Dynamics and Newton's Laws

Newton's First Law — Inertia

Statement

An object at rest remains at rest, and an object in motion continues with constant velocity, unless acted upon by a net external force.

Inertia

Inertia is the tendency of an object to resist changes in its state of motion. It is measured by mass — the greater the mass, the greater the inertia.

Inertial Frames of Reference

A frame of reference in which Newton's first law holds is called an inertial frame. Accelerating frames (like a car rounding a corner) are non-inertial.

Applications

• Passengers lurch forward when a bus brakes suddenly.
• Tablecloth can be pulled out from under dishes if done quickly (dishes have inertia).
• A spacecraft in deep space continues at constant velocity without thrust.

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Newton's Second Law

Forces and Motion: Basics

Explore how net force, mass, and acceleration are related by applying forces to objects. Experiment with friction, applied forces, and different masses to see how they affect motion.

Statement

The net force acting on an object equals the rate of change of momentum:

\vec{F}_{\mathrm{net}} = \frac{d\vec{p}}`\{dt}`

For constant mass:

$$\vec{F}_{\mathrm{net}} = m\vec{a}$$

Key Points

• Force is a vector quantity — direction matters.
• $\vec{F}_{\mathrm{net}}$ is the vector sum of all forces (the resultant force).
• The acceleration is always in the direction of the net force.
• SI unit: newton ($\mathrm{N}$), where $1\mathrm{ N} = 1\mathrm{ kg} \cdot \mathrm{m/s}^2$.

Free-Body Diagrams

A free-body diagram shows all forces acting ON an object:

1. Isolate the object.
2. Draw all forces as arrows (label each force).
3. Do NOT include forces exerted BY the object.
4. Include the weight ($mg$), normal force ($N$), friction ($f$), and any applied forces.

Component Resolution

For forces at angles, resolve into components:

$$F_x = F\cos\theta, \quad F_y = F\sin\theta$$

Apply Newton's second law in each direction:

$$\sum F_x = ma_x, \quad \sum F_y = ma_y$$

Example

A $5\mathrm{ kg}$ block is pulled along a rough horizontal surface by a force of $30\mathrm{ N}$ at $30\degree$ above the horizontal. The coefficient of kinetic friction is $0.3$. Find the acceleration.

Vertical:

$$N + 30\sin 30\degree = mg$$$$N = 5(9.81) - 30(0.5) = 49.05 - 15 = 34.05\mathrm{ N}$$

Friction: $f_k = \mu_k N = 0.3(34.05) = 10.22\mathrm{ N}$.

Horizontal:

$$30\cos 30\degree - f_k = ma$$$$25.98 - 10.22 = 5a$$$$a = \frac{15.76}{5} = 3.15\mathrm{ m/s}^2$$

Connected Bodies

For systems of connected objects (e.g., pulley systems):

• Draw a separate free-body diagram for each object.
• Apply $\vec{F} = m\vec{a}$ to each.
• Use the constraint that connected objects share the same acceleration magnitude.
• The tension in an ideal (massless, frictionless) string is the same throughout.

Example

Two masses $m_1 = 3\mathrm{ kg}$ and $m_2 = 5\mathrm{ kg}$ are connected by a light string over a frictionless pulley ($m_1$ hanging, $m_2$ on a rough table with $\mu_k = 0.4$).

For $m_1$ (hanging): $m_1 g - T = m_1 a$.

For $m_2$ (on table): $T - \mu_k m_2 g = m_2 a$.

Adding: $m_1 g - \mu_k m_2 g = (m_1 + m_2)a$.

$$a = \frac{m_1 g - \mu_k m_2 g}{m_1 + m_2} = \frac{3(9.81) - 0.4(5)(9.81)}{8} = \frac{29.43 - 19.62}{8} = \frac{9.81}{8} = 1.23\mathrm{ m/s}^2$$$$T = m_1(g - a) = 3(9.81 - 1.23) = 25.74\mathrm{ N}$$

Apparent Weight and Elevators

In an elevator accelerating upward at $a$:

$$N - mg = ma \implies N = m(g + a)$$

The apparent weight is $N = m(g + a)$, which is greater than the true weight.

In an elevator accelerating downward at $a$:

$$mg - N = ma \implies N = m(g - a)$$

The apparent weight is $N = m(g - a)$, which is less than the true weight.

