Mechanics

Higher Mechanics

Scalars and Vectors

Physical quantities are either scalars (magnitude only) or vectors (magnitude and Direction).

Scalar	Vector
Mass, speed, energy, time	Displacement, velocity, force, acceleration

Resolving Vectors into Components

Any vector can be resolved into perpendicular components. For a vector of magnitude $F$ at angle $\theta$ to the horizontal:

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

The original vector can be reconstructed: $F = \sqrt{F_x^2 + F_y^2}$At angle $\theta = \arctan(F_y / F_x)$.

This technique is the single most important problem-solving tool in mechanics. Instead of working With forces at awkward angles, resolve every force into components along two perpendicular axes, Apply Newton’s second law to each axis independently, and then recombine the results.

Example: A boat sails north at 5 \mathrm{ m/s in a current flowing east at 3 \mathrm{ m/s. Find the resultant velocity.

|\mathbf{v}| = \sqrt{5^2 + 3^2} = \sqrt{34} \approx 5.83 \mathrm{ m/s

\theta = \arctan\left(\frac{3}{5}\right) \approx 30.96° \mathrm{ east of north

Vector Addition

Vectors can be added using the triangle rule (head-to-tail) or by resolving into components and Adding the components separately:

$\vec{R} = \vec{A} + \vec{B} \implies R_x = A_x + B_x, \quad R_y = A_y + B_y$

Equations of Motion (Kinematics)

For constant acceleration $a$:

Explore the simulation above to develop intuition for this topic.

$v = u + at$

$s = ut + \frac{1}{2}at^2$

$v^2 = u^2 + 2as$

$s = \frac{(u + v)}{2}t$

Where $u$ = initial velocity, $v$ = final velocity, $s$ = displacement, $t$ = time.

Choosing the Right Equation

Each equation relates a different subset of the five variables ($u$, $v$, $a$, $s$, $t$). The Variable that does not appear tells you when to use that equation. If you know $u$, $a$And $s$ And need $v$Use $v^2 = u^2 + 2as$ (it does not contain $t$). If you know $u$, $v$And $t$ and Need $s$Use $s = \frac{1}{2}(u+v)t$ (it does not contain $a$).

Example: A car accelerates uniformly from 10 \mathrm{ m/s to 30 \mathrm{ m/s over a distance Of 200 \mathrm{ m. Find the acceleration and time taken.

Using $v^2 = u^2 + 2as$:

$900 = 100 + 2a(200)$

$800 = 400a$

a = 2 \mathrm{ m/s^2

Using $v = u + at$:

$30 = 10 + 2t$

t = 10 \mathrm{ s

When the Kinematic Equations Fail

These equations assume constant acceleration. If the acceleration changes (a rocket losing mass, a Spring-mass system, a pendulum), these equations do not apply. For variable acceleration, use Calculus: $v = u + \int_0^t a(t')\, dt'$ and $s = \int_0^t v(t')\, dt'$.

Projectile Motion

A projectile follows a parabolic path when air resistance is negligible.

Horizontal: $x = v_0 \cos\theta \cdot t$ (constant velocity)

Vertical: $y = v_0 \sin\theta \cdot t - \frac{1}{2}gt^2$ (uniform acceleration)

Key Results

• Maximum height: $H = \dfrac{(v_0 \sin\theta)^2}{2g}$
• Time of flight: $T = \dfrac{2v_0 \sin\theta}{g}$
• Range: $R = \dfrac{v_0^2 \sin 2\theta}{g}$
• Maximum range at $\theta = 45^\circ$

Why the Path Is a Parabola

Eliminating $t$ from the horizontal and vertical equations:

$t = \frac{x}{v_0 \cos\theta}, \qquad y = x\tan\theta - \frac{gx^2}{2v_0^2\cos^2\theta}$

This is quadratic in $x$ with a negative coefficient, confirming a parabola opening downward.

