9709/43 May/June 2024 — Mathematics Solved Past Paper

Q13MMediumMomentum

Two particles $P$ and $Q$ of masses $0.2\text{ kg}$ and $0.5\text{ kg}$ respectively are at rest on a smooth horizontal plane. Particle $P$ is projected with a speed $6\text{ m s}^{-1}$ directly towards $Q$ . After $P$ and $Q$ collide, $P$ moves with a speed of $1\text{ m s}^{-1}$ .

Find the two possible speeds of $Q$ after the collision.

DifficultyMedium

Worked solution

Approach

Apply conservation of linear momentum for the two possible cases: P continues in the same direction, or P reverses direction.

Working

Take the direction of P's initial motion as positive.

Case 1: P continues in the same direction

0.2 \times 6 + 0.5 \times 0 = 0.2 \times 1 + 0.5v

1.2 = 0.2 + 0.5v

1.0 = 0.5v

v = 2 \text{ m s}^{-1}

Case 2: P reverses direction

0.2 \times 6 + 0.5 \times 0 = 0.2 \times (-1) + 0.5v

1.2 = -0.2 + 0.5v

1.4 = 0.5v

v = 2.8 \text{ m s}^{-1}

Answer

The two possible speeds of Q after the collision are $2 \text{ m s}^{-1}$ and $2.8 \text{ m s}^{-1}$ .

Final answer

2 m/s or 2.8 m/s

Detailed explanation

Walkthrough

We are told two particles collide on a smooth horizontal plane. Since the plane is smooth, there is no friction to worry about, and since the collision involves only the two particles, the total momentum of the system is conserved. Momentum is the product of mass and velocity, and it is a vector quantity, so direction matters.

The key insight is that after the collision, P could move in either direction: it could continue moving in the same direction as before, or it could bounce back and move in the opposite direction. Both are physically possible, so we must consider both cases.

We set up the conservation of momentum equation for each case. Taking the direction of P's initial motion as positive:

Case 1 (P continues forward): initial momentum = final momentum gives $0.2 \times 6 = 0.2 \times 1 + 0.5v$ , which solves to $v = 2 \text{ m s}^{-1}$ .

Case 2 (P reverses): P's final velocity is now $-1 \text{ m s}^{-1}$ , so $0.2 \times 6 = 0.2 \times (-1) + 0.5v$ , which solves to $v = 2.8 \text{ m s}^{-1}$ .

The question asks for speeds, which are magnitudes, so both answers must be positive.

Key Takeaways

Conservation of linear momentum: in the absence of external forces, total momentum before = total momentum after.
Momentum is a vector — direction matters, so we need to consider both possible directions of motion after a collision.
Speed is the magnitude of velocity, so it must be positive.

Common Mistakes

Only considering one direction for P after the collision — both cases are needed for full marks.
Giving negative answers for speed — the mark scheme explicitly says "Do not allow negative."
Using the wrong masses or velocities in the momentum equation.

Things to Be Careful About

Establish a clear sign convention (positive = P's initial direction) and apply it consistently.
Remember that speed is always non-negative; the negative sign in the momentum equation for P's reversed motion reflects direction, not speed.
The mark scheme requires three non-zero terms in the momentum equation for the method mark.

Techniques used

apply conservation of linear momentumconsider both possible directions of motion after collisionsolve linear equations

Approach

The particle is in equilibrium under three forces: its weight $W = 0.2g$ acting vertically downwards, the tension $T$ in the string acting at $30^\circ$ to the vertical wall, and the applied force $X$ acting perpendicular to the string. We resolve these forces horizontally and vertically to form two simultaneous equations, then solve for $X$ and $T$ . An alternative is to resolve along the directions of $X$ and $T$ , or use Lami's theorem.

Working

Method 1: Resolving horizontally and vertically

Let the horizontal direction away from the wall be positive $x$ , and the vertical direction upwards be positive $y$ .

The tension $T$ acts along the string towards the wall, at $30^\circ$ to the vertical. Its components are:

Horizontal: $-T \sin 30^\circ$
Vertical: $T \cos 30^\circ$

The force $X$ acts perpendicular to the string, pointing upwards and away from the wall. Since the string is at $30^\circ$ to the vertical, $X$ is at $60^\circ$ to the vertical (or $30^\circ$ to the horizontal). Its components are:

Horizontal: $X \cos 30^\circ$
Vertical: $X \sin 30^\circ$

The weight $W = 0.2g$ acts vertically downwards.

Resolving horizontally ( $\sum F_x = 0$ ):

X \cos 30^\circ - T \sin 30^\circ = 0

X \frac{\sqrt{3}}{2} - \frac{T}{2} = 0 \implies T = X\sqrt{3} \quad \text{--- (1)}

Resolving vertically ( $\sum F_y = 0$ ):

X \sin 30^\circ + T \cos 30^\circ - 0.2g = 0

\frac{X}{2} + \frac{T\sqrt{3}}{2} = 0.2g \quad \text{--- (2)}

Substitute equation (1) into equation (2):

\frac{X}{2} + \frac{(X\sqrt{3})\sqrt{3}}{2} = 0.2g

\frac{X}{2} + \frac{3X}{2} = 0.2g

2X = 0.2g

Using $g = 10 \text{ m s}^{-2}$ :

2X = 2 \implies X = 1 \text{ N}

Substitute $X = 1$ into equation (1):

T = 1 \times \sqrt{3} = \sqrt{3} \approx 1.73 \text{ N}

Method 2: Resolving along $X$ and $T$ (Alternative)

Resolving along the direction of $X$ :

X - 0.2g \cos 60^\circ = 0 \implies X = 0.2 \times 10 \times 0.5 = 1 \text{ N}

Resolving along the direction of $T$ :

T - 0.2g \sin 60^\circ = 0 \implies T = 0.2 \times 10 \times \frac{\sqrt{3}}{2} = \sqrt{3} \approx 1.73 \text{ N}

Method 3: Lami's Theorem (Alternative)

The angles between the forces are:

Between $T$ and $X$ : $90^\circ$
Between $X$ and $W$ : $120^\circ$
Between $W$ and $T$ : $150^\circ$

Applying Lami's theorem:

\frac{T}{\sin 120^\circ} = \frac{X}{\sin 150^\circ} = \frac{0.2g}{\sin 90^\circ}

\frac{X}{0.5} = \frac{2}{1} \implies X = 1 \text{ N}

\frac{T}{\frac{\sqrt{3}}{2}} = 2 \implies T = \sqrt{3} \approx 1.73 \text{ N}

Answer

X = 1 \text{ N}, \quad T = \sqrt{3} \text{ N} \approx 1.73 \text{ N}

Final answer

X = 1 N, Tension = 1.73 N

Detailed explanation

Walkthrough

The particle is in equilibrium under three concurrent forces: its weight acting vertically downwards, the tension in the string acting along the string towards the wall, and the applied force $X$ acting perpendicular to the string. To find the unknowns $X$ and $T$ , we can use any of three standard equilibrium methods.

