| Exam Board | OCR |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Friction |
| Type | Practical friction scenarios |
| Difficulty | Moderate -0.3 This is a straightforward two-part mechanics question requiring standard SUVAT equations (constant acceleration kinematics) and Newton's second law with friction. The setup is clear, all necessary information is given, and the solution path is direct: find deceleration from u, v, s, t; then use F=ma and μ=F/R. Slightly easier than average due to its routine nature, though it does require connecting multiple concepts correctly. |
| Spec | 3.03d Newton's second law: 2D vectors3.03t Coefficient of friction: F <= mu*R model |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(36 = 0 +/- a \cdot 24^2/2\) | M1 | \(s = vt - at^2/2 = 0 \;+/-\; at^2/2\) OR \(s = ut \;+/-\; at^2/2\) |
| \(a = +/- 0.125 \text{ ms}^{-2}\) | A1 [2] | \(\frac{1}{8}\) |
| OR \(U = \pm 24a\) and \(0^2 = (24a)^2 \pm 2a \cdot 36\) | M1 | Use both \(0 = u \pm 24a\) and \(0^2 = u^2 \pm 2a \cdot 36\), \(U = 3 \text{ ms}^{-1}\) |
| \(a = \pm 0.125 \text{ ms}^{-2} = \pm \frac{1}{8} \text{ ms}^{-2}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((180/g)a = Fr\) | M1 | Mass \(= 18.367\ldots\) kg. Regard \(180a = Fr\) as MR |
| \(Fr = \pm 2.3(0) \text{ N}\) | A1 | May be implied. \(Fr = 22.5\) MR \(-1\) |
| \(\mu = 2.3/180\) | M1 | \(Fr\) and \(R\) both \(+\)ve or both \(-\)ve, \(\mu = 22.5/(180 \times 9.8)\) if MR |
| \(\mu = 0.0128\) | A1 [4] | Award if MR |
# Question 2:
## Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $36 = 0 +/- a \cdot 24^2/2$ | M1 | $s = vt - at^2/2 = 0 \;+/-\; at^2/2$ OR $s = ut \;+/-\; at^2/2$ |
| $a = +/- 0.125 \text{ ms}^{-2}$ | A1 **[2]** | $\frac{1}{8}$ |
| OR $U = \pm 24a$ and $0^2 = (24a)^2 \pm 2a \cdot 36$ | M1 | Use both $0 = u \pm 24a$ and $0^2 = u^2 \pm 2a \cdot 36$, $U = 3 \text{ ms}^{-1}$ |
| $a = \pm 0.125 \text{ ms}^{-2} = \pm \frac{1}{8} \text{ ms}^{-2}$ | A1 | |
## Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(180/g)a = Fr$ | M1 | Mass $= 18.367\ldots$ kg. Regard $180a = Fr$ as MR |
| $Fr = \pm 2.3(0) \text{ N}$ | A1 | May be implied. $Fr = 22.5$ MR $-1$ |
| $\mu = 2.3/180$ | M1 | $Fr$ and $R$ both $+$ve or both $-$ve, $\mu = 22.5/(180 \times 9.8)$ if MR |
| $\mu = 0.0128$ | A1 **[4]** | Award if MR |
---2 In the sport of curling, a heavy stone is projected across a horizontal ice surface. One player projects a stone of weight 180 N , which moves 36 m in a straight line and comes to rest 24 s after the instant of projection. The only horizontal force acting on the stone after its projection is a constant frictional force between the stone and the ice.\\
(i) Calculate the deceleration of the stone.\\
(ii) Find the magnitude of the frictional force acting on the stone, and calculate the coefficient of friction between the stone and the ice.
\hfill \mbox{\textit{OCR M1 2012 Q2 [6]}}