| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Vertical projection: speed at given height |
| Difficulty | Standard +0.8 This is a multi-part mechanics question requiring SUVAT equations, conservation of momentum, and tracking two particles through multiple events. Part (a) is routine, but parts (b) and (c) require careful bookkeeping of velocities and positions after collision and rebound, with the final part demanding simultaneous equation solving to find when particles meet again—significantly above average difficulty. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.03b Conservation of momentum: 1D two particles6.03j Perfectly elastic/inelastic: collisions |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v^2 = 25^2 - 2g \times 20\) | M1 | For use of \(v^2 = u^2 + 2as\) or equivalent to get an equation in \(v\) only with \(u=25\), \(s=20\) and \(a = \pm g\). |
| \(\Rightarrow\) speed \(= 15\ \text{ms}^{-1}\) | A1 | AG. Allow verification – at least one intermediate step from equation of motion to given result. Any errors seen is A0. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.2g \times 20 = \frac{1}{2}\times 0.2 \times 25^2 - \frac{1}{2}\times 0.2\times v^2\) | (M1) | Attempt at energy with \(m=0.2\), \(h=20\), \(u=25\); correct number of terms, allow sign errors. |
| \(\Rightarrow\) speed \(= 15\ \text{ms}^{-1}\) | (A1) | AG. Allow verification - at least one intermediate step from conservation of energy to given result. Any errors seen is A0. |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.2\times 15 - 0.1\times 20 = 0 + 0.1v\) or \(0.2\times 15 - 0.1\times 20 = 0 - 0.1V\) | M1 | OE. Attempt at conservation of momentum; 3 non-zero terms; allow sign errors – use of 25 is M0. |
| \(v = 10\ \text{ms}^{-1}\) upwards | A1 | Must have direction (possibly seen on diagram). If using \(mgv\) for momentum, then M1 A0 max. |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Speed of \(P\) at impact is \(20\ \text{ms}^{-1}\) | B1 | From \(v^2 = 0 + 2\times g\times 20\) (OE) - possibly implied by speed of \(P\) after impact being stated at 10. |
| Time to when \(P\) reaches ground \(= 2\) s | B1 | From \(20 = 0 + \frac{1}{2}\times g\times t^2\) (OE). |
| \(s_P = 10t + \frac{1}{2}\times -g\times t^2\) | M1 | Distance travelled by \(P\) after impact with the ground. Must be using *their* 10 (speed of \(P\) after impact with the ground) and \(a = \pm g\). |
| \(s_Q = 10t + \frac{1}{2}\times g\times t^2\) | M1 | Distance travelled by \(Q\) after \(P\)'s impact with the ground. Must be using their \(\pm 10\) (the speed/vel. of \(Q\) after *their* 2s \((= |
| \(s_P + s_Q = 20 \Rightarrow t = 1\) | A1 | Time after \(P\) hits the ground to next collision. |
| Height \(= 5\) m | A1 | CWO |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(s_Q = 10t + \frac{1}{2} \times -g \times t^2\) | (M1) | Expression for the displacement of \(Q\) after first impact of \(P\) and \(Q\). Must be using *their* 10 (from (b)) and \(a = \pm g\). |
| \(s_P = 10 \times (t-2) + \frac{1}{2} \times -g \times (t-2)^2\) | (M1) | Expression for the displacement of \(P\) (for values of \(t \geqslant 2\)) measured from point of first collision between \(P\) and \(Q\). Must be using *their* 2 (time for \(P\) to reach ground), *their* 10 (speed of \(P\) after impact with the ground), \(a = \pm g\). |
| \(s_Q + 20 = s_P \Rightarrow t = 3\) | (A1) | Correct time between collisions of \(P\) and \(Q\). |
| Height \(= 5\) m | (A1) | CWO |
| 6 |
## Question 7(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v^2 = 25^2 - 2g \times 20$ | **M1** | For use of $v^2 = u^2 + 2as$ or equivalent to get an equation in $v$ only with $u=25$, $s=20$ and $a = \pm g$. |
| $\Rightarrow$ speed $= 15\ \text{ms}^{-1}$ | **A1** | AG. Allow verification – at least one intermediate step from equation of motion to given result. Any errors seen is A0. |