Method 1: Resolving horizontally and vertically
We set up a coordinate system with the $x$ -axis horizontal (away from the wall) and the $y$ -axis vertical (upwards). The tension $T$ makes an angle of $30^\circ$ with the vertical, so its horizontal component is $-T \sin 30^\circ$ (towards the wall) and its vertical component is $T \cos 30^\circ$ (upwards). The force $X$ is perpendicular to the string, so it makes an angle of $60^\circ$ with the vertical (or $30^\circ$ with the horizontal). Its components are $X \cos 30^\circ$ horizontally and $X \sin 30^\circ$ vertically. Setting the sum of horizontal forces to zero gives $X \cos 30^\circ - T \sin 30^\circ = 0$ . Setting the sum of vertical forces to zero gives $X \sin 30^\circ + T \cos 30^\circ - 0.2g = 0$ . Solving these simultaneous equations yields $X = 1$ and $T = \sqrt{3}$ .

Method 2: Resolving along the force directions
Since $X$ and $T$ are perpendicular, we can resolve along their respective directions to eliminate one unknown at a time. Resolving along $X$ eliminates $T$ entirely, leaving $X - 0.2g \cos 60^\circ = 0$ . Resolving along $T$ eliminates $X$ , leaving $T - 0.2g \sin 60^\circ = 0$ . This is the fastest method once the angles are correctly identified.

Method 3: Lami's theorem
Lami's theorem states that for three concurrent forces in equilibrium, each force is proportional to the sine of the angle between the other two. The angles are $90^\circ$ (between $T$ and $X$ ), $120^\circ$ (between $X$ and $W$ ), and $150^\circ$ (between $W$ and $T$ ). Setting up the ratios $\frac{T}{\sin 120^\circ} = \frac{X}{\sin 150^\circ} = \frac{0.2g}{\sin 90^\circ}$ directly gives both answers.

Key Takeaways

Equilibrium requires the vector sum of all forces to be zero in every direction.
Resolving forces along non-perpendicular axes (like $X$ and $T$ ) can simplify calculations when the forces are perpendicular to each other.
Lami's theorem is a powerful shortcut for three-force equilibrium problems where all angles are known.
Always carefully determine the angles each force makes with the chosen axes or with each other.

Common Mistakes

Incorrectly identifying the angles: the string is at $30^\circ$ to the vertical wall, not to the horizontal. This makes $T$ at $30^\circ$ to the vertical, and $X$ at $60^\circ$ to the vertical.
Sign errors when resolving: ensuring that forces pointing in opposite directions have opposite signs in the equilibrium equations.
Forgetting to use the correct value of $g$ : the mark scheme implies $g = 10 \text{ m s}^{-2}$ to yield $X = 1$ . Using $g = 9.8$ would give $X = 0.98$ , which may not match the expected answer format.

Things to Be Careful About

The angle $30^\circ$ is between the string and the vertical wall, not the horizontal. This is a common trap that flips the sine and cosine components.
The force $X$ is perpendicular to the string, meaning the angle between $X$ and the string is $90^\circ$ . This makes resolving along $X$ and $T$ particularly elegant.
When using Lami's theorem, ensure the angles used are the angles between the forces, not the angles the forces make with the axes. The angle between $X$ and the weight is $120^\circ$ , not $60^\circ$ .

Techniques used

resolve forces horizontally and verticallyform simultaneous equationssolve for unknown forcesapply Lami's theorem as alternative

Approach

Let $s$ be the distance $AB$ . The car moves with constant acceleration $a$ , so the suvat equation $s = ut + \frac{1}{2}at^2$ may be used for each stage. The condition $BC = 2s$ connects the two stages, giving two equations that can be solved for $u$ .

Working

Let $s = AB$ .

From $A$ to $B$ , with $t = 8$ :

s = u \times 8 + \frac{1}{2}a \times 8^2 = 8u + 32a

From $B$ to $C$ , the initial speed is the speed at $B$ , which is $u + 8a$ , and $t = 10$ :

2s = (u + 8a) \times 10 + \frac{1}{2}a \times 10^2 = 10u + 130a

Since $BC = 2AB$ , substitute $s = 8u + 32a$ into $2s$ :

2(8u + 32a) = 10u + 130a

Expand and solve:

16u + 64a = 10u + 130a

6u = 66a

u = 11a

Answer

u = 11a

Final answer

u = 11a

Detailed explanation

Walkthrough

This question describes motion with constant acceleration, so the suvat equations are valid for the whole journey. Let $s$ be the distance $AB$ ; then the given information says $BC = 2s$ .

For the first stage, substitute $t = 8$ into $s = ut + \frac{1}{2}at^2$ to get an expression for $s$ .

For the second stage, the car starts with speed $u + 8a$ because it has already accelerated for 8 seconds, and it travels for another 10 seconds. So the distance $BC$ is $(u + 8a) \times 10 + \frac{1}{2}a \times 10^2$ , and this equals $2s$ .

Equating $2(8u + 32a)$ with $10u + 130a$ gives a linear equation in $u$ and $a$ . Solving it eliminates the unknown distance $s$ and gives $u = 11a$ .

Key Takeaways

This question tests the use of the constant acceleration formula $s = ut + \frac{1}{2}at^2$ in two consecutive stages. The important idea is to write each stage's displacement using the correct initial speed and time, then use the distance ratio to connect the stages.

