| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2017 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Travel graphs |
| Type | Two-particle meeting or overtaking |
| Difficulty | Standard +0.3 This is a standard kinematics problem requiring application of SUVAT equations and basic calculus (finding minimum via differentiation). Part (i) is routine setup, part (ii) is straightforward substitution, and part (iii) requires recognizing that minimum distance occurs when velocities are equal. The multi-part structure and need to coordinate two particles' motion makes it slightly above average, but all techniques are standard M1 material with no novel insight required. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.02c Interpret kinematic graphs: gradient and area |
| Answer | Marks | Guidance |
|---|---|---|
| 3(i) | M1 | Attempt s as s = k + 10t (any k) |
| Answer | Marks |
|---|---|
| A | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| B | B1 FT | Allow FT only if s = 10t and |
| Answer | Marks |
|---|---|
| Total: | 3 |
| Answer | Marks |
|---|---|
| 3(ii) | v = 16 – 2t → v = 0, t = 8 |
| B B | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| [= 20 + 10t + t2 – 16t = t 2 – 6t + 20] | M1 | Finding distance between A and B at time t = T ( T > 0 ) |
| Answer | Marks | Guidance |
|---|---|---|
| t = 8, s = 36 (m) | A1 | |
| Total: | 3 | |
| Question | Answer | Marks |
| Answer | Marks |
|---|---|
| 3(iii) | ds |
| Answer | Marks | Guidance |
|---|---|---|
| s = t 2 – 6t + 20 = (t – 3) 2 + 11 | M1 | Either use differentiation or complete the square, or state |
| Answer | Marks | Guidance |
|---|---|---|
| [t = 3] | M1 | Solve for t and evaluate s – s at this value of t |
| Answer | Marks | Guidance |
|---|---|---|
| A B | A1 | |
| Total: | 3 | |
| 4(i)(a) | [P = 850 × 42] | M1 |
| P = 35700 W = 35.7 kW | A1 | Must be in kW to 3sf |
| Total: | 2 | |
| 4(i)(b) | P = 41700 | |
| → [DF = 41700/42] | M1 | Find new power and new DF based on power found in 4(i)(a) |
| [(993 – 850) = 1200a] | M1 | Apply Newton 2, three terms |
| a = 5/42 = 0.119 ms–2 | A1 | |
| Total: | 3 | |
| Question | Answer | Marks |
Question 3:
--- 3(i) ---
3(i) | M1 | Attempt s as s = k + 10t (any k)
A A
s = 20 + 10t
A | A1
s = 16t + ½(–2)t 2 [= 16t – t 2]
B | B1 FT | Allow FT only if s = 10t and
A
s = 16(t – 2) + ½(–2)(t – 2) 2
B
i.e. t measured from when A passes O
Total: | 3
--- 3(ii) ---
3(ii) | v = 16 – 2t → v = 0, t = 8
B B | B1
s = s – s
A B
[= 20 + 10t + t2 – 16t = t 2 – 6t + 20] | M1 | Finding distance between A and B at time t = T ( T > 0 )
found from a valid method for v = 0
B
t = 8, s = 36 (m) | A1
Total: | 3
Question | Answer | Marks | Guidance
--- 3(iii) ---
3(iii) | ds
=2t−6
dt
or
s = t 2 – 6t + 20 = (t – 3) 2 + 11 | M1 | Either use differentiation or complete the square, or state
value of t when speeds are the same
[t = 3] | M1 | Solve for t and evaluate s – s at this value of t
A B
s = s – s = 11 m
A B | A1
Total: | 3
4(i)(a) | [P = 850 × 42] | M1 | Using P = Fv
P = 35700 W = 35.7 kW | A1 | Must be in kW to 3sf
Total: | 2
4(i)(b) | P = 41700
→ [DF = 41700/42] | M1 | Find new power and new DF based on power found in 4(i)(a)
[(993 – 850) = 1200a] | M1 | Apply Newton 2, three terms
a = 5/42 = 0.119 ms–2 | A1
Total: | 3
Question | Answer | Marks | GuidanceA particle $A$ moves in a straight line with constant speed $10$ m s$^{-1}$. Two seconds after $A$ passes a point $O$ on the line, a particle $B$ passes through $O$, moving along the line in the same direction as $A$. Particle $B$ has speed $16$ m s$^{-1}$ at $O$ and has a constant deceleration of $2$ m s$^{-2}$.
\begin{enumerate}[label=(\roman*)]
\item Find expressions, in terms of $t$, for the displacement from $O$ of each particle $t$ s after $B$ passes through $O$.
[3]
\item Find the distance between the particles when $B$ comes to instantaneous rest.
[3]
\item Find the minimum distance between the particles.
[3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2017 Q3 [9]}}