| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2016 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Collision/impulse during SHM |
| Difficulty | Challenging +1.2 This is a comprehensive SHM question with a collision component, requiring multiple standard techniques (showing SHM conditions, finding period/speed/time, applying conservation of momentum, analyzing post-collision motion). While it has many parts and requires careful bookkeeping, each individual step uses routine M3 methods without requiring novel insight—the collision aspect is straightforward momentum conservation and the SHM analysis follows standard procedures. The length and integration of topics makes it moderately above average difficulty. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.03b Conservation of momentum: 1D two particles6.03f Impulse-momentum: relation |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(T = -m\ddot{x}\) | M1 | Using NL2 with \(T\) for tension, acceleration \(a\) or \(\ddot{x}\) |
| \(\frac{15x}{1.2} = -0.5\ddot{x}\) | M1A1 | Using Hooke's Law to obtain equation connecting \(\ddot{x}\) or \(a\) and \(x\); correct equation must have \(\ddot{x}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\ddot{x} = -25x \therefore\) SHM | A1cso | \(\ddot{x} = -25x\) and stating SHM; these 4 marks available without substituting for \(m\), \(\lambda\) or \(l\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Period \(= \frac{2\pi}{5}\) (= 1.256, accept 1.3 or better) | B1ft (5) | Period \(= \frac{2\pi}{\text{their numerical } \omega}\) or decimal equivalent |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(v^2 = \omega^2(a^2 - x^2)\) | ||
| \(x = 0\), \(v = 5 \times 0.8\), \(v = 4 \text{ ms}^{-1}\) | M1A1 (2) | Using \(v^2 = \omega^2(a^2 - x^2)\) with \(x=0\) (or just \(v = a\omega\)) and \(a = 0.8\); ALT: using energy \(\frac{1}{2} \times 0.5v^2 = \frac{15 \times 0.8^2}{2 \times 1.2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(x = a\cos\omega t\) | M1 | Using \(x = a\cos\omega t\) with their \(\omega\) and \(a = 0.8\), \(x = \pm 0.6\) |
| \(x = -0.6\), \(-0.6 = 0.8\cos 5t\) | M1A1ft | Correct equation following through their \(\omega\) |
| \(t = \frac{1}{5}\cos^{-1}\left(-\frac{6}{8}\right) = 0.4837\ldots\text{s}\), accept 0.48 or better | A1cso (3) | \(t = 0.48\) or better |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(T = -m\ddot{y}\) | ||
| \(\frac{15y}{1.2} = -0.8\ddot{y}\) | M1A1 | Using NL2 with tension at new extension and increased mass; acceleration in differential form or just \(a\) |
| \(\ddot{y} = -15.625y\) or \(\ddot{y} = -\frac{125}{8}y \therefore\) SHM | A1 (3) | \(\ddot{y} = -15.625y\) and stating SHM; must have differential form for acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Con of mom: \(0.5 \times 4 = (0.5 + 0.3)V\) | M1 | Using conservation of momentum with their speed of \(P\) at \(B\) |
| \(V = \frac{2}{0.8} = 2.5 \text{ ms}^{-1}\) | A1 | Correct speed for \(R\) at \(B\) (no ft) |
| \((2.5)^2 = 15.625a^2\) | DM1 | Using \(v^2 = \omega^2(a^2 - y^2)\) with \(y=0\) (or just \(v = a\omega\)) with their speed for \(R\) and their \(\omega\); dependent on first M mark |
| \(a^2 = \frac{2.5^2}{15.625}\) | ||
| \(a = 0.6324\ldots\text{m}\), accept 0.63 or better, or \(a = \frac{\sqrt{10}}{5}\) oe | A1cso (4) [17] | \(a = 0.63\) or better or exact answer |
# Question 7:
## Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $T = -m\ddot{x}$ | M1 | Using NL2 with $T$ for tension, acceleration $a$ or $\ddot{x}$ |