Common Mistakes

Treating the distance $BC$ as if its initial speed were still $u$ , instead of $u + 8a$ .
Only doubling one term when substituting $BC = 2AB$ . The whole expression $8u + 32a$ must be doubled.
Using $BC = u \times 10 + \frac{1}{2}a \times 10^2$ and setting it equal to $2AB$ , which leads to $u = -\frac{7}{3}a$ ; the mark scheme withholds the method mark for this.
Using $AC = 2AB$ instead of $BC = 2AB$ , which leads to $u = -49a$ ; this is also not allowed.

Things to Be Careful About

The times are different on each stage: 8 seconds from A to B, 10 seconds from B to C, and 18 seconds from A to C.
When eliminating $s$ , every term on the left-hand side must be doubled before equating to the right-hand side.
Since $a > 0$ , the answer $u = 11a$ is positive, as expected for a car speeding up while moving along the road.

Techniques used

Apply the constant acceleration distance formulaWrite displacement equations for two consecutive intervalsUse the given distance ratio to connect the equationsSolve the resulting simultaneous equation

(b)

Find the speed of the car at $C$ in terms of $a$ .

2M Similar

Q4Medium-EasyKinematics of Motion in a Straight Line

A particle travels in a straight line. The velocity of the particle at time $t\text{ s}$ after leaving a point $O$ is $v\text{ m s}^{-1}$ , where

v = kt^2 - 4t + 3.

The distance travelled by the particle in the first $2\text{ s}$ of its motion is $6\text{ m}$ . You may assume that $v > 0$ in the first $2\text{ s}$ of its motion.

(a)

Find the value of $k$ .

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Since the velocity is given as a function of time, integrate $v$ with respect to $t$ to obtain the displacement. Because $v > 0$ in the first $2$ s, the distance travelled equals the displacement between $t = 0$ and $t = 2$ . Set this displacement equal to $6$ m and solve for $k$ .

Working

Integrate the velocity:

s(t) = \int (kt^2 - 4t + 3)\,dt = \frac{k}{3}t^3 - 2t^2 + 3t + c

The distance travelled in the first $2$ s is $s(2) - s(0)$ :

s(2) - s(0) = \left(\frac{k}{3}(2)^3 - 2(2)^2 + 3(2)\right) - 0 = \frac{8k}{3} - 8 + 6

Simplify:

\frac{8k}{3} - 2 = 6

Solve for $k$ :

\frac{8k}{3} = 8 \quad\Rightarrow\quad k = 3

Answer

k = 3

Final answer

k = 3

Detailed explanation

Walkthrough

We are given the velocity as a function of time and told the distance travelled in the first $2$ seconds. Since $v > 0$ throughout this interval, the particle never reverses direction, so the distance travelled is exactly the displacement from $t = 0$ to $t = 2$ .

To find displacement from velocity, integrate the velocity function. The constant of integration $c$ will cancel when we use the limits $t = 0$ and $t = 2$ . Evaluating the integral at $t = 2$ and subtracting its value at $t = 0$ gives an expression in terms of $k$ . Setting this equal to $6$ m produces a simple linear equation, which we solve to find $k = 3$ .

Key Takeaways

For variable velocity, displacement is the integral of velocity with respect to time.
When the velocity is always positive, distance travelled equals displacement.
A definite integral automatically cancels the constant of integration.

Common Mistakes

Using $s = vt$ , which only applies for constant velocity, instead of integrating.
Forgetting to subtract the value at the lower limit $t = 0$ .
Ignoring the condition $v > 0$ ; if the particle reversed direction, distance would not equal displacement.

Things to Be Careful About

Ensure the integration is correct: $kt^2$ becomes $\frac{k}{3}t^3$ , and $-4t$ becomes $-2t^2$ .
The constant $c$ is not needed because it cancels when subtracting $s(0)$ .
Keep units consistent: displacement is in metres and time in seconds.

Techniques used

integrate velocity to obtain displacementapply limits to find distance travelledsolve the resulting equation for the unknown constant

(b)

Find the value of the minimum velocity of the particle. You do not need to show that this velocity is a minimum.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

To find the minimum velocity, differentiate $v$ with respect to $t$ , set the derivative equal to zero to find the time at which the stationary point occurs, and substitute this time back into the expression for $v$ . The question states that it is not necessary to prove that this velocity is a minimum.

Working

Using $k = 3$ from part (a):

v = 3t^2 - 4t + 3

Differentiate with respect to $t$ :

\frac{dv}{dt} = 6t - 4

Set $\frac{dv}{dt} = 0$ :

6t - 4 = 0 \quad\Rightarrow\quad t = \frac{2}{3}

Substitute $t = \frac{2}{3}$ into $v$ :

v = 3\left(\frac{2}{3}\right)^2 - 4\left(\frac{2}{3}\right) + 3 = \frac{4}{3} - \frac{8}{3} + 3 = \frac{5}{3}

Answer

v = \frac{5}{3}\ \text{m s}^{-1}

Final answer

5/3 m s^-1

Detailed explanation

Walkthrough

From part (a) we know $k = 3$ , so the velocity is $v = 3t^2 - 4t + 3$ .

The minimum velocity occurs at a stationary point of the velocity-time graph, where the derivative $\frac{dv}{dt}$ is zero. Differentiate the quadratic to get $\frac{dv}{dt} = 6t - 4$ . Setting this equal to zero gives $t = \frac{2}{3}$ .

Finally, substitute $t = \frac{2}{3}$ back into the velocity expression to find the minimum velocity:

v = 3\left(\frac{2}{3}\right)^2 - 4\left(\frac{2}{3}\right) + 3 = \frac{5}{3}\ \text{m s}^{-1}

The question explicitly says we do not need to show that this is a minimum, so no second derivative or sign test is required.

Key Takeaways

The minimum or maximum of a velocity function occurs where $\frac{dv}{dt} = 0$ .
For a quadratic with a positive coefficient of $t^2$ , the stationary point is a minimum.
The value of $k$ from part (a) must be used in part (b).

Common Mistakes

Setting $v = 0$ instead of $\frac{dv}{dt} = 0$ when looking for the minimum velocity.
Forgetting to substitute the value of $t$ back into $v$ after finding it.
Using the wrong value of $k$ from part (a).

Things to Be Careful About

If using the formula $t = \frac{-b}{2a}$ , note that $a = 3$ and $b = -4$ , so $t = \frac{4}{6} = \frac{2}{3}$ .
The units of velocity are $\text{m s}^{-1}$ .
The answer $\frac{5}{3}$ is exact; $1.67$ is also accepted but the fraction is preferred.

