FSc Part 1 Physics Chapter 2 Complete Guide Book — Force and Motion

Some quantities in physics only need a number to make sense. If I tell you a bag of flour is 5 kg, you do not ask “5 kg in which direction?” The number, with its unit, tells the whole story. Quantities like that are called scalars. Mass, distance, speed, time, energy and temperature all belong here. A distance of 50 m, for instance, only says how long the path was — it never says whether you walked east or west.

Vectors are fussier. Give them a number alone and they are only half-described. A velocity of 50 km h⁻¹ towards the west is a vector; the same 50 km h⁻¹ without a direction is just speed, a scalar. Displacement, velocity, acceleration and force all need both a size (magnitude) and a direction before they are complete. Push a box with 20 N to the right and it moves right; push with the same 20 N to the left and everything changes, even though the number never changed.

That is the test you can use in any question: ask, “Does direction change the meaning here?” If yes, it is a vector. If the answer stays the same whichever way you turn, it is a scalar. And a small habit worth forming now — in print, vectors are written in bold (like V, F) or with an arrow on top (A⃗); in your own handwriting, always put the arrow, or your teacher cannot tell a vector from a plain number.

A vector pointing diagonally can be a nuisance to work with, so physicists split it into two simpler pieces, one along the x-axis and one along the y-axis, at right angles to each other. Those pieces are its rectangular components. Think of walking diagonally across a field: you could equally say you walked so far east and so far north. The diagonal walk and the two straight walks end at the same place.

Take a vector A making an angle θ with the x-axis. Drop perpendiculars to the two axes and you get a right-angled triangle. Its base is the x-component, A_x = A cos θ, and its height is the y-component, A_y = A sin θ. Together they rebuild the original vector: A = A_x + A_y. Notice how cos goes with the axis the angle is measured from, and sin with the other one — mix those up and both components come out wrong.

Going backwards is just as useful. If somebody hands you the two components, the size of the vector comes from Pythagoras: A = √(A_x² + A_y²), and its direction from tan θ = A_y/A_x, so θ = tan⁻¹(A_y/A_x). The book's worked Example 2.1 uses exactly this idea: two equal forces F whose resultant is also F must be 120° apart, because cos θ = −0.5 gives θ = 120°. Work it once slowly and the method sticks.

Multiplying two vectors is not like multiplying two numbers, because there are two different products, and they give two different kinds of answer. The first one, the scalar or dot product, is written A·B and is defined as A·B = AB cos θ, where θ is the angle between the two vectors. The result is an ordinary number — a scalar — which is why it is called the scalar product.

What does it actually mean? Bring both vectors tail-to-tail, then look at how much of B lies along A. That shadow, B cos θ, times A, is the dot product. So A·B = A × (projection of B on A). The classic example is work: a force F dragging a block a distance d at an angle θ does work W = Fd cos θ, which is just F·d. Only the part of the force along the motion earns any work; the part at right angles earns none.

Five properties are worth memorising. The dot product is commutative (A·B = B·A); it is zero for perpendicular vectors (cos 90° = 0); it equals AB for parallel vectors and −AB for antiparallel ones; a vector dotted with itself gives its magnitude squared (A·A = A²); and in components, A·B = A_xB_x + A_yB_y + A_zB_z. That last form also gives the angle between any two vectors: cos θ = (A_xB_x + A_yB_y + A_zB_z)/AB.

The second product goes the other way: multiply two vectors and you get another vector. The vector or cross product is written A × B and defined as A × B = AB sin θ n̂, where n̂ (n-hat) is a unit vector perpendicular to the plane containing A and B. Its size is AB sin θ; its direction is the tricky part, and that is what the right-hand rule is for.

Put the tails of A and B together. Curl the fingers of your right hand from the first vector towards the second, through the smaller angle, and your erect thumb points along A × B. Turn the order round and the thumb flips: B × A = −(A × B). That sign flip is the whole reason the cross product is non-commutative — order matters here, unlike the dot product. It is also why the product of parallel vectors is zero (sin 0° = 0) and of perpendicular vectors is maximum (sin 90° = 1).

Two examples from the book show why this matters. Torque, the turning effect of a force, is τ = r × F — the position vector crossed with the force. And the magnetic force on a moving charge is F = q(v × B). In both cases the answer has to point out of the plane, so an ordinary number could never describe it. One more neat fact: the magnitude |A × B| equals the area of the parallelogram formed by A and B as adjacent sides.