Independence of Horizontal and Vertical Motion

The horizontal and vertical components are completely independent. There is no horizontal force (ignoring air resistance), so the horizontal velocity is constant. The vertical acceleration is Always $g$ downward, regardless of the horizontal motion. This means you can solve each component Separately using the kinematic equations.

Example: A ball is thrown at 20 \mathrm{ m/s at an angle of $30^\circ$ above the horizontal. Find The maximum height and horizontal range.

Vertical component: v_y = 20\sin 30° = 10 \mathrm{ m/s.

Maximum height (when $v_y = 0$): $v_y^2 = u_y^2 - 2gh$

$0 = 100 - 2(9.8)h$

h = \frac{100}{19.6} \approx 5.10 \mathrm{ m

Time of flight: t = \dfrac{2u_y}{g} = \dfrac{20}{9.8} \approx 2.04 \mathrm{ s.

Horizontal component: v_x = 20\cos 30° = 17.32 \mathrm{ m/s.

Range: R = v_x \cdot t = 17.32 \times 2.04 \approx 35.3 \mathrm{ m.

Newton’s Laws of Motion

First Law: An object remains at rest or moves with constant velocity unless acted upon by a Resultant force.

Second Law: $F = ma$Where $F$ is the resultant force, $m$ is mass, and $a$ is acceleration.

Third Law: If object A exerts a force on object B, then B exerts an equal and opposite force on A.

Why Newton’s Third Law Pairs Act on Different Objects

The normal force and weight on a book resting on a table are not a third law pair. They act on the same object and balance because the book is in equilibrium. The correct third law pairs are: Earth Pulls book down / book pulls Earth up, and table pushes book up / book pushes table down. The fact That the pairs act on different objects is why they cannot cancel.

Example: A block of mass 5 \mathrm{ kg is pulled across a rough horizontal surface by a Horizontal force of 30 \mathrm{ N. The coefficient of friction is $0.4$. Find the acceleration.

Normal force: N = mg = 5 \times 9.8 = 49 \mathrm{ N.

Friction force: f = \mu N = 0.4 \times 49 = 19.6 \mathrm{ N.

Resultant force: F_{\mathrm{net} = 30 - 19.6 = 10.4 \mathrm{ N.

a = \frac{F_{\mathrm{net}}{m} = \frac{10.4}{5} = 2.08 \mathrm{ m/s^2

Free-Body Diagrams and Connected Bodies

Draw a free body diagram for each object separately. Apply Newton’s second law to each object. The Tension in a massless, inextensible string is the same throughout.

Example: Two masses m_1 = 3 \mathrm{ kg and m_2 = 5 \mathrm{ kg are connected by a light Inextensible string over a smooth pulley. Find the acceleration and tension.

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

For $m_2$ (moving down): $m_2 g - T = m_2 a$.

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

a = \frac{(m_2 - m_1)g}{m_1 + m_2} = \frac{2 \times 9.8}{8} = 2.45 \mathrm{ m/s^2

T = m_1(g + a) = 3(9.8 + 2.45) = 36.75 \mathrm{ N

Inclined Planes

On a plane inclined at angle $\theta$Resolve forces parallel and perpendicular to the plane:

• Component of weight down the slope: $mg\sin\theta$
• Normal reaction: $N = mg\cos\theta$
• Friction: $f = \mu N = \mu mg\cos\theta$ (opposing motion)

The block slides when $mg\sin\theta \gt \mu mg\cos\theta$I.e., $\tan\theta \gt \mu$.

Friction on Inclined Planes

If the block is sliding down, friction acts up the slope:

$a = g(\sin\theta - \mu\cos\theta)$

If the block is sliding up, friction acts down the slope:

$a = -g(\sin\theta + \mu\cos\theta)$

The negative sign indicates deceleration.

Work, Energy, and Power

Work Done: $W = Fs\cos\theta$Where $\theta$ is the angle between the force and displacement.

For force in the direction of motion: $W = Fs$.