Techniques used

differentiate velocity with respect to timeset the derivative equal to zero to find the stationary pointsubstitute the stationary time into the velocity expression

Q5MediumNewton's Laws of MotionKinematics of Motion in a Straight Line

A van of mass $4500\text{ kg}$ is towing a trailer of mass $750\text{ kg}$ down a straight hill inclined at an angle of $\theta$ to the horizontal where $\sin\theta = 0.05$ . The van and the trailer are connected by a light rigid tow-bar which is parallel to the road. There are constant resistance forces of $2500\text{ N}$ on the van and $300\text{ N}$ on the trailer.

(a)

It is given that the tension in the tow-bar is $450\text{ N}$ .

Find the acceleration of the trailer and the driving force of the van's engine.

4M Similar

DifficultyMedium-Easy

Worked solution

Approach

Positive direction is taken down the hill. Resolve each weight into a component down the plane, then apply Newton's second law separately to the trailer and to the van. The given tow-bar tension is $450\text{ N}$ ; it pulls the trailer down the slope and pulls the van back up the slope.

Working

Using $g = 10\text{ m s}^{-2}$ and positive down the hill.

For the trailer, the forces down the plane are the tension $450\text{ N}$ and the weight component $750g\sin\theta$ ; the force up the plane is the resistance $300\text{ N}$ . Newton's second law gives

750a = 450 + 750g\sin\theta - 300.

With $\sin\theta = 0.05$ and $g = 10$ :

750a = 450 + 750 \times 10 \times 0.05 - 300 = 525,

so

a = \frac{525}{750} = 0.7\text{ m s}^{-2}.

For the van, the forces down the plane are the driving force $D$ and the weight component $4500g\sin\theta$ ; the forces up the plane are the resistance $2500\text{ N}$ and the tension $450\text{ N}$ . Newton's second law gives

4500a = D + 4500g\sin\theta - 2500 - 450.

Substituting $a=0.7$ , $g=10$ and $\sin\theta=0.05$ :

4500 \times 0.7 = D + 4500 \times 10 \times 0.05 - 2950,

that is

3150 = D - 700.

Therefore

D = 3850\text{ N}.

Answer

a = 0.7\text{ m s}^{-2}, \quad D = 3850\text{ N}.

Final answer

a = 0.7 m/s^2 and D = 3850 N

Detailed explanation

Walkthrough

First choose a positive direction; here down the hill is natural because the van is moving down. On an incline the component of weight along the plane is $mg\sin\theta$ ; this is what affects motion along the slope.

For the trailer, the tow-bar pulls it in the direction of motion, so the tension is positive. Its weight component is also down the slope, while the resistance $300\text{ N}$ opposes the motion and acts up the slope. Putting these into Newton's second law gives a direct equation for $a$ , because the tension is known.

For the van, the driving force and its weight component act down the slope, while the resistance and the tow-bar tension act up the slope. The tow-bar tension is the same size $450\text{ N}$ , but on the van it opposes the motion. With $a$ already found, this equation gives $D$ .

Key Takeaways

The weight on an inclined plane splits into $mg\sin\theta$ along the plane and $mg\cos\theta$ perpendicular to the plane; only the parallel component affects motion along the plane.
Newton's second law must be applied separately to each body, with forces resolved along the direction of motion.
A light rigid tow-bar means the two bodies have the same acceleration.
The tension in the tow-bar is the same at both ends but acts in opposite directions on the two bodies.

Common Mistakes

Using $mg\cos\theta$ as the component along the slope instead of $mg\sin\theta$ .
Forgetting a force term: the tension on the van opposes motion, and the resistance is opposite to the direction of travel.
Missing $g$ when calculating the weight component.
Quoting an answer without showing the step ' $a=$ ' or ' $D=$ '; the mark scheme requires the method to be visible.

Things to Be Careful About

Keep one consistent positive direction. With down the hill positive, forces down the plane are positive and forces up the plane are negative.
Use $g=10\text{ m s}^{-2}$ .
Check units: mass in kilograms, forces in newtons, acceleration in $\text{m s}^{-2}$ .
Because the tow-bar is light and rigid, the trailer and van accelerate together.

Techniques used

resolve weight into components parallel to the planeapply Newton's second law to the trailerapply Newton's second law to the vansolve for the acceleration and driving force

(b)

On another occasion, the van and trailer ascend a straight hill inclined at an angle of $\alpha$ to the horizontal where $\sin\alpha = 0.09$ . The driving force of the van's engine is now $9100\text{ N}$ , and the speed of the van at the bottom of the hill is $20\text{ m s}^{-1}$ . The resistances to motion are unchanged.

7M

(i)

Find the acceleration of the van and the tension in the tow-bar.

5M Similar

DifficultyMedium

Worked solution

Approach

Positive direction is taken up the hill. Form Newton's second law equations for the trailer and the van separately. Both bodies have the same acceleration $a$ , and the tow-bar carries the same tension $T$ ; on the trailer this tension acts up the slope, while on the van it acts back down the slope.

Working

Using $g=10\text{ m s}^{-2}$ and positive up the hill.

For the trailer, the forces up the slope are the tension $T$ ; the forces down the slope are the resistance $300\text{ N}$ and the weight component $750g\sin\alpha$ . Newton's second law gives

750a = T - 300 - 750g\sin\alpha.

With $\sin\alpha = 0.09$ and $g=10$ :

750a = T - 300 - 750 \times 10 \times 0.09 = T - 975.

So

T - 975 = 750a.\qquad(1)

For the van, the force up the slope is the driving force $9100\text{ N}$ ; the forces down the slope are the resistance $2500\text{ N}$ , the tension $T$ and the weight component $4500g\sin\alpha$ . Newton's second law gives

4500a = 9100 - 2500 - T - 4500g\sin\alpha.

With values:

4500a = 9100 - 2500 - T - 4500 \times 10 \times 0.09 = 2550 - T.

So

2550 - T = 4500a.\qquad(2)

From (1), $T = 750a + 975$ . Substituting into (2):

2550 - (750a + 975) = 4500a,

so

1575 = 5250a.

a = 0.3\text{ m s}^{-2}.

Then

T = 750(0.3) + 975 = 225 + 975 = 1200\text{ N}.

