Mathematics 9709/41 — May/June 2011

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

questions

marks

minutes

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

Q1Newton's Laws of MotionEnergy, Work and PowerFree sample

A car of mass $700\text{ kg}$ is travelling along a straight horizontal road. The resistance to motion is constant and equal to $600\text{ N}$ .

(i)

Find the driving force of the car’s engine at an instant when the acceleration is $2\text{ m s}^{-2}$ .

DifficultyMedium-Easy

Worked solution

Approach

The car travels along a straight horizontal road, so the weight and normal reaction are vertical and balance. Apply Newton's second law in the horizontal direction: the resultant force is the driving force $F$ minus the resistance $600\text{ N}$ .

Working

Resolve horizontally:

F - 600 = 700 \times 2

Simplify:

F - 600 = 1400

Solve for $F$ :

F = 1400 + 600 = 2000

Answer

The driving force of the engine is $2000\text{ N}$ .

Final answer

2000 N

Detailed explanation

Walkthrough

The car is on a straight horizontal road, so only horizontal motion matters. The engine pushes the car forward with a driving force $F$ , and there is a constant resistance of $600\text{ N}$ opposing the motion. The weight and normal reaction act vertically and cancel, so they do not affect horizontal acceleration.

By Newton's second law, the resultant horizontal force equals mass times acceleration:

F - 600 = 700 \times 2

The right-hand side is the required resultant force $1400\text{ N}$ . This tells us that the driving force must overcome the $600\text{ N}$ resistance and still leave $1400\text{ N}$ to accelerate the car. Therefore:

F = 1400 + 600 = 2000\text{ N}

Key Takeaways

Newton's second law is $F_{\text{net}} = ma$ .
The resultant horizontal force is the driving force minus the resistance.
When a car accelerates, the driving force must be larger than the resistance.

Common Mistakes

Forgetting the resistance and writing $F = 700 \times 2$ ; the mark scheme requires three terms in the equation.
Using the wrong sign for the resistance, e.g. $600 - F = 1400$ .
Omitting the equation and quoting the answer; the M1 mark is for showing Newton's second law.

Things to Be Careful About

The road is horizontal, so weight does not contribute to horizontal motion.
Assume the driving force is in the direction of motion and resistance opposes it.
Include units: the result is in newtons.

Techniques used

identify the forces acting on the carapply Newton's second law along the direction of motionsolve a linear equation for the driving force

(ii)

Given that the car’s speed at this instant is $15\text{ m s}^{-1}$ , find the rate at which the car’s engine is working.

DifficultyEasy

Worked solution

Approach

The rate at which the engine is working is the power supplied by the driving force. For a force $F$ acting in the direction of motion at speed $v$ , power is $P = Fv$ . Use the driving force from part (i), $F = 2000\text{ N}$ , and the given speed $v = 15\text{ m s}^{-1}$ .

Working

P = 2000 \times 15

P = 30000\text{ W}

Since $1000\text{ W} = 1\text{ kW}$ :

P = 30\text{ kW}

Answer

The rate at which the engine is working is $30000\text{ W}$ , or $30\text{ kW}$ .

Final answer

30000 W or 30 kW

Detailed explanation

Walkthrough

The phrase "rate at which the engine is working" means the power output of the engine. Power is the rate of doing work. In one second the car travels $15\text{ m}$ , and the engine does work equal to the driving force multiplied by this distance, which is $2000 \times 15$ . This is exactly $P = Fv$ .

Using the driving force from part (i), $F = 2000\text{ N}$ , and the speed $v = 15\text{ m s}^{-1}$ :

P = 2000 \times 15 = 30000\text{ W}

So the engine is working at $30000\text{ W}$ , or $30\text{ kW}$ .

Key Takeaways

For a constant force acting in the direction of motion, power is $P = Fv$ .
Power is the rate of doing work; units are watts, where $1\text{ W} = 1\text{ J s}^{-1}$ .
The driving force is the force doing useful work for the engine, not the resultant force.

Common Mistakes

Using the net resultant force ( $2000 - 600 = 1400\text{ N}$ ) instead of the driving force $2000\text{ N}$ for engine power.
Forgetting to use the speed at the instant stated, or mixing up force and acceleration.
Not converting to kilowatts if asked, though giving $30000\text{ W}$ is also accepted.

Things to Be Careful About

The power formula requires speed in the direction of the force; here the car moves horizontally in the direction of the driving force.
The acceleration is not needed in this part; the power is computed at the instant when $v = 15\text{ m s}^{-1}$ .
The mark scheme allows follow-through: if part (i) was wrong but the method $P = Fv$ is correct, the final mark can still be awarded using the student's driving force.

Techniques used

recognise rate of working as powerapply the power formula P = Fv with the given speedconvert power from watts to kilowatts

The rest of this paper

6 more questions

Q2Energy, Work and Power5M
Q3Forces and Equilibrium6M
Q4Forces and Equilibrium7M
Q5Kinematics of Motion in a Straight Line8M
Q6Kinematics of Motion in a Straight Line9M
Q7Newton's Laws of Motion · Energy, Work and Power · Kinematics of Motion in a Straight Line11M

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