Kinetic Energy: $E_k = \dfrac{1}{2}mv^2$

Gravitational Potential Energy: $E_p = mgh$

Work-Energy Principle: The work done by the resultant force equals the change in kinetic energy.

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

Conservation of Energy (mechanical): In the absence of non-conservative forces:

E_k + E_p = \mathrm{constant

Power: $P = \dfrac{W}{t} = Fv$

Why the Work-Energy Theorem Is Powerful

You do not need to know the acceleration, the time, or the detailed path. Given the initial and Final speeds and the forces, you can solve the problem directly. This is especially useful for Variable-force problems and curved paths.

Example: A car of mass 1200 \mathrm{ kg travels at a constant speed of 25 \mathrm{ m/s up a Hill inclined at $5^\circ$ to the horizontal. The total resistive force is 800 \mathrm{ N. Find the Power output of the engine.

Component of weight along the slope: mg\sin 5° = 1200 \times 9.8 \times \sin 5° \approx 1026 \mathrm{ N.

Total force to be overcome: 1026 + 800 = 1826 \mathrm{ N.

P = Fv = 1826 \times 25 = 45650 \mathrm{ W \approx 45.7 \mathrm{ kW

Example: A roller coaster car starts from rest at a height of 30 \mathrm{ m. Neglecting Friction, find its speed at a height of 10 \mathrm{ m.

Conservation of energy: $mgh_1 = mgh_2 + \frac{1}{2}mv^2$

$\frac{1}{2}v^2 = g(h_1 - h_2) = 9.8 \times 20 = 196$

v = \sqrt{392} \approx 19.8 \mathrm{ m/s

Energy with Friction

When friction is present, the work-energy theorem becomes:

W_{\mathrm{nc} = \Delta E_k + \Delta E_p

Where W_{\mathrm{nc} is the work done by non-conservative forces (friction, which is negative).

$-fd = \frac{1}{2}mv^2 - 0 + mg(h_f - h_i)$

Momentum and Impulse

Momentum: $p = mv$

Newton’s Second Law (momentum form): $F = \dfrac{dp}{dt} = \dfrac{\Delta p}{\Delta t}$

Impulse: $J = F\Delta t = \Delta p = m\Delta v$

Conservation of Momentum: For a system with no external forces:

$m_1 u_1 + m_2 u_2 = m_1 v_1 + m_2 v_2$

Why Momentum Is Conserved

Newton’s third law guarantees that internal forces cancel in pairs. If the net external force is Zero, the total momentum of the system cannot change. This is one of the most powerful conservation Laws in physics.

Example: A 2 \mathrm{ kg ball moving at 5 \mathrm{ m/s collides with a stationary 3 \mathrm{ kg ball. If the 2 \mathrm{ kg ball rebounds at 1 \mathrm{ m/sFind the velocity of The 3 \mathrm{ kg ball.

Conservation of momentum:

$2(5) + 3(0) = 2(-1) + 3v$

$10 = -2 + 3v$

v = 4 \mathrm{ m/s

Impulse and Average Force

For a given change in momentum, increasing the impact time decreases the average force:

F_{\mathrm{avg} = \frac{\Delta p}{\Delta t}

This is why car crumple zones, seat belts, air bags, and crash mats all work: they increase the time Over which the momentum changes, reducing the peak force.

Collisions

Elastic collision: Both momentum and kinetic energy are conserved.

Inelastic collision: Momentum is conserved, but kinetic energy is not.

Coefficient of Restitution:

e = \frac{\mathrm{relative speed of separation}{\mathrm{relative speed of approach}

For a perfectly elastic collision, $e = 1$. For a perfectly inelastic collision, $e = 0$.

Elastic Collision Formulas (Stationary Target)

When $m_2$ is initially at rest:

$v_1 = \frac{m_1 - m_2}{m_1 + m_2} u_1, \qquad v_2 = \frac{2m_1}{m_1 + m_2} u_1$

Special case: If $m_1 = m_2$The objects exchange velocities.

