Mathematics 9709/43 — May/June 2013

Cambridge A-Level · Mechanics · worked solutions for every part, with the mark scheme

questions

marks

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Topics Kinematics of Motion in a Straight Line · Newton's Laws of Motion · Energy, Work and Power · Forces and Equilibrium

Q14MNewton's Laws of MotionKinematics of Motion in a Straight LineFree sample

A straight ice track of length $50\text{ m}$ is inclined at $14^{\circ}$ to the horizontal. A man starts at the top of the track, on a sledge, with speed $8\text{ m s}^{-1}$ . He travels on the sledge to the bottom of the track. The coefficient of friction between the sledge and the track is $0.02$ . Find the speed of the sledge and the man when they reach the bottom of the track.

DifficultyMedium

Worked solution

Approach

The motion is down a straight inclined track, so we resolve the forces along and perpendicular to the slope. Newton’s second law along the slope gives the constant acceleration. With that acceleration, the suvat equation $v^2 = u^2 + 2as$ gives the final speed at the bottom.

Working

Let the combined weight of the man and sledge be $W = mg$ , and let $R$ be the normal reaction. Perpendicular to the slope, the sledge does not accelerate, so:

R = W\cos 14^{\circ}

The friction force acts up the slope and has magnitude

F = \mu R = 0.02W\cos 14^{\circ}

The component of weight down the slope is $W\sin 14^{\circ}$ . Therefore the resultant force down the slope is

W\sin 14^{\circ} - 0.02W\cos 14^{\circ}

Applying Newton’s second law, $F = ma$ , with mass $m = \frac{W}{g}$ :

\frac{W}{g}a = W\sin 14^{\circ} - 0.02W\cos 14^{\circ}

Cancel $W$ and use $g = 10\text{ m s}^{-2}$ :

a = g\left(\sin 14^{\circ} - 0.02\cos 14^{\circ}\right) = 2.225\ldots\text{ m s}^{-2}

Now use $v^2 = u^2 + 2as$ with $u = 8\text{ m s}^{-1}$ and $s = 50\text{ m}$ :

v^2 = 8^2 + 2(2.225\ldots)(50) = 286.5\ldots

Therefore

v = \sqrt{286.5\ldots} = 16.9\text{ m s}^{-1}\quad (3\text{ s.f.})

Answer

16.9\text{ m s}^{-1}

Final answer

16.9 m s^-1

Detailed explanation

Walkthrough

The track is inclined at $14^{\circ}$ and the sledge slides down it. The weight $W$ must be resolved into two components: $W\sin 14^{\circ}$ parallel to the slope, pulling the sledge down the slope, and $W\cos 14^{\circ}$ perpendicular to the slope, pushing it into the track. There is no motion perpendicular to the track, so the normal reaction $R$ balances this perpendicular component: $R = W\cos 14^{\circ}$ .

Because the sledge is moving down the slope, friction acts up the slope and its magnitude is $F = \mu R = 0.02W\cos 14^{\circ}$ . The resultant force down the slope is therefore

W\sin 14^{\circ} - 0.02W\cos 14^{\circ}.

Newton’s second law is $F = ma$ . Here the mass is $m = \frac{W}{g}$ , so

\frac{W}{g}a = W\sin 14^{\circ} - 0.02W\cos 14^{\circ}.

Since $W$ appears on both sides, it cancels, giving

a = g\left(\sin 14^{\circ} - 0.02\cos 14^{\circ}\right) = 2.225\ldots\text{ m s}^{-2}.

The acceleration is constant, so the suvat equation $v^2 = u^2 + 2as$ can be used. Let the direction of motion be positive. With $u = 8\text{ m s}^{-1}$ and $s = 50\text{ m}$ :

v^2 = 8^2 + 2(2.225\ldots)(50) = 286.5\ldots

Taking the positive square root gives the final speed

v = 16.9\text{ m s}^{-1}.

Key Takeaways

This question combines force analysis on an inclined plane with kinematics. The key idea is to resolve the weight into components along and perpendicular to the slope, use $F = \mu R$ for friction, and apply Newton’s second law to find the acceleration. Once the acceleration is known and is constant, the suvat equation $v^2 = u^2 + 2as$ gives the final speed without needing to find the time.

Common Mistakes

Using $R = W$ instead of $R = W\cos \alpha$ . This is the most common error when dealing with inclined planes.
Drawing friction down the slope. Since the sledge is moving down the track, friction opposes the motion and acts up the slope.
Forgetting that the mass is $W/g$ when applying $F = ma$ , which can lead to an incorrect equation.
Using the wrong value of $g$ . The mark scheme value corresponds to $g = 10\text{ m s}^{-2}$ , so use the value specified in the paper.
Substituting the vertical height instead of the distance along the slope into $s$ . Here $s = 50\text{ m}$ , the length of the track.
Not showing the Newton’s second law equation; the mark scheme awards a method mark for this key step.

Things to Be Careful About

Make sure your calculator is in degree mode, because the slope angle is given as $14^{\circ}$ .
Choose a clear positive direction: here down the slope is positive, so the acceleration is positive and the final speed is greater than the initial speed.
When taking the square root, only the positive value is physically meaningful for speed.
The answer should be rounded to $16.9\text{ m s}^{-1}$ , correct to 3 significant figures.
The friction coefficient is small, but it still slightly reduces the acceleration from $g\sin 14^{\circ}$ .

Techniques used

resolve forces along and perpendicular to the slopeapply Newton's second law to find the accelerationuse the constant acceleration formula to find final speed

The rest of this paper

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