Mathematics 9709/42 — February/March 2020

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

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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 · Momentum

Q1Energy, Work and PowerNewton's Laws of MotionFree sample

A lorry of mass $16\,000\text{ kg}$ is travelling along a straight horizontal road. The engine of the lorry is working at constant power. The work done by the driving force in $10\text{ s}$ is $750\,000\text{ J}$ .

(a)

Find the power of the lorry's engine.

DifficultyEasy

Worked solution

Approach

Power is the rate at which work is done. Since the engine works at constant power, divide the total work done by the time taken.

Working

P = \frac{W}{t} = \frac{750000}{10} = 75000\text{ W}

So the power is $75000\text{ W}$ , or $75\text{ kW}$ .

Answer

P = 75000\text{ W} = 75\text{ kW}

Final answer

75000 W (75 kW)

Detailed explanation

Walkthrough

The question gives the work done by the driving force over a fixed time interval. Power tells us how quickly work is done, so we divide the work by the time. Here $W = 750000\text{ J}$ and $t = 10\text{ s}$ , so $P = 750000/10 = 75000\text{ W}$ . Since $1000\text{ W} = 1\text{ kW}$ , this is $75\text{ kW}$ .

Key Takeaways

Power is the rate of doing work. When power is constant, the work done in a time interval is simply power multiplied by time, so power can be found by dividing work by time.

Common Mistakes

Forgetting to divide work by time and quoting the work done as the power.
Not converting watts to kilowatts when asked for the power in kW.

Things to Be Careful About

The units must be consistent: work in joules and time in seconds give power in watts. The mass of the lorry is not needed in part (a).

Techniques used

use rate of doing work to find powersubstitute given work and time

(b)

There is a constant resistance force acting on the lorry of magnitude $2400\text{ N}$ .

Find the acceleration of the lorry at an instant when its speed is $25\text{ m s}^{-1}$ .

DifficultyMedium-Easy

Worked solution

Approach

Use the relationship between power, driving force and speed to find the driving force at the instant when the speed is $25\text{ m s}^{-1}$ . Then apply Newton's second law, remembering that the resistance acts against the motion.

Working

At speed $v = 25\text{ m s}^{-1}$ and power $P = 75000\text{ W}$ :

F = \frac{P}{v} = \frac{75000}{25} = 3000\text{ N}

The resistance is $2400\text{ N}$ and opposes the motion, so the resultant force is:

F_{\text{net}} = 3000 - 2400 = 600\text{ N}

Using Newton's second law, $F_{\text{net}} = ma$ :

600 = 16000a

a = \frac{600}{16000} = 0.0375\text{ m s}^{-2}

Answer

a = 0.0375\text{ m s}^{-2} \quad \left(\text{or } \frac{3}{80}\text{ m s}^{-2}\right)

Final answer

0.0375 m s^{-2} (or 3/80 m s^{-2})

Detailed explanation

Walkthrough

In part (a) we found the engine power is $75000\text{ W}$ . At the instant the lorry is moving at $25\text{ m s}^{-1}$ , the driving force is related to power and speed by $P = Fv$ , so $F = P/v = 75000/25 = 3000\text{ N}$ .

The resistance force of $2400\text{ N}$ acts in the opposite direction to the motion. Therefore the resultant force on the lorry is $3000 - 2400 = 600\text{ N}$ forward.

Newton's second law says resultant force equals mass times acceleration. The lorry's mass is $16000\text{ kg}$ , so $600 = 16000a$ , giving $a = 600/16000 = 0.0375\text{ m s}^{-2}$ .

Key Takeaways

When an engine works at constant power, the driving force changes with speed: $P = Fv$ . To find acceleration, first find the resultant force by combining the driving force and the resistance, then use Newton's second law.

Common Mistakes

Using the power from part (a) as the driving force instead of dividing by the speed.
Adding the resistance instead of subtracting it. The resistance opposes motion, so it reduces the resultant force.
Forgetting to include the mass of the lorry in Newton's second law.

Things to Be Careful About

The driving force is not constant because the power is constant and the speed changes. The value $3000\text{ N}$ is the driving force only at the instant when $v = 25\text{ m s}^{-1}$ . Use consistent units: newtons, kilograms and metres per second squared.

Techniques used

relate power, driving force and speedapply Newton's second law with resistance opposing motionsolve for acceleration

The rest of this paper

6 more questions

Q2Kinematics of Motion in a Straight Line · Forces and Equilibrium · Newton's Laws of Motion6M
Q3Energy, Work and Power6M
Q4Kinematics of Motion in a Straight Line7M
Q5Forces and Equilibrium8M
Q6Newton's Laws of Motion · Kinematics of Motion in a Straight Line · Momentum9M
Q7Kinematics of Motion in a Straight Line10M

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