| $\frac{15x}{1.2} = -0.5\ddot{x}$ | M1A1 | Using Hooke's Law to obtain equation connecting $\ddot{x}$ or $a$ and $x$; correct equation must have $\ddot{x}$ |
### Part (a)(i):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\ddot{x} = -25x \therefore$ SHM | A1cso | $\ddot{x} = -25x$ **and** stating SHM; these 4 marks available without substituting for $m$, $\lambda$ or $l$ |
### Part (a)(ii):
| Working | Marks | Guidance |
|---------|-------|----------|
| Period $= \frac{2\pi}{5}$ (= 1.256, accept 1.3 or better) | B1ft (5) | Period $= \frac{2\pi}{\text{their numerical } \omega}$ or decimal equivalent |
## Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $v^2 = \omega^2(a^2 - x^2)$ | | |
| $x = 0$, $v = 5 \times 0.8$, $v = 4 \text{ ms}^{-1}$ | M1A1 (2) | Using $v^2 = \omega^2(a^2 - x^2)$ with $x=0$ (or just $v = a\omega$) and $a = 0.8$; ALT: using energy $\frac{1}{2} \times 0.5v^2 = \frac{15 \times 0.8^2}{2 \times 1.2}$ |
## Part (c):
| Working | Marks | Guidance |
|---------|-------|----------|
| $x = a\cos\omega t$ | M1 | Using $x = a\cos\omega t$ with their $\omega$ and $a = 0.8$, $x = \pm 0.6$ |
| $x = -0.6$, $-0.6 = 0.8\cos 5t$ | M1A1ft | Correct equation following through their $\omega$ |
| $t = \frac{1}{5}\cos^{-1}\left(-\frac{6}{8}\right) = 0.4837\ldots\text{s}$, accept 0.48 or better | A1cso (3) | $t = 0.48$ or better |
## Part (d):
| Working | Marks | Guidance |
|---------|-------|----------|
| $T = -m\ddot{y}$ | | |
| $\frac{15y}{1.2} = -0.8\ddot{y}$ | M1A1 | Using NL2 with tension at new extension **and** increased mass; acceleration in differential form or just $a$ |
| $\ddot{y} = -15.625y$ or $\ddot{y} = -\frac{125}{8}y \therefore$ SHM | A1 (3) | $\ddot{y} = -15.625y$ **and** stating SHM; must have differential form for acceleration |
## Part (e):
| Working | Marks | Guidance |
|---------|-------|----------|
| Con of mom: $0.5 \times 4 = (0.5 + 0.3)V$ | M1 | Using conservation of momentum with their speed of $P$ at $B$ |
| $V = \frac{2}{0.8} = 2.5 \text{ ms}^{-1}$ | A1 | Correct speed for $R$ at $B$ (no ft) |
| $(2.5)^2 = 15.625a^2$ | DM1 | Using $v^2 = \omega^2(a^2 - y^2)$ with $y=0$ (or just $v = a\omega$) with their speed for $R$ and their $\omega$; dependent on first M mark |
| $a^2 = \frac{2.5^2}{15.625}$ | | |
| $a = 0.6324\ldots\text{m}$, accept 0.63 or better, or $a = \frac{\sqrt{10}}{5}$ oe | A1cso (4) [17] | $a = 0.63$ or better or exact answer |7. A particle $P$ of mass 0.5 kg is attached to one end of a light elastic spring, of natural length 1.2 m and modulus of elasticity 15 N . The other end of the spring is attached to a fixed point $A$ on a smooth horizontal table. The particle is placed on the table at the point $B$ where $A B = 1.2 \mathrm {~m}$. The particle is pulled away from $B$ to the point $C$, where $A B C$ is a straight line and $B C = 0.8 \mathrm {~m}$, and is then released from rest.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Show that $P$ moves with simple harmonic motion with centre $B$.
\item Find the period of this motion.
\end{enumerate}\item Find the speed of $P$ when it reaches $B$.
The point $D$ is the midpoint of $A B$.
\item Find the time taken for $P$ to move directly from $C$ to $D$.
When $P$ first comes to instantaneous rest a particle $Q$ of mass 0.3 kg is placed at $B$. When $P$ reaches $B$ again, $P$ strikes and adheres to $Q$ to form a single particle $R$.
\item Show that $R$ also moves with simple harmonic motion.
\item Find the amplitude of this motion.\\
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\hfill \mbox{\textit{Edexcel M3 2016 Q7 [17]}}