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Newton's Third Law

Statement

For every action, there is an equal and opposite reaction:

$$\vec{F}_{A \mathrm{ on } B} = -\vec{F}_{B \mathrm{ on } A}$$

Key Points

• Action-reaction pairs act on different objects.
• They are equal in magnitude and opposite in direction.
• They are the same type of force (both gravitational, both normal, etc.).
• Action-reaction forces do NOT cancel because they act on different bodies.

Common Action-Reaction Pairs

Action	Reaction
Book pushes down on table	Table pushes up on book (normal force)
Earth pulls on apple (weight)	Apple pulls on Earth
Foot pushes backward on ground	Ground pushes forward on foot
Rocket pushes exhaust gases down	Exhaust gases push rocket up

Exam Tip

A common mistake is to confuse Newton's third law pairs with balanced forces on the same object. The weight and normal force on a stationary book are NOT an action-reaction pair — they act on the SAME object and balance. The reaction to the book's weight is the book pulling the Earth upward.

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Friction

Static Friction

Static friction prevents a stationary object from starting to move. It varies from $0$ up to a maximum value:

$$f_s \le \mu_s N$$

where $\mu_s$ is the coefficient of static friction and $N$ is the normal force.

Kinetic Friction

Kinetic friction opposes the motion of a sliding object:

$$f_k = \mu_k N$$

where $\mu_k$ is the coefficient of kinetic friction.

Key Relationships

• $\mu_s \gt \mu_k$ in general (it is harder to start moving than to keep moving).
• Friction is independent of the area of contact.
• Friction is proportional to the normal force.
• Kinetic friction is approximately constant (independent of speed).

Angle of Repose

The angle at which an object on an inclined plane begins to slide:

$$\tan\theta = \mu_s$$

Example

A block rests on an inclined plane at $35\degree$. Does it slide if $\mu_s = 0.6$?

$$\tan 35\degree \approx 0.700$$

Since $\tan 35\degree \gt \mu_s = 0.6$, the block slides.

Friction on an Inclined Plane

For a block on an incline at angle $\theta$:

Force	Expression
Component of weight along plane	$mg\sin\theta$
Component of weight perpendicular to plane	$mg\cos\theta$
Normal force	$N = mg\cos\theta$
Friction force	$f = \mu N = \mu mg\cos\theta$

The block slides down if $mg\sin\theta \gt \mu_s mg\cos\theta$, i.e., $\tan\theta \gt \mu_s$.

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Momentum and Impulse

Linear Momentum

$$\vec{p} = m\vec{v}$$

Momentum is a vector quantity with SI unit $\mathrm{kg} \cdot \mathrm{m/s}$.

Newton's Second Law (Momentum Form)

\vec{F}_{\mathrm{net}} = \frac{d\vec{p}}`\{dt}`

Impulse

Impulse equals the change in momentum:

$$\vec{J} = \vec{F}_{\mathrm{net}} \cdot \Delta t = \Delta\vec{p} = m\vec{v} - m\vec{u}$$

For a constant force: $J = F \Delta t$.

The unit of impulse is $\mathrm{N} \cdot \mathrm{s} = \mathrm{kg} \cdot \mathrm{m/s}$.

Impulse and Force-Time Graphs

The area under a force-time graph equals the impulse.

Example

A $0.15\mathrm{ kg}$ cricket ball moving at $30\mathrm{ m/s}$ is hit back at $20\mathrm{ m/s}$ in the opposite direction in $0.01\mathrm{ s}$. Find the average force.

$$J = \Delta p = 0.15(20 - (-30)) = 0.15 \times 50 = 7.5\mathrm{ N}\cdot\mathrm{s}$$$$F = \frac{J}{\Delta t} = \frac{7.5}{0.01} = 750\mathrm{ N}$$

Conservation of Linear Momentum

In a closed system (no external forces), the total momentum is conserved:

$$m_1\vec{u}_1 + m_2\vec{u}_2 = m_1\vec{v}_1 + m_2\vec{v}_2$$

Elastic and Inelastic Collisions

Type	Kinetic Energy	Momentum
Perfectly elastic	Conserved	Conserved
Inelastic	NOT conserved	Conserved
Perfectly inelastic	Maximum loss	Conserved (objects stick together)

Example

A $2\mathrm{ kg}$ object moving at $5\mathrm{ m/s}$ collides with a $3\mathrm{ kg}$ object at rest. After the collision, they stick together. Find the velocity after collision.

$$m_1 u_1 + m_2 u_2 = (m_1 + m_2)v$$$$2(5) + 3(0) = 5v$$$$v = 2\mathrm{ m/s}$$

KE before: $\dfrac{1}{2}(2)(25) = 25\mathrm{ J}$.