Circular Motion (Higher)

Centripetal Acceleration and Force

For an object moving in a circle of radius $r$ at speed $v$:

$a = \frac{v^2}{r} = \omega^2 r, \qquad F = \frac{mv^2}{r} = m\omega^2 r$

The centripetal force always points towards the centre of the circle. It is not a separate force; it Is the net radial force.

Why Objects Do Not Fly Off in Circular Motion

Newton’s first law says an object continues in a straight line unless acted upon by a force. In Circular motion, the centripetal force continuously deflects the object from its straight-line path. If the centripetal force disappears (the string breaks, friction is insufficient), the object Continues in a straight line tangent to the circle at the point of release. It does not fly radially Outward.

Example: A car of mass 1200 \mathrm{ kg travels around a roundabout of radius 25 \mathrm{ m at 12 \mathrm{ m/s. Find the centripetal force.

F = \frac{mv^2}{r} = \frac{1200 \times 144}{25} = \frac{172800}{25} = 6912 \mathrm{ N

Vertical Circular Motion

At the top of a vertical circle: $F_c = mg + T$ (both weight and tension point toward the centre).

At the bottom: $F_c = T - mg$ (tension points toward the centre, weight points away).

The minimum speed at the top to maintain circular motion: $v_{\min} = \sqrt{gr}$ (when $T = 0$).

---

Common Pitfalls

1. Confusing mass and weight: Mass is measured in kg; weight is a force measured in N ($W = mg$).

2. Forgetting to resolve forces: On an inclined plane, always resolve forces parallel and perpendicular to the plane.

3. Sign conventions in momentum: Be consistent with positive and negative directions.

4. Assuming all collisions are elastic: Always check whether kinetic energy is conserved.

5. Units in power: Power is measured in watts (W). Ensure force is in N and velocity in m/s.

6. Using the wrong kinematic equation: Choose the equation that does not contain the unknown variable you do not have.

7. Confusing centripetal force with a separate force: The centripetal force is the resultant radial force, not an additional interaction.

8. Forgetting that the range formula only applies when launch and landing heights are equal.

---

Practice Questions

1. A stone is thrown horizontally from a cliff 40 \mathrm{ m high at 15 \mathrm{ m/s. Find the time to hit the ground and the horizontal distance travelled.

2. A 60 \mathrm{ kg skier accelerates down a slope inclined at $20^\circ$ to the horizontal. The coefficient of friction is $0.1$. Find the acceleration.

3. A car of mass 800 \mathrm{ kg accelerates from rest to 20 \mathrm{ m/s in 8 \mathrm{ s. Calculate the work done and the average power.

4. A 4 \mathrm{ kg object moving at 6 \mathrm{ m/s collides with a 6 \mathrm{ kg object moving at 2 \mathrm{ m/s in the same direction. They stick together. Find the velocity after the collision and the kinetic energy lost.

5. A ball is projected at 25 \mathrm{ m/s at $40^\circ$ above the horizontal from the top of a 15 \mathrm{ m cliff. Find the speed when it hits the ground.

6. Two particles of masses 2 \mathrm{ kg and 3 \mathrm{ kg are connected by a string over a pulley. The system is released from rest. Find the speed after the 3 \mathrm{ kg mass has fallen 2 \mathrm{ m.

7. A crane lifts a 500 \mathrm{ kg load from rest with constant acceleration. In the first 3 \mathrm{ sIt rises 9 \mathrm{ m. Find the tension in the cable.

8. A bullet of mass 10 \mathrm{ g travelling at 400 \mathrm{ m/s embeds itself in a wooden block of mass 2 \mathrm{ kg at rest. Find the velocity of the block immediately after impact.

9. A car of mass 1000 \mathrm{ kg rounds a bend of radius 80 \mathrm{ m at 15 \mathrm{ m/s. Find the minimum coefficient of static friction required.