Answer

a = 0.3\text{ m s}^{-2}, \quad T = 1200\text{ N}.

Final answer

a = 0.3 m/s^2 and T = 1200 N

Detailed explanation

Walkthrough

For the ascent, take up the hill as positive. The weight components now pull both bodies back down the slope.

For the trailer, the tow-bar tension $T$ pulls it up the slope, while the resistance $300\text{ N}$ and its weight component $750g\sin\alpha$ act down the slope. This gives an equation involving $T$ and $a$ .

For the van, the driving force $9100\text{ N}$ acts up the slope, while the resistance $2500\text{ N}$ , the weight component $4500g\sin\alpha$ , and the tow-bar tension $T$ act down the slope. The tension has the same magnitude in both equations because it is the same light tow-bar, but on the van it points opposite to the motion.

Both equations contain the same two unknowns $a$ and $T$ , so solve them simultaneously. Express $T$ from the trailer equation and substitute into the van equation; this eliminates $T$ and gives $a$ . Substituting back gives $T$ .

Key Takeaways

Connected bodies moving together have the same acceleration, so one common $a$ can be used in both equations.
The tension in a light connector is the same at both ends, but each body feels it in opposite directions.
On an incline, the component of weight along the slope is $mg\sin\alpha$ .
Two equations in two unknowns can be solved by substitution.

Common Mistakes

Using the same direction for $T$ in both equations. The tension pulls the trailer up the slope but pulls the van back down.
Forgetting one of the down-slope forces on the van, such as the tension or the weight component.
Using $\cos\alpha$ instead of $\sin\alpha$ for the component along the slope.
Omitting $g$ from the weight components.
Not showing the step to ' $a=$ ' or ' $T=$ '; the mark scheme requires visible method.

Things to Be Careful About

With positive up the hill, the acceleration is positive in the direction of motion.
The resistances are unchanged: $2500\text{ N}$ on the van and $300\text{ N}$ on the trailer.
The driving force is now $9100\text{ N}$ , different from part (a).
g
If the van and trailer were treated as a system, the tension would be internal and cancel; here separate equations with the same $T$ were used.

Techniques used

resolve weight components on the inclineform equations of motion for trailer and van with an unknown common tensionsolve simultaneous equations for acceleration and tension

(ii)

Find the speed of the van when it has travelled a distance of $375\text{ m}$ up the hill.

2M Similar

DifficultyEasy

Worked solution

Approach

The acceleration is constant, so use the constant-acceleration formula $v^2 = u^2 + 2as$ with $u = 20\text{ m s}^{-1}$ , $a = 0.3\text{ m s}^{-2}$ and $s = 375\text{ m}$ .

Working

v^2 = 20^2 + 2(0.3)(375) = 400 + 225 = 625.

Since the van continues to move up the hill, take the positive root:

v = \sqrt{625} = 25\text{ m s}^{-1}.

Answer

v = 25\text{ m s}^{-1}.

Final answer

25 m/s

Detailed explanation

Walkthrough

After part (i), the acceleration up the hill is constant because all forces remain constant. The initial speed at the bottom is $20\text{ m s}^{-1}$ , and we want the speed after travelling $375\text{ m}$ up the hill.

Time is not involved, so the most direct suvat equation is $v^2 = u^2 + 2as$ . Substitute $u=20\text{ m s}^{-1}$ , $a=0.3\text{ m s}^{-2}$ and $s=375\text{ m}$ .

This gives $v^2=625$ . Because the van is moving up the hill, take the positive square root, $v=25\text{ m s}^{-1}$ .

Key Takeaways

The suvat equations apply only when acceleration is constant; here it is.
$v^2 = u^2 + 2as$ is useful when displacement is given and time is not needed.
Speed is the positive root of $v^2$ .

Common Mistakes

Using the acceleration from part (a), $0.7\text{ m s}^{-2}$ , instead of the new acceleration from (b)(i).
g
Forgetting to square $20$ or forgetting the factor $2$ in $2as$ .
Taking the negative square root and reporting a negative speed.
Mixing distance with displacement; here the distance up the slope is exactly the displacement $375\text{ m}$ .

Things to Be Careful About

The mark scheme follows through on their value of $a$ from part (i), provided it does not give a negative root.
Units must be consistent: speeds in $\text{m s}^{-1}$ , acceleration in $\text{m s}^{-2}$ , distance in metres.
The final speed is $25\text{ m s}^{-1}$ .

Techniques used

apply constant acceleration formula connecting v, u, a and sevaluate and take the positive root for the final speed

Q6MediumNewton's Laws of MotionEnergy, Work and PowerForces and Equilibrium

A cyclist is travelling along a straight horizontal road. The total mass of the cyclist and her bicycle is $80\text{ kg}$ . There is a constant resistance force of magnitude $32\text{ N}$ to the cyclist's motion. At an instant when she is travelling at $7\text{ m s}^{-1}$ , her acceleration is $0.1\text{ m s}^{-2}$ .

(a)

Find the power output of the cyclist.

3M Similar

DifficultyMedium-Easy

Worked solution

Approach

Let $D$ be the driving force of the cyclist. Use Newton's second law to find $D$ , then use $P = Dv$ to find the power output.

Working

Resolving horizontally, with motion taken as positive:

D - 32 = 80 \times 0.1

D - 32 = 8

D = 40\text{ N}

The power output is the product of the driving force and the speed:

P = Dv = 40 \times 7

P = 280\text{ W}

Answer

280\text{ W}

Final answer

280 W

Detailed explanation

Walkthrough

The cyclist is moving horizontally, so we need to consider only horizontal forces. Let $D$ be the driving force. Resistance is $32\text{ N}$ opposing motion. Since the acceleration is $0.1\text{ m s}^{-2}$ , the resultant horizontal force is $80 \times 0.1 = 8\text{ N}$ . Newton's second law gives

D - 32 = 80 \times 0.1.

This step corresponds to the M1 mark for using Newton's second law. Solving, $D = 40\text{ N}$ . Power is the rate at which the driving force does work. When the force and velocity are in the same direction, $P = Fv$ , so substituting $D = 40$ and $v = 7$ gives $P = 280\text{ W}$ .