KE after: $\dfrac{1}{2}(5)(4) = 10\mathrm{ J}$.

$15\mathrm{ J}$ of kinetic energy is lost (inelastic collision).

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Work-Energy Theorem

Work Done

Work is done when a force causes displacement:

$$W = Fd\cos\theta$$

where $\theta$ is the angle between the force and the displacement.

• Work is a scalar quantity (SI unit: joule, $\mathrm{J}$).
• Work done by friction is always negative (opposes motion).
• If $F$ and $d$ are in the same direction: $W = Fd$.
• If $F \perp d$: $W = 0$ (e.g., normal force does no work on horizontal motion).

Work-Energy Theorem

$$W_{\mathrm{net}} = \Delta E_k = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$$

The net work done on an object equals its change in kinetic energy.

Kinetic Energy

$$E_k = \frac{1}{2}mv^2$$

Power

$$P = \frac{W}{t} = \vec{F} \cdot \vec{v}$$

SI unit: watt ($\mathrm{W}$), where $1\mathrm{ W} = 1\mathrm{ J/s}$.

Efficiency

$$\mathrm{Efficiency} = \frac{\mathrm{useful energy output}}{\mathrm{total energy input}} \times 100\%$$

Alternatively:

$$\mathrm{Efficiency} = \frac{\mathrm{useful power output}}{\mathrm{total power input}} \times 100\%$$

Example

A car of mass $1200\mathrm{ kg}$ accelerates from $15\mathrm{ m/s}$ to $25\mathrm{ m/s}$ in $6\mathrm{ s}$. The engine provides a constant driving force of $3000\mathrm{ N}$. Find the average friction force.

$$\Delta E_k = \frac{1}{2}(1200)(625 - 225) = \frac{1}{2}(1200)(400) = 240000\mathrm{ J}$$$$W_{\mathrm{net}} = W_{\mathrm{engine}} + W_{\mathrm{friction}} = 3000(6 \times 20) + W_f$$

Wait, let me recalculate properly:

$$W_{\mathrm{net}} = (F_{\mathrm{engine}} - f) \times d$$$$a = \frac{25-15}{6} = \frac{5}{3}\mathrm{ m/s}^2$$$$d = 15(6) + \frac{1}{2}\left(\frac{5}{3}\right)(36) = 90 + 30 = 120\mathrm{ m}$$$$240000 = (3000 - f)(120)$$$$3000 - f = 2000 \implies f = 1000\mathrm{ N}$$

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IB Exam-Style Questions

Question 1 (Paper 1 style)

A $10\mathrm{ kg}$ box sits on a rough surface. A horizontal force of $60\mathrm{ N}$ is applied. The box accelerates at $4\mathrm{ m/s}^2$. Find the coefficient of kinetic friction.

$$F_{\mathrm{net}} = F - f_k = ma$$ $$60 - \mu_k \times 10(9.81) = 10(4)$$ $$60 - 98.1\mu_k = 40$$ $$\mu_k = \frac{20}{98.1} \approx 0.204$$

Question 2 (Paper 2 style)

Two ice skaters, one of mass $60\mathrm{ kg}$ moving at $3\mathrm{ m/s}$ and the other of mass $80\mathrm{ kg}$ at rest, collide and move off together on frictionless ice.

(a) Find their common velocity after collision.

$$60(3) + 80(0) = (60 + 80)v$$ $$180 = 140v \implies v = 1.29\mathrm{ m/s}$$

(b) Calculate the kinetic energy lost.

Before: $\dfrac{1}{2}(60)(9) = 270\mathrm{ J}$.

After: $\dfrac{1}{2}(140)\left(\dfrac{9}{7}\right)^2 = \dfrac{1}{2}(140)\left(\dfrac{81}{49}\right) = 115.7\mathrm{ J}$.

Lost: $270 - 115.7 = 154.3\mathrm{ J}$.

Question 3 (Paper 2 style)

A block of mass $4\mathrm{ kg}$ is placed on a rough inclined plane at $30\degree$ to the horizontal. The coefficient of static friction is $0.5$ and the coefficient of kinetic friction is $0.3$.