10. A 3 \mathrm{ kg block slides 4 \mathrm{ m down a $30^\circ$ incline with $\mu_k = 0.2$. Find the work done by friction, the work done by gravity, and the final speed if the block started from rest.

11. Derivation of the Kinematic Equations from Calculus (Advanced Higher)

For constant acceleration $a$Starting from the definition of acceleration:

$a = \frac{dv}{dt}$

Integrating with respect to time:

$\int_{u}^{v} dv = \int_0^t a\, dt' \implies v - u = at \implies v = u + at$

This is the first kinematic equation. Now integrate velocity to get displacement:

$v = \frac{ds}{dt} = u + at$

$\int_0^s ds = \int_0^t (u + at')\, dt' \implies s = ut + \frac{1}{2}at^2$

This is the second kinematic equation. To eliminate $t$Substitute $t = (v - u)/a$ into the second Equation:

$s = u\frac{v - u}{a} + \frac{1}{2}a\left(\frac{v - u}{a}\right)^2 = \frac{uv - u^2}{a} + \frac{v^2 - 2uv + u^2}{2a}$

$2as = 2uv - 2u^2 + v^2 - 2uv + u^2 = v^2 - u^2$

$v^2 = u^2 + 2as$

This is the third kinematic equation. The fourth follows from the average velocity:

$s = \bar{v}t = \frac{u + v}{2}t$

12. Worked Example: Projectile from a Height

A ball is thrown from the top of a 40 \mathrm{ m cliff at 15 \mathrm{ m/s at $25^\circ$ above the Horizontal. Find the time to hit the ground and the horizontal distance.

v_x = 15\cos 25° = 13.59 \mathrm{ m/s, \qquad v_y = 15\sin 25° = 6.34 \mathrm{ m/s

The vertical motion: $y = 40 + 6.34t - 4.9t^2$. At impact, $y = 0$:

$4.9t^2 - 6.34t - 40 = 0$

$t = \frac{6.34 \pm \sqrt{6.34^2 + 4 \times 4.9 \times 40}}{2 \times 4.9} = \frac{6.34 \pm \sqrt{40.20 + 784}}{9.8} = \frac{6.34 \pm 28.72}{9.8}$

Taking the positive root: t = \frac{35.06}{9.8} = 3.58 \mathrm{ s.

Horizontal distance: R = v_x t = 13.59 \times 3.58 = 48.7 \mathrm{ m.

13. Worked Example: Energy with Friction on an Incline

A 4 \mathrm{ kg block slides 5 \mathrm{ m down a $30^\circ$ rough incline with $\mu_k = 0.2$Starting From rest.

Work done by gravity: W_g = mg\sin\theta \times d = 4 \times 9.8 \times \sin 30° \times 5 = 98 \mathrm{ J

Normal force: N = mg\cos\theta = 4 \times 9.8 \times \cos 30° = 33.95 \mathrm{ N

Friction force: f = \mu_k N = 0.2 \times 33.95 = 6.79 \mathrm{ N

Work done by friction: W_f = -fd = -6.79 \times 5 = -33.95 \mathrm{ J

Net work: W_{\mathrm{net} = 98 - 33.95 = 64.05 \mathrm{ J

Final speed: W_{\mathrm{net} = \frac{1}{2}mv^2 \implies v = \sqrt{\frac{2 \times 64.05}{4}} = \sqrt{32.03} = 5.66 \mathrm{ m/s

14. Vertical Circular Motion: Detailed Analysis

For a ball of mass $m$ on a string of length $L$ swung in a vertical circle:

At the bottom: $T - mg = \frac{mv^2}{L}$So $T = mg + \frac{mv^2}{L}$ (tension is maximum Here).

At the top: $T + mg = \frac{mv^2}{L}$So $T = \frac{mv^2}{L} - mg$ (tension is minimum here).

Minimum speed at the top: Set $T = 0$: $v_{\min} = \sqrt{gL}$.