Key Takeaways

Newton's second law connects resultant force with mass and acceleration.
Power output is calculated from the driving force and the speed, not from the resultant force.
Resistive forces oppose motion and are subtracted from the driving force.

Common Mistakes

Using only the resistance $32\text{ N}$ as the force in $P = Fv$ without finding the driving force.
Using the resultant force $8\text{ N}$ instead of the driving force in the power formula.
Sign errors in the equation of motion.

Things to Be Careful About

The driving force is $40\text{ N}$ , obtained by adding the resistance to the resultant force.
Units: power is in watts, force in newtons, speed in metres per second.
The mark scheme allows any force term in the expression $P = DF \times 7$ as long as the relationship is seen.

Techniques used

apply Newton's second law to find the driving forcesubstitute force and speed into P = Fv

(b)

Find the steady speed that the cyclist can maintain if her power output and the resistance force are both unchanged.

2M Similar

(c)

The cyclist later descends a straight hill of length $32.2\text{ m}$ , inclined at an angle of $\sin^{-1}\left(\frac{1}{20}\right)$ to the horizontal. Her power output is now $120\text{ W}$ , and the resistance force now has variable magnitude such that the work done against this force in descending the hill is $1128\text{ J}$ . The time taken to descend the hill is $4\text{ s}$ .

Given that the speed of the cyclist at the top of the hill is $7.5\text{ m s}^{-1}$ , find her speed at the bottom of the hill.

6M Similar

DifficultyMedium

Worked solution

Approach

Use the work-energy principle. The work done by the cyclist and the loss in gravitational potential energy increase the kinetic energy, while the work done against resistance removes energy from the cyclist.

Working

Work done by the cyclist over $4$ seconds:

120 \times 4 = 480\text{ J}

The initial kinetic energy is

\frac{1}{2} \times 80 \times 7.5^2 = 2250\text{ J}

and the final kinetic energy is

\frac{1}{2} \times 80v^2 = 40v^2

Loss in gravitational potential energy, using $g = 10\text{ m s}^{-2}$ and $\sin\theta = \frac{1}{20}$ :

80g \times \frac{1}{20} \times 32.2 = 80 \times 10 \times 1.61 = 1288\text{ J}

Work-energy equation, with work against resistance $1128\text{ J}$ :

480 + 1288 - 1128 = 40v^2 - 2250

Simplify the left-hand side:

640 = 40v^2 - 2250

40v^2 = 2890

v^2 = 72.25

Since speed is positive:

v = 8.5\text{ m s}^{-1}

Answer

8.5\text{ m s}^{-1}

Final answer

8.5 m s^-1

Detailed explanation

Walkthrough

The descent is not at constant acceleration, and the resistance is variable, so use energy. The energy balance is

\text{work by cyclist} + \text{loss in PE} - \text{work against resistance} = \text{gain in KE}.

Here the cyclist's power output is $120\text{ W}$ for $4$ seconds, so the work done by the cyclist is $120 \times 4 = 480\text{ J}$ . The hill has length $32.2\text{ m}$ and angle $\theta$ with $\sin\theta = \frac{1}{20}$ , so the vertical drop is $32.2 \times \frac{1}{20} = 1.61\text{ m}$ . Taking $g = 10$ , the loss in gravitational potential energy is

80 \times 10 \times 1.61 = 1288\text{ J}.

The initial kinetic energy is $\frac{1}{2} \times 80 \times 7.5^2 = 2250\text{ J}$ and the final kinetic energy is $\frac{1}{2} \times 80v^2 = 40v^2$ . The work against resistance is $1128\text{ J}$ . Substituting into the work-energy equation:

480 + 1288 - 1128 = 40v^2 - 2250.

Simplifying gives $640 = 40v^2 - 2250$ , so $40v^2 = 2890$ , $v^2 = 72.25$ and $v = 8.5$ .

Key Takeaways

Work-energy principle links work input, potential energy, resistance, and kinetic energy.
Work done by a constant power over time is power multiplied by time.
For an inclined descent, the change in height is the length of the slope times $\sin\theta$ .

Common Mistakes

Using constant-acceleration (suvat) equations; the resistance is variable and the acceleration is not constant, so this loses method marks.
Getting the sign of the potential energy term wrong: descending means PE decreases, releasing energy.
Forgetting to include initial kinetic energy on the right-hand side.
Adding rather than subtracting the work done against resistance.

Things to Be Careful About

The five terms in the work-energy equation are work done by cyclist, loss in PE, work against resistance, final KE and initial KE.
Use the vertical height $1.61\text{ m}$ , not the slope length $32.2\text{ m}$ , when calculating PE.
The final speed is positive because it is a speed.
If $g$ were specified as $9.8\text{ m s}^{-2}$ , the numerical result would differ; the mark scheme here uses $g = 10$ .

Techniques used

calculate work done from power and timecompute gravitational potential energy lostapply the work-energy principle including resistance worksolve for the final speed

Q7MediumEnergy, Work and PowerNewton's Laws of Motion

Fig. 7.1

The diagram shows a track $ABCD$ which lies in a vertical plane. The section $AB$ is a straight line inclined at an angle of $30^\circ$ to the horizontal and is smooth. The section $BC$ is a horizontal straight line and is rough. The section $CD$ is a straight line inclined at an angle of $30^\circ$ to the horizontal and is rough. The lengths $AB$ , $BC$ and $CD$ are each $2\text{ m}$ .

A particle is released from rest at $A$ . The coefficient of friction between the particle and both $BC$ and $CD$ is $\mu$ . There is no change in the speed of the particle when it passes through either of the points $B$ or $C$ .

(a)

It is given that $\mu = 0.1$ .

Find the distance which the particle has moved up the section $CD$ when its speed is $1\text{ m s}^{-1}$ .

5M Similar

DifficultyMedium

Worked solution

Approach

Use the work-energy principle from the release point A to the point on section CD where the speed is 1 m/s. The initial gravitational potential energy is converted into final potential energy, kinetic energy, and work done against friction on the rough sections BC and CD.

Working

Take the horizontal section BC as the reference level for gravitational potential energy.

At point A, the height above BC is $2 \sin 30° = 1$ m, so the initial PE is $mg \times 1 = mg$ .

