Mathematics 9709/43 — October/November 2014

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

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

Q1Energy, Work and PowerNewton's Laws of MotionFree sample

A car of mass $1400\text{ kg}$ moves on a horizontal straight road. The resistance to the car’s motion is constant and equal to $800\text{ N}$ and the power of the car’s engine is constant and equal to $P\text{ W}$ . At an instant when the car’s speed is $18\text{ m s}^{-1}$ its acceleration is $0.5\text{ m s}^{-2}$ .

(i)

Find the value of $P$ .

DifficultyMedium-Easy

Worked solution

Approach

At speed $v$ the driving force is given by $\text{DF} = \frac{P}{v}$ . Along the horizontal road, the resultant force is the driving force minus the constant resistance, and this equals $ma$ .

Working

At $v = 18\ \text{m s}^{-1}$ ,

\text{DF} = \frac{P}{18}

Using $\text{DF} - R = ma$ :

\frac{P}{18} - 800 = 1400(0.5) = 700

Therefore

\frac{P}{18} = 700 + 800 = 1500

P = 1500 \times 18 = 27000\ \text{W}

Answer

P = 27000\ \text{W}

Final answer

P = 27000 W

Detailed explanation

Walkthrough

This part asks for the constant engine power $P$ . The key link is that the engine power is the rate at which the driving force does work, so at a given speed $v$ the instantaneous driving force is $\text{DF} = \frac{P}{v}$ . Once the driving force is written in terms of $P$ , Newton's second law gives the equation of motion: the driving force pushes the car forward, while the $800\ \text{N}$ resistance opposes it, so the resultant horizontal force is $\text{DF} - 800$ . This resultant equals the car's mass times its acceleration. Substituting $v = 18$ and $a = 0.5$ gives a single linear equation in $P$ , which is then solved.

Key Takeaways

The engine power of a moving vehicle is related to the driving force and the speed by $P = Fv$ . The driving force depends on the speed even when the power is constant. Newton's second law is applied by taking the vector sum of horizontal forces and equating it to $ma$ .

Common Mistakes

A common mistake is to confuse the engine power $P$ with the driving force. The power is fixed, but the driving force at an instant is $P/v$ . Another common mistake is to add the resistance to the driving force instead of subtracting it, since resistance opposes the motion.

Things to Be Careful About

The resistance is constant, but the driving force is not constant because the speed changes. The values given are instantaneous, so the equation $\text{DF} - R = ma$ is used at the single instant speed is $18\ \text{m s}^{-1}$ . Keep the units consistent: mass in kg, speed in m s $^{-1}$ , acceleration in m s $^{-2}$ , and force in N.

Techniques used

convert engine power into driving force using the relationship between power, force and speedapply Newton's second law along the direction of motionsolve the resulting equation for the unknown power

(ii)

The car continues and passes through another point with speed $25\text{ m s}^{-1}$ .

Find the car’s acceleration at this point.

DifficultyMedium-Easy

Worked solution

Approach

Since the engine power is constant, use $\text{DF} = \frac{P}{v}$ at the new speed $v = 25\ \text{m s}^{-1}$ . Then apply Newton's second law, $\text{DF} - R = ma$ , to solve for $a$ .

Working

The driving force at $25\ \text{m s}^{-1}$ is

\text{DF} = \frac{27000}{25} = 1080\ \text{N}

Newton's second law gives

1080 - 800 = 1400a

280 = 1400a

a = \frac{280}{1400} = 0.2\ \text{m s}^{-2}

Answer

a = 0.2\ \text{m s}^{-2}

Final answer

a = 0.2 m s^-2

Detailed explanation

Walkthrough

Since the power is constant, the value $P = 27000\ \text{W}$ obtained in part (i) is still valid at the later point. At the new speed $v = 25\ \text{m s}^{-1}$ , the driving force is $\text{DF} = \frac{27000}{25} = 1080\ \text{N}$ . The resistance is still $800\ \text{N}$ , so the resultant forward force is $1080 - 800 = 280\ \text{N}$ . Applying Newton's second law gives $280 = 1400a$ , so $a = 0.2\ \text{m s}^{-2}$ . The positive sign shows the car is still accelerating forward.

Key Takeaways

A constant engine power means the driving force changes as speed changes: it is smaller at higher speeds. To find acceleration, compute the instantaneous driving force from $P = Fv$ , subtract the resistance, and divide by the mass.

Common Mistakes

Forgetting to use the previously found value of $P$ and instead treating the driving force as $800\ \text{N}$ or as $P$ . Another common mistake is using the initial speed $18\ \text{m s}^{-1}$ instead of the new speed $25\ \text{m s}^{-1}$ . Some candidates also divide $P$ by the mass instead of using the driving force.

Things to Be Careful About

The acceleration is positive and the units are m s $^{-2}$ . Do not round prematurely; $280/1400$ is exactly $0.2$ . The resistance force remains $800\ \text{N}$ throughout because it is stated to be constant.

Techniques used

use the constant power value to calculate the driving force at the new speedapply Newton's second law with the driving force and constant resistancesolve for the acceleration

The rest of this paper

6 more questions

Q2Newton's Laws of Motion5M
Q3Forces and Equilibrium6M
Q4Kinematics of Motion in a Straight Line7M
Q5Forces and Equilibrium · Newton's Laws of Motion8M
Q6Newton's Laws of Motion · Kinematics of Motion in a Straight Line8M
Q7Newton's Laws of Motion · Energy, Work and Power · Forces and Equilibrium · Kinematics of Motion in a Straight Line11M

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