Mathematics 9709/43 — October/November 2013

Walkthrough

The particle is moving up a rough inclined plane, so while it is in contact with the plane it experiences three forces: its weight vertically downwards, the normal reaction perpendicular to the plane, and friction acting down the slope because it opposes the upward motion.

First find the normal reaction. Because there is no acceleration perpendicular to the plane, the normal reaction balances the component of weight perpendicular to the plane, so $R = W\cos\alpha$.

The friction force when the particle slides is $F = \mu R$. Substituting $\mu = \frac{1}{3}$ and $R = W\cos\alpha$ gives the friction force down the plane.

We are told $\sin\alpha = 0.28$. To use $\cos\alpha$ in the normal reaction, use the identity $\sin^2\alpha + \cos^2\alpha = 1$; this gives $\cos\alpha = 0.96$.

Now apply Newton's second law parallel to the plane. Choose up the slope as positive. The forces along the slope are friction $F$ down the slope and the component $W\sin\alpha$ of the weight down the slope, so the total force is $-F - W\sin\alpha$. This equals $ma$, where $m = W/g$.

Substitute the numbers and cancel $W$: the acceleration is $(-0.32 - 0.28)g = -0.6g$. Using $g = 10\text{ m s}^{-2}$ gives $-6\text{ m s}^{-2}$, as required. The negative sign tells us the acceleration is down the slope, so the particle is slowing down while moving up.

Key Takeaways

• On an inclined plane, the weight has component $W\sin\alpha$ down the slope and $W\cos\alpha$ perpendicular to the slope.
• For a moving particle on a rough plane, friction has magnitude $\mu R$ and acts opposite to the direction of motion.
• Newton's second law can be applied along the slope once the normal reaction is found from the perpendicular direction.
• The mass cancels, so the acceleration is independent of the mass of the particle.

Common Mistakes

• Using $R = W$ instead of $R = W\cos\alpha$.
• Putting friction up the slope. Because the particle is moving up, friction must act down the slope.
• Forgetting that both friction and the component of weight act down the slope, so their signs are both negative in the chosen positive-up direction.
• Confusing $\sin\alpha$ and $\cos\alpha$ when resolving the weight.

Things to Be Careful About

• Use the identity $\sin^2\alpha + \cos^2\alpha = 1$ correctly: with $\sin\alpha = 0.28$, $\cos\alpha = 0.96$.
• The question expects $g = 10\text{ m s}^{-2}$; otherwise the numerical value would not be exactly $-6$.
• This is an AG (answer given) result, so every step must be shown clearly; an unsupported answer earns no marks.

Walkthrough

For the second part we only need the one-dimensional constant-acceleration equation linking initial velocity, final velocity, acceleration and displacement.

At the instant it reaches its highest point, the particle's speed is zero, so $v = 0$. It started with $u = 5.4\text{ m s}^{-1}$ and, from part (i), the acceleration down the slope is $6\text{ m s}^{-2}$. Using up-slope as positive, this acceleration is written as $a = -6\text{ m s}^{-2}$.

Substitute into $v^2 = u^2 + 2as$:

Then $12s = 5.4^2$, so $s = 2.43\text{ m}$.

An equivalent energy method gives the same result: initial kinetic energy equals the gain in gravitational potential energy plus the work done against friction.

Key Takeaways

• At maximum height on a slope, the instantaneous velocity is zero.
• For constant acceleration problems, $v^2 = u^2 + 2as$ is the efficient formula when time is not involved.
• The same result can be obtained using energy: loss of kinetic energy = gain in gravitational potential energy + work done against friction.

Common Mistakes

• Forgetting that $v = 0$ at the highest point.
• Using positive acceleration because the magnitude is $6\text{ m s}^{-2}$; since up-slope is positive and the particle is slowing down, acceleration must be $-6\text{ m s}^{-2}$.
• Dropping units or reporting a negative distance; distance must be positive.

Things to Be Careful About

• Choose one sign convention (up the slope positive) and apply it consistently.
• The final answer must be a distance in metres, so it is positive: $2.43\text{ m}$.
• If using the energy method, the work done against friction is $\frac{1}{3}mg\cos\alpha \times s$, and $\cos\alpha = 0.96$ must be used.