| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Collision or meeting problems |
| Difficulty | Standard +0.3 This is a straightforward M1 question requiring standard differentiation/integration of polynomials and solving simple equations. Parts (i)-(iii) are routine calculus, and part (iv) involves integrating Q's velocity and equating positions—all standard techniques with clear signposting. Slightly above average due to the multi-part nature and algebraic manipulation required, but no novel insight needed. |
| Spec | 1.07i Differentiate x^n: for rational n and sums1.08d Evaluate definite integrals: between limits3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v = \frac{dx}{dt} = 0.3t^2 - 0.6t + 0.2\) m s\(^{-1}\) | M1 A1 | Differentiate displacement |
| \(a = \frac{dv}{dt} = 0.6t - 0.6\) m s\(^{-2}\) | M1 A1 | Differentiate velocity |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(a = 0 \Rightarrow 0.6t - 0.6 = 0 \Rightarrow t = 1\) | M1 | |
| \(x = 0.1(1)^3 - 0.3(1)^2 + 0.2(1) = 0.1 - 0.3 + 0.2 = 0\) | M1 A1 | Hence P is at O |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(v = 0\): \(0.3t^2 - 0.6t + 0.2 = 0\) | M1 | |
| \(t = \frac{0.6 \pm \sqrt{0.36 - 0.24}}{0.6} = \frac{0.6 \pm \sqrt{0.12}}{0.6}\) | M1 | |
| \(t = 0.423\) s or \(t = 1.577\) s | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Displacement of Q from O: integrate \(v_Q = 0.2t^2 - 0.4t\): \(x_Q = \frac{0.2t^3}{3} - 0.2t^2\) | M1 | |
| At collision, \(x_P = x_Q\): \(0.1T^3 - 0.3T^2 + 0.2T = \frac{0.2T^3}{3} - 0.2T^2\) | M1 A1 | |
| \(0.1T^3 - \frac{0.2T^3}{3} - 0.1T^2 + 0.2T = 0\) | M1 | |
| Divide by \(T\) (first collision so \(T \neq 0\)): \(\frac{T^2}{30} - \frac{T}{10} + \frac{1}{5} = 0\)... simplify | M1 | |
| Multiply through: \(T^2 - 9T + 18 = 0\) | A1 | shown |
| \((T-3)(T-6) = 0\), \(T = 3\) or \(T = 6\); first collision so \(T = 3\) s | A1 |
# Question 7:
**(i)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = \frac{dx}{dt} = 0.3t^2 - 0.6t + 0.2$ m s$^{-1}$ | M1 A1 | Differentiate displacement |
| $a = \frac{dv}{dt} = 0.6t - 0.6$ m s$^{-2}$ | M1 A1 | Differentiate velocity |
**(ii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $a = 0 \Rightarrow 0.6t - 0.6 = 0 \Rightarrow t = 1$ | M1 | |
| $x = 0.1(1)^3 - 0.3(1)^2 + 0.2(1) = 0.1 - 0.3 + 0.2 = 0$ | M1 A1 | Hence P is at O |
**(iii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $v = 0$: $0.3t^2 - 0.6t + 0.2 = 0$ | M1 | |
| $t = \frac{0.6 \pm \sqrt{0.36 - 0.24}}{0.6} = \frac{0.6 \pm \sqrt{0.12}}{0.6}$ | M1 | |
| $t = 0.423$ s or $t = 1.577$ s | A1 | |
**(iv)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Displacement of Q from O: integrate $v_Q = 0.2t^2 - 0.4t$: $x_Q = \frac{0.2t^3}{3} - 0.2t^2$ | M1 | |
| At collision, $x_P = x_Q$: $0.1T^3 - 0.3T^2 + 0.2T = \frac{0.2T^3}{3} - 0.2T^2$ | M1 A1 | |
| $0.1T^3 - \frac{0.2T^3}{3} - 0.1T^2 + 0.2T = 0$ | M1 | |
| Divide by $T$ (first collision so $T \neq 0$): $\frac{T^2}{30} - \frac{T}{10} + \frac{1}{5} = 0$... simplify | M1 | |
| Multiply through: $T^2 - 9T + 18 = 0$ | A1 | shown |
| $(T-3)(T-6) = 0$, $T = 3$ or $T = 6$; first collision so $T = 3$ s | A1 | |7 A particle $P$ is projected from a fixed point $O$ on a straight line. The displacement $x$ m of $P$ from $O$ at time $t \mathrm {~s}$ after projection is given by $x = 0.1 t ^ { 3 } - 0.3 t ^ { 2 } + 0.2 t$.\\
(i) Express the velocity and acceleration of $P$ in terms of $t$.\\
(ii) Show that when the acceleration of $P$ is zero, $P$ is at $O$.\\
(iii) Find the values of $t$ when $P$ is stationary.
At the instant when $P$ first leaves $O$, a particle $Q$ is projected from $O$. $Q$ moves on the same straight line as $P$ and at time $t \mathrm {~s}$ after projection the velocity of $Q$ is given by $\left( 0.2 t ^ { 2 } - 0.4 \right) \mathrm { ms } ^ { - 1 } . P$ and $Q$ collide first when $t = T$.\\
(iv) Show that $T$ satisfies the equation $t ^ { 2 } - 9 t + 18 = 0$, and hence find $T$.
\hfill \mbox{\textit{OCR M1 2011 Q7 [17]}}