Section BC (rough, horizontal):
Normal reaction: $R = mg$
Friction force: $F = \mu R = 0.1mg$
Work done against friction over 2 m: $0.1mg \times 2 = 0.2mg$

Section CD (rough, inclined at 30° to horizontal):
Normal reaction: $R = mg \cos 30°$
Friction force: $F = \mu R = 0.1mg \cos 30°$
Work done against friction over distance $d$ : $0.1mg \cos 30° \times d = 0.1mg \frac{\sqrt{3}}{2} d$
Change in PE over distance $d$ up the slope: $mg \times d \sin 30° = \frac{1}{2}mgd$

Work-energy equation (initial PE = final PE + work done against friction + final KE):

mg = 0.2mg + 0.1mg \frac{\sqrt{3}}{2} d + \frac{1}{2}mgd + \frac{1}{2}m(1)^2

Dividing through by $m$ and using $g = 10$ :

10 = 2 + \frac{\sqrt{3}}{2} d + 5d + 0.5

10 = 2.5 + d\left(5 + \frac{\sqrt{3}}{2}\right)

7.5 = d\left(\frac{10 + \sqrt{3}}{2}\right)

d = \frac{15}{10 + \sqrt{3}}

Rationalising the denominator:

d = \frac{15(10 - \sqrt{3})}{(10 + \sqrt{3})(10 - \sqrt{3})} = \frac{15(10 - \sqrt{3})}{100 - 3} = \frac{15(10 - \sqrt{3})}{97}

d \approx 1.28 \text{ m}

Answer

d = 1.28 \text{ m}

Final answer

1.28 m

Detailed explanation

Walkthrough

We apply the work-energy principle from point A (where the particle is released from rest) to a point on section CD at distance $d$ from C (where the speed is 1 m/s).

Step 1: Set up the reference level. Take BC as the zero level for gravitational PE. Point A is at height $2\sin 30° = 1$ m above BC, so the initial PE is $mg \times 1 = mg$ .

Step 2: Work done against friction on BC. The section BC is horizontal and rough. The normal reaction equals the weight: $R = mg$ . The friction force is $F = \mu R = 0.1mg$ . Over the 2 m length of BC, the work done against friction is $0.1mg \times 2 = 0.2mg$ .

Step 3: Work done against friction and PE change on CD. Section CD is inclined at 30° and rough. The normal reaction is $R = mg\cos 30°$ . The friction force is $F = 0.1mg\cos 30° = 0.1mg\frac{\sqrt{3}}{2}$ . Over distance $d$ , the work done against friction is $0.1mg\frac{\sqrt{3}}{2}d$ . The particle rises by $d\sin 30° = \frac{d}{2}$ , so the gain in PE is $\frac{1}{2}mgd$ .

Step 4: Final kinetic energy. At the point of interest, the speed is 1 m/s, so the final KE is $\frac{1}{2}m(1)^2 = \frac{1}{2}m$ .

Step 5: Apply the work-energy principle. Initial PE = final PE + total work done against friction + final KE:

mg = 0.2mg + 0.1mg\frac{\sqrt{3}}{2}d + \frac{1}{2}mgd + \frac{1}{2}m

Dividing by $m$ and using $g = 10$ :

10 = 2 + \frac{\sqrt{3}}{2}d + 5d + 0.5

7.5 = d\left(5 + \frac{\sqrt{3}}{2}\right) = d\left(\frac{10 + \sqrt{3}}{2}\right)

d = \frac{15}{10 + \sqrt{3}} = \frac{15(10 - \sqrt{3})}{97} \approx 1.28 \text{ m}

Key Takeaways

The work-energy principle is a powerful tool for problems involving changing speed, height, and friction across multiple sections.
On an inclined plane, the normal reaction is $mg\cos\theta$ and the friction force is $\mu mg\cos\theta$ .
Work done against friction is always positive (energy lost from the system), while the change in PE depends on whether the particle moves up or down.
The mass $m$ cancels out in work-energy equations for this type of problem.

Common Mistakes

Forgetting that the normal reaction on the inclined section CD is $mg\cos 30°$ , not $mg$ .
Using $\sin 30°$ instead of $\cos 30°$ for the normal reaction on CD.
Sign errors in the work-energy equation: work done against friction should be on the same side as the final PE (both represent energy that the initial PE must supply).
Not dividing through by $m$ correctly, or forgetting to use $g = 10$ .
Computing the height of A above BC incorrectly as $2\cos 30°$ instead of $2\sin 30°$ .

Things to Be Careful About

The problem states there is no change in speed at B or C, so we can treat the motion continuously across these points without worrying about energy losses at the junctions.
The distance $d$ is measured along the slope CD, not horizontally. The vertical rise is $d\sin 30°$ and the normal reaction involves $\cos 30°$ .
Allow sign errors in the work-energy equation as long as the equation is dimensionally correct and consistent (the mark scheme permits this).
The answer $d = 1.28$ m is the distance up CD from C, not the total distance from A.

Techniques used

apply work-energy principleresolve forces on inclined planecalculate normal reactioncompute friction forcesolve for distance

(b)

It is given instead that with a different value of $\mu$ the particle travels $1\text{ m}$ up the track from $C$ before it comes instantaneously to rest.

Find the value of $\mu$ and the speed of the particle at the instant that it passes $C$ for the second time.

4M Similar

DifficultyMedium

Worked solution

Approach

Use the work-energy principle in two stages. First, apply it from A to the point where the particle comes instantaneously to rest (1 m up CD) to find $\mu$ . Second, apply it from the stopping point back to C to find the speed on the second pass through C.

Working

Part 1: Find $\mu$

Take BC as the reference level for PE. The particle starts at A (height $2\sin 30° = 1$ m) and stops 1 m up CD (height $1\sin 30° = 0.5$ m).

Initial PE at A: $mg \times 2\sin 30° = mg$

Work done against friction on BC (distance 2 m): $\mu mg \times 2 = 2\mu mg$

Work done against friction on CD (distance 1 m): $\mu mg\cos 30° \times 1 = \mu mg\frac{\sqrt{3}}{2}$

Final PE at stopping point: $mg \times 1\sin 30° = \frac{1}{2}mg$

Final KE = 0 (particle is at rest).

