A-Level Mathematics May/June 2024 Q5(b): (b) It is given instead that the coefficient of friction between the bobsled and the sl…

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  Let the mass of the bobsled be $m$ kg. The slope is inclined at an angle $\theta = 12^\circ$. The coefficient of friction is $\mu = 0.03$. The initial speed is $u = 5 \text{ m s}^{-1}$ and the distance is $s = 60 \text{ m}$.

  First, we find the acceleration, $a$, of the bobsled down the slope.

  Resolve forces perpendicular to the slope: $R - mg \cos(12^\circ) = 0$ $R = mg \cos(12^\circ)$

  The frictional force, $F$, is given by $F = \mu R$. $F = 0.03 \times mg \cos(12^\circ)$

  Apply Newton's second law parallel to the slope, in the direction of motion: The component of weight down the slope is $mg \sin(12^\circ)$. The net force is $mg \sin(12^\circ) - F$. $mg \sin(12^\circ) - F = ma$

  Substitute the expression for $F$: $mg \sin(12^\circ) - 0.03 mg \cos(12^\circ) = ma$

  The mass $m$ cancels from all terms: $a = g \sin(12^\circ) - 0.03 g \cos(12^\circ)$ $a = g (\sin(12^\circ) - 0.03 \cos(12^\circ))$ Using $g = 10 \text{ m s}^{-2}$: $a = 10 (\sin(12^\circ) - 0.03 \cos(12^\circ))$ $a = 10 (0.207911... - 0.03 \times 0.978147...)$ $a = 10 (0.207911... - 0.029344...)$ $a = 10 (0.178567...)$ $a = 1.78567... \text{ m s}^{-2}$

  Now, use a constant acceleration (suvat) equation to find the time, $t$. We have: $s = 60$, $u = 5$, $a = 1.78567...$, $t = ?$ Using $s = ut + \frac{1}{2}at^2$: $60 = 5t + \frac{1}{2}(1.78567...)t^2$ $60 = 5t + 0.89283...t^2$

  Rearrange into a quadratic equation $Ax^2 + Bx + C = 0$: $0.89283...t^2 + 5t - 60 = 0$

  Solve for $t$ using the quadratic formula, $t = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}$: $t = \frac{-5 \pm \sqrt{5^2 - 4(0.89283...)(-60)}}{2(0.89283...)}$ $t = \frac{-5 \pm \sqrt{25 + 214.280...}}{1.78567...}$ $t = \frac{-5 \pm \sqrt{239.280...}}{1.78567...}$ $t = \frac{-5 \pm 15.468...}{1.78567...}$

  Since time must be positive: $t = \frac{-5 + 15.468...}{1.78567...} = \frac{10.468...}{1.78567...}$ $t = 5.8626...$

  The time taken is $5.86$ s (to 3 s.f.).

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  How the marks are awarded

  • B1 — Correctly resolving forces perpendicular to the plane to find the normal reaction force, shown by the equation $R = mg \cos(12^\circ)$.
  • M1 — Applying Newton's second law down the slope, with three terms representing the weight component, friction, and the resultant force: $mg \sin(12^\circ) - F = ma$.
  • DM1 — Substituting the friction formula $F = 0.03R$ and the expression for $R$ into the equation of motion, then solving to find a value for acceleration, $a \approx 1.79$.
  • DM1 — Using a complete kinematic method, specifically $s = ut + \frac{1}{2}at^2$, with $s=60$, $u=5$ and the calculated value of $a$ to form a quadratic equation for time $t$.
  • A1 — Obtaining the correct final answer for the time, $t = 5.86$ s, after correctly solving the quadratic equation.

  Common mistakes

  • Confusing sine and cosine components, for example using $mg \cos(12^\circ)$ for the force down the slope and $mg \sin(12^\circ)$ for the normal reaction.
  • Making a sign error in Newton's second law, such as adding the frictional force instead of subtracting it ($mg \sin(12^\circ) + F = ma$), implying friction assists motion.
  • Forgetting the initial speed and using the simpler kinematic formula $s = \frac{1}{2}at^2$, which is only valid for objects starting from rest.
  • Using a prematurely rounded value for acceleration (e.g., $a=1.8$) which can lead to a final answer outside the acceptable accuracy range if not handled carefully.

  Examiner tip: Master the standard procedure for inclined plane problems: resolve forces perpendicular and parallel to the slope, then use the resulting acceleration in the appropriate constant acceleration (suvat) equations.

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