| Exam Board | CAIE |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Momentum and Collisions |
| Type | Vertical collision or projection |
| Difficulty | Challenging +1.2 This is a multi-step mechanics problem requiring kinematics equations, conservation of momentum, and case analysis (two possible mass ratios). While it involves several concepts and careful bookkeeping, the techniques are standard M1 material with no novel insights required—moderately above average difficulty for A-level. |
| Spec | 3.02h Motion under gravity: vector form6.03b Conservation of momentum: 1D two particles |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(s_A = \pm(30t - 5t^2)\) or \(s_B = \pm 5t^2\) | B1 | Use of constant acceleration equations to find expressions for displacements of \(A\) or \(B\). |
| \(s_A + s_B = 15\) leading to \(15 = 30t\) leading to \(t = 0.5\) | B1 | Use \(s_A + s_B = 15\) to find time at which particles collide. |
| \(t = 0.5\) leading to \(v_A = \pm 25\) and \(v_B = \pm 5\) | B1 | Find speed of particles at \(t = 0.5\) before collision. |
| \(t = 0.5\) leading to \(h_A = \pm\left(30 \times 0.5 - \frac{1}{2}g \times 0.5^2\right) = \pm 13.75\) | B1 | Find position of \(A\) or \(B\) at which collision occurs at \(t = 0.5\). Alternatively allow \(h_B = \pm 1.25\) as displacement of \(B\). |
| \(25 \times (2m) - 5(m) = (3m)v \rightarrow v_1 = 15\) \(25(m) - 5 \times (2m) = (3m)v \rightarrow v_2 = 5\) | M1 | Use of conservation of momentum, either case, using *their* \(v_A\) and \(v_B \neq 0\) or 30, with 3 terms. |
| A1 | Both values of \(v\) correct |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Particle \(C_1\): \(-13.75 = 15t - 5t^2\) | M1 | Use of \(s = ut + \frac{1}{2}at^2\) OE to find \(t\), using either their numerical \(v_1\) or numerical \(v_2\) from a relevant conservation of momentum equation |
| Particle \(C_2\): \(-13.75 = 5t - 5t^2\) | ||
| \(t_{C_1}, t_{C_2} = 3.74, 2.23\) leading to \(T = 1 + \sqrt{5} - \sqrt{3} = 1.50\) | A1 | Find \(T = t_{C_1} - t_{C_2}\) from \(t_{C_1} = 3.736\) and \(t_{C_2} = 2.232\) |
| 8 | Subscripts 1 and 2 refer to the two cases |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0 = 15 - gt_1\), \(0 = 5 - gt_2 \rightarrow t_1 = 1.5\), \(t_2 = 0.5\); Total heights \(h_1 = 13.75 + 11.25 = 25\); Or \(h_2 = 13.75 + 1.25 = 15\); \(25 = 5T_1^2\) and \(15 = 5T_2^2 \rightarrow T_1 = \sqrt{5}\), \(T_2 = \sqrt{3}\) | M1 | Use of \(v = u - gt\) to find time to highest point; use of \(v^2 = u^2 - 2gs\) to find total height for either case, using either their numerical \(v_1\) or \(v_2\) from a relevant conservation of momentum equation; Use \(s = 0 + \frac{1}{2}gT^2\) to find time to reach ground (either case) |
| \(T = 1.5 + \sqrt{5} - (0.5 + \sqrt{3}) = 1 + \sqrt{5} - \sqrt{3} = 1.50\) | A1 | Find difference in total times \(T = (t_1 + T_1) - (t_2 + T_2)\) |
## Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| $s_A = \pm(30t - 5t^2)$ or $s_B = \pm 5t^2$ | B1 | Use of constant acceleration equations to find expressions for displacements of $A$ or $B$. |
| $s_A + s_B = 15$ leading to $15 = 30t$ leading to $t = 0.5$ | B1 | Use $s_A + s_B = 15$ to find time at which particles collide. |