Work-energy equation:

mg = 2\mu mg + \mu mg\frac{\sqrt{3}}{2} + \frac{1}{2}mg

Dividing by $mg$ :

1 = 2\mu + \frac{\sqrt{3}}{2}\mu + \frac{1}{2}

\frac{1}{2} = \mu\left(2 + \frac{\sqrt{3}}{2}\right) = \mu\left(\frac{4 + \sqrt{3}}{2}\right)

\mu = \frac{1}{4 + \sqrt{3}}

Rationalising:

\mu = \frac{4 - \sqrt{3}}{(4 + \sqrt{3})(4 - \sqrt{3})} = \frac{4 - \sqrt{3}}{16 - 3} = \frac{4 - \sqrt{3}}{13}

\mu \approx 0.174

Part 2: Speed at C on the second pass

The particle slides back down 1 m from the stopping point to C. Apply the work-energy principle for this downward journey.

Initial PE at stopping point (1 m up CD): $mg \times 1\sin 30° = \frac{1}{2}mg$

Work done against friction on CD (distance 1 m, going down): $\mu mg\cos 30° \times 1 = \mu mg\frac{\sqrt{3}}{2}$

Final PE at C: 0

Final KE at C: $\frac{1}{2}mv^2$

Work-energy equation:

\frac{1}{2}mg - \mu mg\frac{\sqrt{3}}{2} = \frac{1}{2}mv^2

Dividing by $m$ and using $g = 10$ :

5 - 5\sqrt{3}\mu = \frac{1}{2}v^2

Substituting $\mu = \frac{4 - \sqrt{3}}{13}$ :

5 - 5\sqrt{3}\left(\frac{4 - \sqrt{3}}{13}\right) = \frac{1}{2}v^2

5 - \frac{20\sqrt{3} - 15}{13} = \frac{1}{2}v^2

\frac{65 - 20\sqrt{3} + 15}{13} = \frac{1}{2}v^2

\frac{80 - 20\sqrt{3}}{13} = \frac{1}{2}v^2

v^2 = \frac{160 - 40\sqrt{3}}{13}

v \approx 2.64 \text{ m s}^{-1}

Answer

\mu = 0.174, \quad v = 2.64 \text{ m s}^{-1}

Final answer

μ = 0.174, speed = 2.64 m/s

Detailed explanation

Walkthrough

Part 1: Finding $\mu$

The particle is released from rest at A and travels up CD, coming to rest 1 m from C. We apply the work-energy principle from A to the stopping point.

At A, the height above BC is $2\sin 30° = 1$ m, so initial PE = $mg$ .

The particle travels 2 m along rough horizontal BC, losing energy to friction: work done = $\mu mg \times 2 = 2\mu mg$ .

Then it travels 1 m up rough inclined CD. The normal reaction is $mg\cos 30°$ , so friction = $\mu mg\cos 30°$ . Work done against friction = $\mu mg\cos 30° \times 1 = \mu mg\frac{\sqrt{3}}{2}$ .

At the stopping point, the height above BC is $1\sin 30° = 0.5$ m, so final PE = $\frac{1}{2}mg$ . Final KE = 0.

By work-energy: initial PE = final PE + total work done against friction:

mg = \frac{1}{2}mg + 2\mu mg + \mu mg\frac{\sqrt{3}}{2}

Dividing by $mg$ and solving:

\frac{1}{2} = \mu\left(2 + \frac{\sqrt{3}}{2}\right) = \mu\left(\frac{4 + \sqrt{3}}{2}\right)

\mu = \frac{2}{4 + \sqrt{3}} = \frac{4 - \sqrt{3}}{13} \approx 0.174

Part 2: Speed at C on the return journey

The particle now slides back down 1 m from the stopping point to C. Apply work-energy for this downward motion.

Initial PE (at stopping point): $\frac{1}{2}mg$

Work done against friction (going down 1 m on CD): $\mu mg\cos 30° \times 1 = \mu mg\frac{\sqrt{3}}{2}$

Final PE at C: 0

Final KE at C: $\frac{1}{2}mv^2$

Work-energy: initial PE - work done against friction = final KE:

\frac{1}{2}mg - \mu mg\frac{\sqrt{3}}{2} = \frac{1}{2}mv^2

Dividing by $m$ and using $g = 10$ :

5 - 5\sqrt{3}\mu = \frac{1}{2}v^2

Substituting $\mu = \frac{4 - \sqrt{3}}{13}$ :

5 - \frac{5\sqrt{3}(4 - \sqrt{3})}{13} = 5 - \frac{20\sqrt{3} - 15}{13} = \frac{65 - 20\sqrt{3} + 15}{13} = \frac{80 - 20\sqrt{3}}{13}

v^2 = \frac{160 - 40\sqrt{3}}{13} \approx 6.98

v \approx 2.64 \text{ m s}^{-1}

Key Takeaways

When a particle comes to rest and then reverses direction, you can apply the work-energy principle to each leg of the journey separately.
The coefficient of friction $\mu$ can be found from one application of work-energy, then used in a second application for the return journey.
On the return journey down an inclined plane, the weight component does positive work (helps motion) while friction does negative work (opposes motion).

Common Mistakes

Using the wrong height for the stopping point: it is 1 m up CD, so the height is $1\sin 30° = 0.5$ m, not 1 m.
Forgetting that friction always opposes motion, so on the downward return journey, friction still does negative work (energy is still lost to friction going down).
Not rationalising $\mu$ correctly: $\frac{1}{4+\sqrt{3}} = \frac{4-\sqrt{3}}{13}$ , not $\frac{4+\sqrt{3}}{13}$ .
Using $g = 9.8$ or $9.81$ instead of $g = 10$ (Cambridge A-Level typically uses $g = 10$ unless stated otherwise).

Things to Be Careful About

The question asks for the speed when the particle passes C for the second time. The first pass is on the way from B to up CD. The second pass is on the way back down from the stopping point to C.
When the particle returns down CD, it does NOT pass through C again after this (it would continue along BC, but the question only asks for the speed at C on the return).
Make sure to use the correct value of $\mu$ (to at least 2 significant figures) in the second work-energy equation, or use the exact form $\frac{4-\sqrt{3}}{13}$ to avoid rounding errors.
The work done against friction on the return journey is the same as on the upward journey over the same distance, because friction magnitude is the same and distance is the same.

Techniques used

apply work-energy principlefind coefficient of frictionapply work-energy principle for return motioncalculate speed on return

Mathematics 9709/43 — May/June 2024

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