Mathematics 9709/43 — October/November 2020

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

questions

marks

minutes

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

Q1Kinematics of Motion in a Straight LineFree sample

A particle $P$ is projected vertically upwards with speed $v\text{ m s}^{-1}$ from a point on the ground. $P$ reaches its greatest height after $3\text{ s}$ .

(a)

Find $v$ .

DifficultyEasy

Worked solution

Approach

At the greatest height the particle's velocity is zero. Use the suvat equation $v = u + at$ with $v = 0$ , $a = -g$ and $t = 3$ to find the initial speed $u$ .

Working

Take upwards as positive and $g = 10\text{ m s}^{-2}$ .

At greatest height, $v = 0$ , $a = -10$ , $t = 3$ .

v = u + at

0 = u + (-10)(3)

u = 30

Answer

v = 30\text{ m s}^{-1}

Final answer

v = 30 m s^-1

Detailed explanation

Walkthrough

At the highest point of a vertical projection, the particle instantaneously stops before it starts falling back down, so its velocity is zero. We are told it takes 3 s to reach that point. The only force acting is gravity, so the acceleration is $g$ downwards. Taking upwards as positive, the acceleration is $-g$ . With $g = 10$ , substitute $v = 0$ , $a = -10$ and $t = 3$ into the suvat equation $v = u + at$ and solve for the initial speed $u$ .

Key Takeaways

At maximum height in vertical motion, the final velocity is $0$ .
The suvat equation $v = u + at$ links initial speed, final speed, acceleration and time.
Choose a positive direction and keep the sign of acceleration consistent throughout.

Common Mistakes

Forgetting that $v = 0$ at the greatest height.
Using $a = +10$ instead of $-10$ when upwards is positive.
Not stating the assumed value of $g$ ; the mark scheme uses $g = 10$ .

Things to Be Careful About

The initial speed is positive when upwards is taken as positive, so $u = 30$ is correct.
The mark scheme awards B1 for $v = 30$ ; showing the substitution is still recommended because later parts rely on this value.
If a different value of $g$ is used, the answer changes; in this course $g = 10$ is assumed unless stated otherwise.

Techniques used

apply the suvat equation v = u + atuse v = 0 at greatest heightsolve for initial speed

(b)

Find the greatest height of $P$ above the ground.

DifficultyMedium-Easy

Worked solution

Approach

Use the suvat equation $v^2 = u^2 + 2as$ with $v = 0$ , $u = 30$ and $a = -10$ to find the displacement $s$ to the greatest height.

Working

At greatest height, $v = 0$ , $u = 30$ , $a = -10$ .

v^2 = u^2 + 2as

0^2 = 30^2 + 2(-10)s

0 = 900 - 20s

20s = 900

s = 45

Answer

\text{Greatest height} = 45\text{ m}

Final answer

45 m

Detailed explanation

Walkthrough

From part (a), the initial speed is $30\text{ m s}^{-1}$ . At the greatest height the velocity is again $0$ . The acceleration is still $-10\text{ m s}^{-2}$ if upwards is positive. Since we want the displacement and do not need time, use $v^2 = u^2 + 2as$ . Substitute $v = 0$ , $u = 30$ and $a = -10$ : $0 = 30^2 + 2(-10)s$ , so $0 = 900 - 20s$ , giving $s = 45$ . This displacement from the ground is the greatest height.

Key Takeaways

The suvat equation $v^2 = u^2 + 2as$ is ideal when time is not involved.
The result from part (a) becomes the initial speed in part (b).
Displacement is measured from the starting point on the ground, so the greatest height is $45\text{ m}$ .

Common Mistakes

Using $u = 0$ instead of $u = 30$ .
Using $a = +10$ instead of $-10$ , which would give the wrong sign for $s$ .
Confusing $u$ and $v$ : at greatest height the final velocity is $0$ , not the initial velocity.
Attempting part (b) without using the value from part (a).

Things to Be Careful About

The greatest height is a distance, so give a positive value: $45\text{ m}$ .
The mark scheme requires a method mark (M1) for using the correct suvat equation; an unsupported answer may not earn full marks.
Equivalent methods such as $s = ut + \frac12 at^2$ with $t = 3$ are acceptable and give the same result.

Techniques used

apply the suvat equation v^2 = u^2 + 2assubstitute known valuessolve for displacement

The rest of this paper

6 more questions

Q2Energy, Work and Power4M
Q3Forces and Equilibrium6M
Q4Momentum · Energy, Work and Power6M
Q5Kinematics of Motion in a Straight Line10M
Q6Forces and Equilibrium · Energy, Work and Power · Newton's Laws of Motion10M
Q7Forces and Equilibrium · Newton's Laws of Motion · Kinematics of Motion in a Straight Line11M

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