Worked Example: A ball of mass 0.5 \mathrm{ kg on a string of length 1.2 \mathrm{ m has speed 7 \mathrm{ m/s at the bottom. Find the tension at the bottom and the tension at the top.

At the bottom: T_b = mg + \frac{mv_b^2}{L} = 0.5 \times 9.8 + \frac{0.5 \times 49}{1.2} = 4.9 + 20.42 = 25.3 \mathrm{ N

Energy conservation to find speed at the top:

$\frac{1}{2}mv_b^2 = \frac{1}{2}mv_t^2 + mg(2L)$

$v_t^2 = v_b^2 - 4gL = 49 - 4 \times 9.8 \times 1.2 = 49 - 47.04 = 1.96$

v_t = 1.4 \mathrm{ m/s

At the top: T_t = \frac{mv_t^2}{L} - mg = \frac{0.5 \times 1.96}{1.2} - 4.9 = 0.817 - 4.9 = -4.08 \mathrm{ N

A negative tension means the string would go slack. The ball does not complete the full circle. For The ball to just complete the circle, we need v_t = \sqrt{gL} = \sqrt{11.76} = 3.43 \mathrm{ m/s.

15. Summary Table: Key Mechanics Formulas

Topic	Formula	Variables	Conditions
Kinematics	$v = u + at$	$u$, $v$, $a$, $t$	Constant acceleration
Kinematics	$v^2 = u^2 + 2as$	$u$, $v$, $a$, $s$	Constant acceleration
Projectile range	$R = u^2\sin 2\theta / g$	u$$\theta$$g	Launch = landing height
Newton’s 2nd law	$F = ma$	F$$m$$a	Vector equation
Weight	$W = mg$	m$$g	Near surface
Friction	$f = \mu N$	\mu$$N	Sliding
Work	$W = Fs\cos\theta$	F$$s$$\theta	Constant force
Kinetic energy	$E_k = \frac{1}{2}mv^2$	m$$v	Always positive
Potential energy	$E_p = mgh$	m$$g$$h	Near surface
Power	$P = Fv$	F$$v	Constant velocity
Momentum	$p = mv$	m$$v	Vector
Impulse	$J = F\Delta t = \Delta p$	F$$\Delta t$$\Delta p	—
Centripetal force	$F = mv^2/r$	m$$v$$r	Uniform circular motion

16. Practice Questions (Additional)

11. A stone is thrown at 18 \mathrm{ m/s at $50^\circ$ to the horizontal from the top of a 25 \mathrm{ m cliff. Calculate the speed at which it hits the ground.

12. A car of mass 900 \mathrm{ kg travels at constant speed around a circular bend of radius 60 \mathrm{ m that is banked at $12^\circ$. Calculate the minimum coefficient of static friction required.

13. A bullet of mass 8 \mathrm{ g travelling at 600 \mathrm{ m/s passes through a wooden block of mass 2 \mathrm{ kgEmerging at 200 \mathrm{ m/s. Calculate the velocity of the block and the kinetic energy lost.

14. A 5 \mathrm{ kg box is pushed up a $35^\circ$ incline at constant speed by a force of 50 \mathrm{ N applied parallel to the incline. Calculate the coefficient of kinetic friction.

15. A ball of mass 0.3 \mathrm{ kg is attached to a string of length 0.8 \mathrm{ m and swung in a vertical circle. Calculate the minimum speed at the lowest point for the ball to complete the full circle.

16. Two objects of masses 3 \mathrm{ kg and 5 \mathrm{ kgConnected by a light string, are placed on a smooth table with a pulley at the edge. The 5 \mathrm{ kg mass hangs over the edge. Find the speed of the system after the 5 \mathrm{ kg mass has fallen 1.5 \mathrm{ m using energy methods.

17. A satellite of mass 500 \mathrm{ kg orbits at a height of 600 \mathrm{ km above Earth’s surface. Calculate the orbital speed, period, and gravitational potential energy of the satellite.