| $t = 0.5$ leading to $v_A = \pm 25$ and $v_B = \pm 5$ | B1 | Find speed of particles at $t = 0.5$ before collision. |
| $t = 0.5$ leading to $h_A = \pm\left(30 \times 0.5 - \frac{1}{2}g \times 0.5^2\right) = \pm 13.75$ | B1 | Find position of $A$ or $B$ at which collision occurs at $t = 0.5$. Alternatively allow $h_B = \pm 1.25$ as displacement of $B$. |
| $25 \times (2m) - 5(m) = (3m)v \rightarrow v_1 = 15$ $25(m) - 5 \times (2m) = (3m)v \rightarrow v_2 = 5$ | M1 | Use of conservation of momentum, either case, using *their* $v_A$ and $v_B \neq 0$ or 30, with 3 terms. |
| | A1 | Both values of $v$ correct |
## Question 6:
| Answer | Marks | Guidance |
|--------|-------|----------|
| Particle $C_1$: $-13.75 = 15t - 5t^2$ | M1 | Use of $s = ut + \frac{1}{2}at^2$ OE to find $t$, using either their numerical $v_1$ or numerical $v_2$ from a relevant conservation of momentum equation |
| Particle $C_2$: $-13.75 = 5t - 5t^2$ | | |
| $t_{C_1}, t_{C_2} = 3.74, 2.23$ leading to $T = 1 + \sqrt{5} - \sqrt{3} = 1.50$ | A1 | Find $T = t_{C_1} - t_{C_2}$ from $t_{C_1} = 3.736$ and $t_{C_2} = 2.232$ |
| | **8** | Subscripts 1 and 2 refer to the two cases |
**Alternative method for final two marks:**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0 = 15 - gt_1$, $0 = 5 - gt_2 \rightarrow t_1 = 1.5$, $t_2 = 0.5$; Total heights $h_1 = 13.75 + 11.25 = 25$; Or $h_2 = 13.75 + 1.25 = 15$; $25 = 5T_1^2$ and $15 = 5T_2^2 \rightarrow T_1 = \sqrt{5}$, $T_2 = \sqrt{3}$ | M1 | Use of $v = u - gt$ to find time to highest point; use of $v^2 = u^2 - 2gs$ to find total height for either case, using either their numerical $v_1$ or $v_2$ from a relevant conservation of momentum equation; Use $s = 0 + \frac{1}{2}gT^2$ to find time to reach ground (either case) |
| $T = 1.5 + \sqrt{5} - (0.5 + \sqrt{3}) = 1 + \sqrt{5} - \sqrt{3} = 1.50$ | A1 | Find difference in total times $T = (t_1 + T_1) - (t_2 + T_2)$ |
---6 A particle $A$ is projected vertically upwards from level ground with an initial speed of $30 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. At the same instant a particle $B$ is released from rest 15 m vertically above $A$. The mass of one of the particles is twice the mass of the other particle. During the subsequent motion $A$ and $B$ collide and coalesce to form particle $C$.
Find the difference between the two possible times at which $C$ hits the ground.\\
$7 \quad$ A particle $P$ moving in a straight line starts from rest at a point $O$ and comes to rest 16 s later. At time $t \mathrm {~s}$ after leaving $O$, the acceleration $a \mathrm {~m} \mathrm {~s} ^ { - 2 }$ of $P$ is given by
$$\begin{array} { l l }
a = 6 + 4 t & 0 \leqslant t < 2 , \\
a = 14 & 2 \leqslant t < 4 , \\
a = 16 - 2 t & 4 \leqslant t \leqslant 16 .
\end{array}$$
There is no sudden change in velocity at any instant.
\begin{enumerate}[label=(\alph*)]
\item Find the values of $t$ when the velocity of $P$ is $55 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\item Complete the sketch of the velocity-time diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{41e63d05-d109-47dc-80a6-927953e3e607-11_511_1054_351_584}
\item Find the distance travelled by $P$ when it is decelerating.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.
\end{enumerate}
\hfill \mbox{\textit{CAIE M1 2021 Q6 [8]}}