(a) Determine whether the block slides.

$$mg\sin 30\degree = 4(9.81)(0.5) = 19.62\mathrm{ N}$$ $$\mu_s mg\cos 30\degree = 0.5(4)(9.81)(0.866) = 17.0\mathrm{ N}$$

Since $19.62 \gt 17.0$, the block slides.

(b) Find the acceleration down the plane.

$$a = g(\sin\theta - \mu_k\cos\theta) = 9.81(0.5 - 0.3 \times 0.866) = 9.81(0.240) = 2.36\mathrm{ m/s}^2$$

Question 4 (Paper 1 style)

A motor lifts a $500\mathrm{ kg}$ elevator through a height of $20\mathrm{ m}$ in $10\mathrm{ s}$. Find the power output of the motor.

P = \frac`\{mgh}`{t} = \frac{500 \times 9.81 \times 20}{10} = 9810\mathrm{ W} = 9.81\mathrm{ kW}

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Summary

Law	Statement
Newton's First	An object continues at rest or constant velocity unless acted on by net force
Newton's Second	$\vec{F}_{\mathrm{net}} = m\vec{a}$ or $\vec{F} = \dfrac{d\vec{p}}{dt}$
Newton's Third	Every action has an equal and opposite reaction

Quantity	Formula	Unit
Momentum	$\vec{p} = m\vec{v}$	$\mathrm{kg} \cdot \mathrm{m/s}$
Impulse	$\vec{J} = \vec{F}\Delta t = \Delta\vec{p}$	$\mathrm{N} \cdot \mathrm{s}$
Work	$W = Fd\cos\theta$	$\mathrm{J}$
Kinetic energy	$E_k = \dfrac{1}{2}mv^2$	$\mathrm{J}$
Power	$P = \dfrac{W}{t} = Fv$	$\mathrm{W}$

Exam Strategy

Always draw free-body diagrams. Resolve forces into components along the chosen axes. For momentum questions, clearly define the positive direction. For efficiency questions, remember that efficiency is always less than 100% in real systems.

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Advanced Applications

Motion in Lifts (Elevators)

Situation	Equation	Apparent Weight
Stationary or constant velocity	$N = mg$	Normal ($mg$)
Accelerating upward	$N = m(g + a)$	Greater than normal
Accelerating downward	$N = m(g - a)$	Less than normal
Free fall ($a = g$)	$N = 0$	Weightless

Example

A person of mass $70\mathrm{ kg}$ stands on a scale in a lift. Find the scale reading when the lift accelerates upward at $2\mathrm{ m/s}^2$.

$$N = m(g + a) = 70(9.81 + 2) = 70 \times 11.81 = 826.7\mathrm{ N}$$

The scale reads $826.7\mathrm{ N}$ (equivalent to $84.3\mathrm{ kg}$).

Connected Bodies on Inclined Planes

For two masses connected by a string over a pulley on an inclined plane, draw separate free-body diagrams and apply Newton's second law to each body.

Example

Mass $m_1 = 5\mathrm{ kg}$ hangs vertically. Mass $m_2 = 8\mathrm{ kg}$ is on a $30\degree$ incline with $\mu_k = 0.2$.

For $m_1$: $m_1 g - T = m_1 a$

For $m_2$: $T - m_2 g\sin\theta - \mu_k m_2 g\cos\theta = m_2 a$

Adding: $m_1 g - m_2 g\sin\theta - \mu_k m_2 g\cos\theta = (m_1 + m_2)a$

$$a = \frac{5(9.81) - 8(9.81)(0.5) - 0.2(8)(9.81)(0.866)}{13}$$$$= \frac{49.05 - 39.24 - 13.60}{13} = \frac{-3.79}{13}$$

The negative value means the system accelerates in the opposite direction to what was assumed (i.e., $m_2$ slides down and $m_1$ goes up).

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Friction: Extended Analysis

Static Friction Graph

As the applied force increases from zero:

1. The static friction matches the applied force (up to $\mu_s N$).
2. At the limiting friction point, the object begins to move.
3. Once moving, kinetic friction ($\mu_k N$) applies, which is less than the maximum static friction.
4. Kinetic friction is approximately constant regardless of speed.

Rolling Friction

Rolling friction is much smaller than sliding friction, which is why wheels are so effective. It arises from deformation of the rolling object and the surface.

Drag Force

At low speeds: $F_d \propto v$ (viscous drag, e.g., in oil).

At high speeds: $F_d \propto v^2$ (turbulent drag, e.g., air resistance on a car).

