Mathematics 9709/41 — May/June 2018

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

questions

marks

minutes

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

Q13MKinematics of Motion in a Straight LineFree sample

A particle $P$ is projected vertically upwards with speed $24\text{ m s}^{-1}$ from a point $5\text{ m}$ above ground level. Find the time from projection until $P$ reaches the ground.

DifficultyMedium-Easy

Worked solution

Approach

Take upward as the positive direction. The displacement from the point of projection to the ground is $s = -5$ m, the initial velocity is $u = 24\text{ m s}^{-1}$ , and the acceleration is $a = -g = -10\text{ m s}^{-2}$ . Use the constant-acceleration formula $s = ut + \frac{1}{2}at^2$ , form the resulting quadratic in $t$ , solve it, and reject the negative root because $t$ is the time after projection.

Working

Using $s = ut + \frac{1}{2}at^2$ :

-5 = 24t + \frac{1}{2}(-10)t^2

-5 = 24t - 5t^2

Rearrange:

5t^2 - 24t - 5 = 0

Factorise:

(5t + 1)(t - 5) = 0

So $t = -0.2$ or $t = 5$ . Since time after projection cannot be negative, reject $t = -0.2$ .

Answer

t = 5\text{ s}

Final answer

t = 5 s

Detailed explanation

Walkthrough

The particle starts 5 m above ground and is projected vertically upwards. It rises, stops, then falls back past its starting point and continues down to ground level. We use displacement, not distance: taking upward as positive, the displacement from the point of projection to the ground is $s = -5$ m because it ends 5 m below where it started.

Choose the constant-acceleration (suvat) formula $s = ut + \frac{1}{2}at^2$ . Substitute $u = 24\text{ m s}^{-1}$ , $a = -10\text{ m s}^{-2}$ and $s = -5$ m:

-5 = 24t - 5t^2

Rearrange to $5t^2 - 24t - 5 = 0$ . Factorising gives $(5t + 1)(t - 5) = 0$ , so $t = -0.2$ or $t = 5$ . The negative root is impossible because $t$ is measured after the instant of projection, so $t = 5$ s.

Key Takeaways

Motion under gravity with no air resistance can be solved with constant-acceleration (suvat) equations.
Displacement is signed: it depends on the chosen positive direction.
Quadratic equations in motion often have one non-physical root; identify and reject it.
The same final answer can be found by splitting the motion at the highest point; both methods must be shown clearly for method marks.

Common Mistakes

Using $s = 5$ instead of $s = -5$ with upward chosen as positive.
Using $a = +10\text{ m s}^{-2}$ when upward has been chosen positive.
Forgetting to solve the 3-term quadratic, or giving only $t = 5$ without showing the equation, which would lose method marks.
Accepting the negative solution as a valid time.

Things to Be Careful About

Set a consistent sign convention and use it for every vector quantity ( $s$ , $u$ , $a$ ).
The mark scheme awards one method mark for forming $s = ut + \frac{1}{2}at^2$ and one for solving the quadratic; full working is required for full marks.
If the alternative method is used, remember that the particle reaches 33.8 m above the ground at its highest point, then falls that distance to the ground; the separate times are 2.4 s and 2.6 s, giving 5 s in total.
Check units: time is measured in seconds.

Techniques used

apply constant-acceleration (suvat) formula with correct sign conventionform and solve a quadratic in timereject the physically impossible negative root

The rest of this paper

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Q7Newton's Laws of Motion · Energy, Work and Power · Forces and Equilibrium12M

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