Three equations carry almost the whole chapter, and they only work under one condition: the acceleration must stay constant, and the motion must be along a straight line. They pair the quantities velocity, displacement and time in three ways — velocity–time, position–time, and velocity–position — which is why they are called the first, second and third equations in that order.

The first equation comes straight from the definition of acceleration. If velocity grows from v_i to v_f in time t, then a = (v_f − v_i)/t, and rearranging gives v_f = v_i + at. The second equation uses average velocity: displacement is average velocity × time, so S = ((v_i + v_f)/2) × t, and putting v_f = v_i + at into that gives S = v_it + ½at². The third equation eliminates time altogether: from the same average-velocity step, 2S = (v_i + v_f)t, and t = (v_f − v_i)/a, so 2aS = v_f² − v_i².

Choosing the right one is half the exam. Time is given or asked? Start with the first or second. Time is nowhere in the question, like a car stopping over a known distance? Reach for the third. And remember the sign rule the book repeats: take the initial direction as positive, and anything opposite to it — a velocity backwards, a deceleration — gets a negative sign.

The same three equations can be read off a velocity–time graph, and seeing them there makes them much harder to forget. On such a graph, the slope of the line is the acceleration, and the area under the line is the displacement. For the first equation, the graph rises from v_i to v_f; the rise over the run is a, so the extra velocity is at, giving v_f = v_i + at. For the second, the area under the line is a rectangle plus a triangle: v_it + ½ × t × at, which is v_it + ½at². The third follows the same area, this time as a trapezium: S = ½(v_i + v_f) × t, with t replaced from the first equation.

Now the book's three worked examples, because the method matters more than the answers. Example 2.2: a car at 10 m s⁻¹ accelerates at 2 m s⁻² for 5 s. First equation: v_f = 10 + 2 × 5 = 20 m s⁻¹. Example 2.3: a car starts at 15 m s⁻¹, accelerates at 2 m s⁻² for 4 s. Second equation: S = 15 × 4 + ½ × 2 × 16 = 60 + 16 = 76 m. Example 2.4: a car starts from rest, reaches 300 km h⁻¹ over 0.45 km. First convert: 300 km h⁻¹ = 300 × 1000 / 3600 = 83.33 m s⁻¹, and 0.45 km = 450 m. Third equation: a = (83.33² − 0) / (2 × 450) = 6943.88 / 900 = 7.72 m s⁻². Note the unit conversion first — most lost marks in this chapter come from skipping it, not from the formula.

A body falling freely is the most familiar example of uniformly accelerated motion, and Galileo's point still surprises students: in a vacuum, a heavy iron ball and a light feather, dropped from the same height, hit the ground together. The acceleration due to gravity, g, is about 9.8 m s⁻² downwards at the Earth's surface, and it does not depend on the mass at all. Air resistance is what makes the feather flutter in real life — take the air away and the difference vanishes.

The equations of motion still apply; you only replace a by g and displacement S by height h: v_f = v_i + gt, h = v_it + ½gt², and v_f² − v_i² = 2gh. The sign rule needs care here. For a falling body, g is taken positive when the initial velocity is zero. For a body thrown upwards, g is negative, because gravity now opposes the motion. Getting that sign right is the whole question in half the free-fall problems.

Example 2.5 ties it together. A 1 kg iron ball is dropped from a tower and reaches the ground in 3.34 s. (a) Its striking velocity: v_f = 0 + 9.8 × 3.34 = 32.7 m s⁻¹. (b) The tower's height, from the third equation: (32.7)² − 0 = 2 × 9.8 × h, so h = 1069.29 / 19.6 = 54.56 m. Notice the mass, 1 kg, was never used — a good check that you understood Galileo's point, not just the arithmetic.

Until now everything moved along one straight line. A projectile breaks that pattern: a ball thrown from a height travels forward and falls downward at the same time, tracing a curve. The trick the book teaches is to split that curve into two independent motions. Horizontally, nothing pushes the ball — ignore air friction and there is no horizontal force, so by Newton's first law there is no horizontal acceleration, and the horizontal velocity stays exactly what it started as. Vertically, gravity pulls all the time, so the vertical motion is just free fall.