Terminal velocity is reached when drag equals the driving force (e.g., weight for a falling object):

mg = kv_{\mathrm{terminal}}^2 \implies v_{\mathrm{terminal}} = \sqrt{\frac`\{mg}`{k}}

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Momentum: Extended Applications

Rockets and Thrust

A rocket expels mass (exhaust gases) at high velocity. By conservation of momentum:

$$\mathrm{Thrust} = v_e \frac{\Delta m}{\Delta t}$$

where $v_e$ is the exhaust velocity and $\dfrac{\Delta m}{\Delta t}$ is the mass flow rate.

Rocket Equation (Tsiolkovsky)

$$\Delta v = v_e \ln\!\left(\frac{m_i}{m_f}\right)$$

where $m_i$ is the initial mass and $m_f$ is the final mass.

Impulse-Momentum in Two Dimensions

Momentum is conserved separately in each direction:

$$\sum m\vec{v}_{\mathrm{initial}} = \sum m\vec{v}_{\mathrm{final}}$$

Resolve into $x$ and $y$ components and apply conservation in each direction independently.

Example

A $3\mathrm{ kg}$ object moving at $4\mathrm{ m/s}$ collides with a stationary $2\mathrm{ kg}$ object. After the collision, the $3\mathrm{ kg}$ object moves at $2\mathrm{ m/s}$ at $30\degree$ above the original direction. Find the velocity of the $2\mathrm{ kg}$ object.

x-direction: $3(4) = 3(2)\cos 30\degree + 2v_x$

$$12 = 5.196 + 2v_x \implies v_x = 3.402\mathrm{ m/s}$$

y-direction: $0 = 3(2)\sin 30\degree + 2v_y$

$$0 = 3 + 2v_y \implies v_y = -1.5\mathrm{ m/s}$$$$v = \sqrt{3.402^2 + (-1.5)^2} = \sqrt{11.57 + 2.25} = \sqrt{13.82} = 3.72\mathrm{ m/s}$$

Direction: $\theta = \arctan\!\left(\dfrac{-1.5}{3.402}\right) = -23.8\degree$ (below the original direction).

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Additional IB Exam-Style Questions

Question 5 (Paper 2 style)

A $60\mathrm{ kg}$ skier starts from rest at the top of a $30\degree$ slope that is $100\mathrm{ m}$ long. The coefficient of kinetic friction is $0.1$.

(a) Find the acceleration down the slope.

$$a = g(\sin\theta - \mu_k\cos\theta) = 9.81(\sin 30\degree - 0.1\cos 30\degree) = 9.81(0.5 - 0.0866) = 9.81(0.4134) = 4.06\mathrm{ m/s}^2$$

(b) Find the speed at the bottom of the slope.

$$v^2 = u^2 + 2as = 0 + 2(4.06)(100) = 812$$ $$v = 28.5\mathrm{ m/s}$$

(c) The skier then travels across level ground with the same coefficient of friction. How far do they slide before stopping?

$$\mu_k mg \cdot d = \frac{1}{2}mv^2$$ $$d = \frac{v^2}{2\mu_k g} = \frac{812}{2(0.1)(9.81)} = \frac{812}{1.962} = 414\mathrm{ m}$$

Question 6 (Paper 2 style)

Two objects, one of mass $m$ and the other of mass $3m$, collide head-on. The lighter object is moving at $6\mathrm{ m/s}$ and the heavier one at $2\mathrm{ m/s}$ in the opposite direction. After the collision, the lighter object moves at $2\mathrm{ m/s}$ in the opposite direction to its original motion.

(a) Find the velocity of the heavier object after the collision.

Taking the direction of $m$ as positive:

$$m(6) + 3m(-2) = m(-2) + 3m v$$ $$6m - 6m = -2m + 3mv$$ $$0 = -2 + 3v \implies v = \frac{2}{3}\mathrm{ m/s} \mathrm{ (in the positive direction)}$$

(b) Is the collision elastic?

KE before $= \dfrac{1}{2}m(36) + \dfrac{1}{2}(3m)(4) = 18m + 6m = 24m$.

KE after $= \dfrac{1}{2}m(4) + \dfrac{1}{2}(3m)\!\left(\dfrac{4}{9}\right) = 2m + \dfrac{2m}{3} = \dfrac{8m}{3}$.

Since $24m \neq \dfrac{8m}{3}$, the collision is inelastic.