18. A 2 \mathrm{ kg object slides from rest down a frictionless curved ramp of total height 4 \mathrm{ m. At the bottom, it enters a rough horizontal surface with $\mu_k = 0.3$. How far does it slide before stopping?

19. A crane lifts a 300 \mathrm{ kg load with an upward acceleration of 1.5 \mathrm{ m/s^2 for 4 \mathrm{ sThen travels at constant speed for 3 \mathrm{ sThen decelerates at 2 \mathrm{ m/s^2 to rest. Calculate the total work done by the tension in the cable.

20. Explain why the range of a projectile is the same for complementary angles (e.g., $30^\circ$ and $60^\circ$). Include a mathematical derivation.

Extended Worked Examples

Example 21: Work-Energy with Variable Force

A force $F(x) = 3x^2 - 2x$ (in N) acts on a 2 \mathrm{ kg object moving along the x-axis. Calculate The work done by this force as the object moves from $x = 0$ to x = 3 \mathrm{ mAnd find the Speed at x = 3 \mathrm{ m if the object starts from rest.

Step 1: Calculate the work done by integration

W = \int_0^3 F(x) \, dx = \int_0^3 (3x^2 - 2x) \, dx = \left[ x^3 - x^2 \right]_0^3 = (27 - 9) - 0 = 18 \mathrm{ J

Step 2: Apply the work-energy theorem

$W = \Delta KE = \frac{1}{2}mv^2 - \frac{1}{2}mu^2$

$18 = \frac{1}{2}(2)v^2 - 0$

v = \sqrt{18} = 4.24 \mathrm{ m/s

:::info The work-energy theorem is one of the most powerful tools in mechanics. It works even when Forces are variable (requiring calculus) and does not require knowledge of the path taken, only the Initial and final speeds. :::

Example 22: Oblique Collision in 2D

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 $30^\circ$ to its original Direction. Find the velocity of the 2 \mathrm{ kg object after the collision.

Take the original direction as the x-axis.

Step 1: Conservation of momentum (x-component)

$3 \times 4 + 0 = 3 \times 2\cos 30° + 2v_{2x}$

$12 = 3 \times 1.732 + 2v_{2x} = 5.196 + 2v_{2x}$

v_{2x} = \frac{12 - 5.196}{2} = 3.402 \mathrm{ m/s

Step 2: Conservation of momentum (y-component)

$0 = 3 \times 2\sin 30° + 2v_{2y}$

$0 = 3 \times 1 + 2v_{2y} = 3 + 2v_{2y}$

v_{2y} = -1.5 \mathrm{ m/s

Step 3: Find magnitude and direction of $v_2$

|v_2| = \sqrt{3.402^2 + (-1.5)^2} = \sqrt{11.57 + 2.25} = \sqrt{13.82} = 3.72 \mathrm{ m/s

$\theta = \arctan\left(\frac{-1.5}{3.402}\right) = -23.8^\circ$

The 2 \mathrm{ kg object moves at 3.72 \mathrm{ m/s at $23.8^\circ$ below the original direction.

Example 23: Connected Bodies on an Inclined Plane

Two masses, m_A = 8 \mathrm{ kg on a $30^\circ$ incline and m_B = 5 \mathrm{ kg hanging vertically, are Connected by a light inextensible string over a smooth pulley at the top of the incline. The Coefficient of friction between $m_A$ and the incline is $\mu = 0.2$. Find the acceleration of the System and the tension in the string. Take g = 9.8 \mathrm{ m/s^2.

Step 1: Identify the direction of motion

Weight component of $m_A$ down the slope: m_A g \sin 30° = 8 \times 9.8 \times 0.5 = 39.2 \mathrm{ N

Friction force (opposing motion): f = \mu m_A g \cos 30° = 0.2 \times 8 \times 9.8 \times 0.866 = 13.6 \mathrm{ N

Weight of $m_B$: m_B g = 5 \times 9.8 = 49 \mathrm{ N

Since m_B g = 49 \mathrm{ N \gt m_A g \sin 30° + f = 39.2 + 13.6 = 52.8 \mathrm{ N… Actually $49 \lt 52.8$.

