M1 | Apply work-energy equation where $d$ is the extra distance travelled by $A$ leading to a positive value for $d$ |
| Total distance $= 2.1$ m | A1 | Distance $= d + 1$ |
---
## Question 7(ii) Alternative [Work-Energy applied to $A$ only]:
| Answer | Mark | Guidance |
|--------|------|----------|
| Friction $= 0.2 \times 1.6g\ [= 3.2]$ | B1 | |
| $2.4g\sin 30 - T = 2.4a$; $T - F = 1.6a$ | M1 | Apply Newton's 2nd law to $A$ and $B$ and solve for $T$ |
| $T = 6.72$ | A1 | |
| $[\frac{1}{2} \times 1.6 \times v^2]$ | M1 | Attempt KE for $A$ only |
| | A1 | Correct KE for $A$ |
| $[6.72 \times 1 = \frac{1}{2} \times 1.6 \times v^2 + 3.2 \times 1]$ | M1 | Use work-energy equation for $A$ |
---
## Question 7(ii) Alternative [Work-Energy applied to $B$ only]:
| Answer | Mark | Guidance |
|--------|------|----------|
| Friction $= 0.2 \times 1.6g\ [= 3.2]$ | B1 | |
| $2.4g\sin 30 - T = 2.4a$; $T - F = 1.6a$ | M1 | Apply Newton's 2nd law to $A$ and $B$ and solve for $T$ |
| $T = 6.72$ | A1 | |
| | M1 | Find energy loss/gain for $B$; allow either term |
| $\pm(\frac{1}{2} \times 2.4 \times v^2 - 2.4g\sin 30)$ | A1 | |
| $2.4g\sin 30 = \frac{1}{2} \times 2.4 \times v^2 + 6.72 \times 1$ | M1 | Use work-energy equation for $B$ |7\\
\begin{tikzpicture}[scale=1]
% Key coordinates
\coordinate (A) at (0.5, 0.09);
\coordinate (P) at (5, 0);
\coordinate (Pa) at (5.03, 0.12);
\coordinate (bottom) at ($(P)+(-30:3.5)$);
\coordinate (bottoma) at ($(bottom)+(0,0.09)$);
\coordinate (B) at ($(bottom)+(150:2)+(0,0.09)$);
% Horizontal plane
\draw[thick] (-0.3, 0) -- (P);
\foreach \x in {-0.1,0.2,0.5,0.8,1.1,1.4,1.7,2.0,2.3,2.6,2.9,3.2,3.5,3.8,4.1,4.4,4.7,5.0}
{\draw (\x,0) -- ++(210:0.2);}
% Inclined plane
\draw[thick] (P) -- (bottom);
\foreach \f in {0.07,0.14,0.21,0.28,0.35,0.42,0.49,0.56,0.63,0.70,0.77,0.84,0.91,0.98}
{\draw ($(P)!\f!(bottom)$) -- ++(240:0.2);}
% Pulley at P
\draw[fill=white] (P) circle (3pt);
\node[above=4pt] at (P) {$P$};
% Particle A
\fill (A) circle (2.5pt);
\node[above=2pt] at (A) {$A$};
\node[above left=-1pt] at (A) {1.6\,kg};
% String A to P
\draw (A) -- (Pa) -- (B);
\node[above=4pt] at ($(A)!0.5!(P)$) {2.5\,m};
% Particle B
\fill (B) circle (2.5pt);
\node[above right=1pt] at (B) {$B$};
\node[right=5pt] at (B) {2.4\,kg};
% Label 1 m from B to bottom
\node[right=6pt] at ($(B)!0.5!(bottom)$) {1\,m};
% Barrier at bottom of incline
% Dashed horizontal for angle marking
\draw[dashed] (bottom) -- ++(-1.2,0);
\draw [<->] ($(B)+(0.1,0)$) -- ($(bottom)+(0.1,0.09)$);
% 30-degree arc
\draw ($(bottom)+(-0.7,0)$) arc (0:-30:-0.7);
\node at ($(bottom)+(-1,0.22)$) {$30^\circ$};
\end{tikzpicture}
As shown in the diagram, a particle $A$ of mass 1.6 kg lies on a horizontal plane and a particle $B$ of mass 2.4 kg lies on a plane inclined at an angle of $30 ^ { \circ }$ to the horizontal. The particles are connected by a light inextensible string which passes over a small smooth pulley $P$ fixed at the top of the inclined plane. The distance $A P$ is 2.5 m and the distance of $B$ from the bottom of the inclined plane is 1 m . There is a barrier at the bottom of the inclined plane preventing any further motion of $B$. The part $B P$ of the string is parallel to a line of greatest slope of the inclined plane. The particles are released from rest with both parts of the string taut.\\
(i) Given that both planes are smooth, find the acceleration of $A$ and the tension in the string.\\
(ii) It is given instead that the horizontal plane is rough and that the coefficient of friction between $A$ and the horizontal plane is 0.2 . The inclined plane is smooth. Find the total distance travelled by $A$.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE M1 2018 Q7 [14]}}