Question 7 (Paper 1 style)

A force of $F = 3t^2\mathrm{ N}$ acts on a $2\mathrm{ kg}$ object initially at rest. Find the velocity after $4\mathrm{ s}$.

$$F = ma \implies a = \frac{3t^2}{2} = 1.5t^2$$ $$v = \int_0^4 1.5t^2\,dt = \left[0.5t^3\right]_0^4 = 0.5(64) = 32\mathrm{ m/s}$$

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Advanced Dynamics Topics

Systems with Friction on Multiple Surfaces

When a system involves multiple surfaces with different coefficients of friction, draw separate free-body diagrams for each object and apply Newton's second law individually.

Motion on Curved Paths

For an object moving along a curved path (not necessarily circular), the normal force provides the centripetal component of acceleration:

$$N - mg\cos\theta = \frac{mv^2}{r}$$

where $\theta$ is the angle of the surface with the horizontal.

Friction on a Banked Curve with Speed Different from Ideal

On a banked curve designed for speed $v_0$, if a car travels at speed $v \neq v_0$, friction provides the additional centripetal force:

• $v \gt v_0$: friction acts down the slope (adds to centripetal force).
• $v \lt v_0$: friction acts up the slope (reduces centripetal force).

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Additional IB Exam-Style Questions

Question 8 (Paper 2 style)

A block of mass $3\mathrm{ kg}$ is on a rough horizontal surface ($\mu_s = 0.4$, $\mu_k = 0.3$). A force of $15\mathrm{ N}$ is applied at $25\degree$ above the horizontal.

(a) Determine whether the block moves.

Vertical: $N + 15\sin 25\degree = 3(9.81)$.

$$N = 29.43 - 6.34 = 23.09\mathrm{ N}$$

Maximum static friction: $f_{s,\max} = 0.4 \times 23.09 = 9.24\mathrm{ N}$.

Horizontal applied force: $15\cos 25\degree = 13.59\mathrm{ N}$.

Since $13.59 \gt 9.24$, the block moves.

(b) Find the acceleration.

$$a = \frac{15\cos 25\degree - 0.3(23.09)}{3} = \frac{13.59 - 6.93}{3} = \frac{6.66}{3} = 2.22\mathrm{ m/s}^2$$

(c) If the applied force is removed, how far does the block slide before stopping?

$$\mu_k mg \cdot d = \frac{1}{2}mv^2$$

Need to find $v$ at the moment force is removed. Assuming the force was applied from rest:

This depends on how long the force was applied. If the question implies the block has some velocity $v_0$ when the force is removed:

$$d = \frac{v_0^2}{2\mu_k g} = \frac{v_0^2}{5.886}$$

Question 9 (Paper 1 style)

A $5\mathrm{ kg}$ object experiences a force that varies with time: $F = 6t - 2\mathrm{ N}$ for $0 \le t \le 5\mathrm{ s}$. If the object starts from rest, what is its momentum at $t = 5\mathrm{ s}$?

$$\Delta p = \int_0^5 F\,dt = \int_0^5 (6t - 2)\,dt = [3t^2 - 2t]_0^5 = 75 - 10 = 65\mathrm{ kg}\cdot\mathrm{m/s}$$

Since $p_0 = 0$: $p = 65\mathrm{ kg}\cdot\mathrm{m/s}$.

Question 10 (Paper 2 style)

Two trolleys, A ($2\mathrm{ kg}$) and B ($3\mathrm{ kg}$), are held together by a compressed spring between them on a frictionless surface. When released, trolley A moves at $4\mathrm{ m/s}$ to the left.

(a) Find the velocity of trolley B.

$$0 = m_A v_A + m_B v_B$$ $$0 = 2(-4) + 3v_B \implies v_B = \frac{8}{3} = 2.67\mathrm{ m/s} \mathrm{ (to the right)}$$

(b) Calculate the elastic potential energy stored in the spring.

$$E_e = \frac{1}{2}m_A v_A^2 + \frac{1}{2}m_B v_B^2 = \frac{1}{2}(2)(16) + \frac{1}{2}(3)\!\left(\frac{64}{9}\right) = 16 + 10.67 = 26.67\mathrm{ J}$$

(c) If the spring has spring constant $k = 2000\mathrm{ N/m}$, find the initial compression.

$$E_e = \frac{1}{2}kx^2 \implies 26.67 = 1000x^2 \implies x = 0.163\mathrm{ m}$$

For the A-Level treatment of this topic, see Dynamics.

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