**Alternative method for Question 7(a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.2g \times 20 = \frac{1}{2}\times 0.2 \times 25^2 - \frac{1}{2}\times 0.2\times v^2$ | **(M1)** | Attempt at energy with $m=0.2$, $h=20$, $u=25$; correct number of terms, allow sign errors. |
| $\Rightarrow$ speed $= 15\ \text{ms}^{-1}$ | **(A1)** | AG. Allow verification - at least one intermediate step from conservation of energy to given result. Any errors seen is A0. |
| | **2** | |
---
## Question 7(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.2\times 15 - 0.1\times 20 = 0 + 0.1v$ or $0.2\times 15 - 0.1\times 20 = 0 - 0.1V$ | **M1** | OE. Attempt at conservation of momentum; 3 non-zero terms; allow sign errors – use of 25 is M0. |
| $v = 10\ \text{ms}^{-1}$ upwards | **A1** | Must have direction (possibly seen on diagram). If using $mgv$ for momentum, then M1 A0 max. |
| | **2** | |
---
## Question 7(c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Speed of $P$ at impact is $20\ \text{ms}^{-1}$ | **B1** | From $v^2 = 0 + 2\times g\times 20$ (OE) - possibly implied by speed of $P$ after impact being stated at 10. |
| Time to when $P$ reaches ground $= 2$ s | **B1** | From $20 = 0 + \frac{1}{2}\times g\times t^2$ (OE). |
| $s_P = 10t + \frac{1}{2}\times -g\times t^2$ | **M1** | Distance travelled by $P$ after impact with the ground. Must be using *their* 10 (speed of $P$ after impact with the ground) and $a = \pm g$. |
| $s_Q = 10t + \frac{1}{2}\times g\times t^2$ | **M1** | Distance travelled by $Q$ after $P$'s impact with the ground. Must be using their $\pm 10$ (the speed/vel. of $Q$ after *their* 2s $(=|{'}10{'}+(-g)\times 2|$ where '10' is the value from **(b)**)) and $a = \pm g$. |
| $s_P + s_Q = 20 \Rightarrow t = 1$ | **A1** | Time after $P$ hits the ground to next collision. |
| Height $= 5$ m | **A1** | CWO |
## Question 7(c) — Alternative Method for last 4 marks:
| Answer | Mark | Guidance |
|--------|------|----------|
| $s_Q = 10t + \frac{1}{2} \times -g \times t^2$ | **(M1)** | Expression for the displacement of $Q$ after first impact of $P$ and $Q$. Must be using *their* 10 (from **(b)**) and $a = \pm g$. |
| $s_P = 10 \times (t-2) + \frac{1}{2} \times -g \times (t-2)^2$ | **(M1)** | Expression for the displacement of $P$ (for values of $t \geqslant 2$) measured from point of first collision between $P$ and $Q$. Must be using *their* 2 (time for $P$ to reach ground), *their* 10 (speed of $P$ after impact with the ground), $a = \pm g$. |
| $s_Q + 20 = s_P \Rightarrow t = 3$ | **(A1)** | Correct time between collisions of $P$ and $Q$. |
| Height $= 5$ m | **(A1)** | CWO |
| | **6** | |7 A particle $P$ of mass 0.2 kg is projected vertically upwards from horizontal ground with speed $25 \mathrm {~ms} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the speed of $P$ when it reaches 20 m above the ground is $15 \mathrm {~ms} ^ { - 1 }$.\\
When $P$ reaches 20 m above the ground it collides with a second particle $Q$ of mass 0.1 kg which is moving downwards at $20 \mathrm {~ms} ^ { - 1 } . P$ is brought to instantaneous rest in the collision.
\item Find the velocity of $Q$ immediately after the collision.\\
When $P$ reaches the ground it rebounds back directly upwards with half of the speed that it had immediately before hitting the ground.
\item Find the height above the ground at which $P$ and $Q$ next collide.\\
If you use the following page to complete the answer to any question, the question number must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2024 Q7 [10]}}