## Question 7(a):

| Answer | Marks | Guidance |
|--------|-------|----------|
| For $Q$: $-2mg\sin\alpha - F = 2ma$ giving $[-16m - 7.2m = 2ma]$; $R = 2mg\cos\alpha$ $[= 12m]$ | M1 | Apply Newton's 2nd law along or perpendicular to the plane to particle $Q$. Must use values for $\alpha$ or $\sin\alpha$ or $\cos\alpha$ |
| Both correct | A1 | Both correct |
| $F = 0.6 \times 2mg\cos\alpha = 0.6 \times 0.6 \times 20m$ $[= 7.2m]$; $[2(m)a = -2(m)g(0.8) - 0.6 \times 2(m)g(0.6)]$ | M1 | Using $F = 0.6R$ where $R$ is a component of $2mg$ only |
| Acceleration of $Q$ up the plane while moving up the plane is $a = -11.6 \text{ ms}^{-2}$ | A1 | AG |

---

## Question 7(b):

| Answer | Marks | Guidance |
|--------|-------|----------|
| For $P$: $mg\sin\alpha - 0.6R = ma$, leading to $8m - 3.6m = ma$; $[R = mg\cos\alpha = 6m,\ a = 4.4 \text{ ms}^{-2}]$ | M1 | Apply Newton's 2nd law to attempt to find acceleration of particle $P$. Must use values for $\alpha$ or $\sin\alpha$ |
| $Q$ comes to rest when $10 - 11.6T_1 = 0$, $\left[T_1 = \frac{25}{29} = 0.862\right]$ | M1 | For using constant acceleration equations to attempt to determine when $v_Q = 0$ |
| For $P$: $s_{P(\text{down})} = \frac{1}{2} \times 4.4 \times T_1^2$ $[= 1.635]$; For $Q$: $s_{Q(\text{up})} = 10T_1 + \frac{1}{2} \times (-11.6) \times T_1^2$ $[= 4.31]$ | M1 | Use constant acceleration equations to attempt to find either $s_{P(\text{down})}$ or $s_{Q(\text{up})}$ at time $T_1$ |
| $d = 6.4 - s_{P(\text{down})} - s_{Q(\text{up})}$ $[= 0.455]$; find $T_2\ [= 0.12]$ by using $d = s_{P2} - s_{Q2} = (4.4T_1) \times T_2$; [$s_{P2}$ and $s_{Q2}$ are distances travelled by $P$ and $Q$ in time $T_2$] | M1 | For attempting to find extra distance $d\ [= 0.455]$ needed to reach 6.4 m and using $u_P = 4.4T_1$ at $T_1$ to find $T_2$ as $d = (4.4T_1)T_2 + \frac{1}{2} \times 4.4T_2^2 - \frac{1}{2} \times 4.4T_2^2$ |
| Time before collision $= [t = T_1 + T_2 = 0.862 + 0.12 =]\ 0.982$ | A1 | $t = 0.98194357\ldots$ |

**Alternative method for 7(b):**

| Answer | Marks | Guidance |
|--------|-------|----------|
| For $P$: $s_{P(\text{down})} = \frac{1}{2} \times 4.4 \times t^2$; For $Q$: $s_{Q(\text{up})} = 10T_1 + \frac{1}{2} \times (-11.6)T_1^2 - \frac{1}{2} \times 4.4(t-T_1)^2$ | M1 | Use constant acceleration equations to attempt to find either $s_{P(\text{down})}$ or $s_{Q(\text{up})}$ at time $t$ where $t$ is total time before collision |
| $\frac{1}{2} \times 4.4t^2 + 10T_1 + \frac{1}{2} \times (-11.6)T_1^2 - \frac{1}{2} \times 4.4(t-T_1)^2 = 6.4$ | M1 | For using $s_{P(\text{down})} + s_{Q(\text{up})} = 6.4$ and solving for $t$ |
| Time before collision is $t = 0.982$ s | A1 | $t = 0.98194357\ldots$ |

**Special case (Q does not come to rest/change direction):**

| Answer | Marks | Guidance |
|--------|-------|----------|
| For $P$: $s_{p(\text{down})} = (\pm)\frac{1}{2} \times 4.4t^2$; For $Q$: $s_{q(\text{up})} = (\pm)10t + \frac{1}{2} \times (-11.6)t^2$ | M1 | For using constant acceleration equations to find either $s_{p(\text{down})}$ or $s_{q(\text{up})}$ |
| $s_p + s_q = 6.4$ leading to $\frac{1}{2} \times 4.4t^2 + 10t + \frac{1}{2} \times (-11.6)t^2 = 6.4$ | M1 | For applying $(\pm)s_p + (\pm)s_q = 6.4$ using expressions for $s_p$ and $s_q$ to set up and solve a 3-term quadratic in $t$ to obtain at least 1 solution |
| Time that particles are in motion before collision $= t = 1$ s | A1 | Must reject $t = 16/9$. **Maximum mark 4 out of 5** |

---

## Question 7(c):

| Answer | Marks | Guidance |
|--------|-------|----------|
| $u_{p(\text{down})} = 0 + 4.4 \times 0.982\ [= 4.3208]$ | B1 FT | Allow $\pm 4.4$. FT on *their* 4.4 and *their* 0.982 |
| $u_{q(\text{down})} = 4.4 \times 0.12\ [= 0.528]$ | B1 FT | Allow $\pm 4.4$. FT on *their* 4.4 and *their* 0.12 |
| $\pm m \times 4.3208 \pm 2m \times 0.528 = \pm(m+2m)v$; [Correct equation is $m \times 4.3208 + 2m \times 0.528 = \pm(m+2m)v$] | M1 | Apply conservation of momentum, 4 terms, using *their* $u_p$ and $u_q$ values with $m$ and $2m$ respectively. Velocity of $P$ and $Q$ after impact must be equal |
| Speed of combined particle immediately after impact $= v = 1.79 \text{ ms}^{-1}$ | A1 | Must be positive |

**Special case (7c):**

| Answer | Marks | Guidance |
|--------|-------|----------|
| $u_{p(\text{down})} = 0 + 4.4 \times 1\ [= 4.4]$ | B1 FT | Allow $\pm 4.4$. FT on *their* 1 and *their* 4.4 |
| $u_{q(\text{up})} = 10 - 11.6 \times 1\ [= -1.6]$ so $u_{q(\text{down})} = 1.6$ | B1 FT | Allow $\pm(10 - 11.6 \times 1)$, FT on *their* 1 |
| $\pm m \times 4.4 \pm 2m \times 1.6 = \pm(m+2m)v$ | M1 | Apply conservation of momentum, 4 terms, using *their* $u_p$ and $u_q$ values with $m$ and $2m$ respectively. Velocity of $P$ and $Q$ after impact must be equal |
| Speed of combined particle immediately after impact $= v = 2.53 \text{ ms}^{-1}$ | A1 | Allow $v = \frac{38}{15}$. Must be positive |