So the system does not move. Let me verify: if we assume $m_B$ moves down:

$m_B g - m_A g \sin 30° - \mu m_A g \cos 30° = (m_A + m_B)a$

$49 - 39.2 - 13.6 = 13a$

-3.8 = 13a \implies a = -0.29 \mathrm{ m/s^2

The negative acceleration means our assumption was wrong. The system would try to move with $m_A$ Sliding down, but let us check:

$m_A g \sin 30° - m_B g - \mu m_A g \cos 30° = (m_A + m_B)a$

$39.2 - 49 - 13.6 = 13a$

-23.4 = 13a \implies a = -1.8 \mathrm{ m/s^2

This is also negative. Neither direction produces positive acceleration, so the system remains in Equilibrium. The static friction adjusts to prevent motion.

Correct answer: The system is in equilibrium. The tension equals m_B g = 49 \mathrm{ N (since The string supports $m_B$ in equilibrium), and the static friction on $m_A$ adjusts to balance the Net force.

:::caution Always check whether motion actually occurs before applying the equations of motion. If The net force in every possible direction is negative (accounting for the maximum static friction), The system remains stationary. :::

Common Pitfalls Extended

Pitfall 6: Using Kinematic Equations for Non-Constant Acceleration

The standard kinematic equations (v = u + at$$s = ut + \frac{1}{2}at^2Etc.) assume constant Acceleration. For variable acceleration, you must use calculus:

$a = \frac{dv}{dt}, \quad v = \frac{ds}{dt}, \quad a = v\frac{dv}{ds}$

Pitfall 7: Incorrect Free Body Diagrams for Connected Bodies

When drawing free body diagrams for connected bodies (e.g., two masses on a pulley), each body gets Its own diagram. The tension in the string appears on both diagrams, and the acceleration is the Same for both bodies (assuming an inextensible string). Never combine forces from different bodies On one diagram.

Pitfall 8: Forgetting That Normal Force Is Not Always $mg$

On an incline: $N = mg\cos\theta$. In an accelerating lift: $N = m(g \pm a)$. The normal force Adjusts to prevent motion perpendicular to the surface, and is only equal to $mg$ on a horizontal Surface with no vertical acceleration.

Additional Practice Problems

21. A 0.5 \mathrm{ kg ball is thrown horizontally from a cliff 40 \mathrm{ m high at 15 \mathrm{ m/s. Calculate (a) the time to hit the ground, (b) the vertical velocity at impact, (c) the total speed at impact, and (d) the horizontal distance from the base of the cliff.

22. A car of mass 1200 \mathrm{ kg travels at 20 \mathrm{ m/s on a level road. The engine provides a driving force of 3000 \mathrm{ N and the total resistive force is 800 \mathrm{ N. Calculate the acceleration, the distance to reach 30 \mathrm{ m/sAnd the power developed at 30 \mathrm{ m/s.

23. A 60 \mathrm{ kg person stands on a spring scale in a lift. Calculate the scale reading when the lift (a) accelerates upward at 2 \mathrm{ m/s^2(b) travels at constant velocity, (c) decelerates at 3 \mathrm{ m/s^2 while moving upward, and (d) is in free fall.

24. Two particles of masses 0.3 \mathrm{ kg and 0.5 \mathrm{ kg approach each other with speeds 4 \mathrm{ m/s and 2 \mathrm{ m/s respectively. After a perfectly elastic head-on collision, calculate the velocities of both particles.

25. Derive an expression for the period of a simple pendulum using energy considerations. A pendulum of length 1.5 \mathrm{ m is displaced $10^\circ$ from the vertical. Calculate the maximum speed of the bob and the period of oscillation.