For a ball thrown horizontally with velocity v_x, the horizontal distance is simply x = v_xt, and the vertical drop, starting from zero vertical velocity, is y = ½gt². Put the two together and the path is a parabola — the same shape whether you throw hard or gently, only stretched further. For a ball thrown at an angle θ with speed v_i, resolve first: v_ix = v_icos θ stays constant, while v_iy = v_isin θ is the part gravity fights. At any instant, v_x = v_ix, and v_y = v_isin θ − gt, so the speed is v = √(v_x² + v_y²) at an angle tan φ = v_y/v_x to the horizontal.

📌 Interesting information (book): a photograph in the book shows two balls released together from the same level — one dropped straight down, one projected horizontally. Both reach the ground at the same instant, because their vertical motions are identical. Horizontal motion never speeds up or slows down the fall. If that feels wrong, it is exactly the misconception the photograph is there to break.

Three questions come up for every projectile: how high does it go, how long does it stay up, and how far does it land? All three fall out of the same splitting trick. At the top of the flight the vertical velocity is momentarily zero, so from 2aS = v_f² − v_i² with a = −g, the maximum height is h = v_i²sin²θ / 2g. For the time of flight, the body returns to the same level, so total vertical displacement is zero: 0 = v_isin θ × t − ½gt², giving t = 2v_isin θ / g. And the range is just horizontal speed × time: R = v_icos θ × t, which simplifies to R = (v_i²/g) sin 2θ.

That last formula carries the chapter's most examined fact. Since sin 2θ is largest, 1, when 2θ = 90°, the range is maximum at θ = 45°. It also explains why complementary angles give the same range: sin 2θ and sin 2(90° − θ) = sin(180° − 2θ) are equal, so a throw at 30° and one at 60° land in the same place — though the 60° throw flies higher and longer. Do not confuse this with maximum height, which keeps growing up to 90°.

Example 2.6 puts numbers on it. A ball is thrown at 30 m s⁻¹ at 30° above the horizontal. Its components are v_ix = 30 × cos 30° = 25.98 m s⁻¹ and v_iy = 30 × sin 30° = 15 m s⁻¹. Time of flight t = 2 × 15 / 9.8 = 3.1 s. Height h = (30² × 0.5²) / (2 × 9.8) = 225 / 19.6 = 11.5 m. Range R = (30² × sin 60°) / 9.8 = (900 × 0.866) / 9.8 = 79.53 m.

📌 Do you know? (book): for an angle below 45°, both the height and the range are smaller; above 45°, the height grows but the range shrinks again. 📌 For your information (book): with air friction, a fast projectile falls short of its ideal parabolic path — the trajectory is skewed, steeper on the way down than on the way up, because air resistance slows the body throughout, and slows it less as it slows down.

Why is a fast cricket ball harder to stop than a slow one, and a heavy truck harder to stop than a bicycle at the same speed? Newton gave the property a name: momentum. Linear momentum is p = mv, mass times velocity. It is a vector, pointing wherever the velocity points, and its SI unit is kg m s⁻¹, which is the same as newton-second (N s). Faster or heavier — either way the momentum is bigger, and the stop is harder.

Newton's second law can be written in momentum form, and the book shows why that form is more general. A force F acting for time t changes the velocity from v_i to v_f, so a = (v_f − v_i)/t and F = ma give F × t = mv_f − mv_i. In words: the time rate of change of momentum equals the applied force, F = (mv_f − mv_i)/t. This version still works when the mass itself changes — a rocket burning fuel, for instance — where F = ma alone would mislead you.

When a force acts for a very short, changing time — a bat hitting a ball — it is easier to talk about its total effect, the impulse: Impulse = F × t = mv_f − mv_i, the change in momentum. Example 2.7: a 1500 kg car slows from 20 to 15 m s⁻¹ in 3.0 s. F × 3.0 = 1500 × 15 − 1500 × 20 = −7500, so F = −2500 N = −2.5 kN. The minus sign is the point: the force opposes the motion, so it is a retarding force.

📌 Point to ponder! (book): a kick on a stone hurts more than the same kick on a football — the stone stops in about 1/100 s while the football gives for about 1/10 s, so the same change of momentum is squeezed into a far shorter time and the force is far larger. 📌 Point to ponder! (book): can a moving object experience impulse? Yes — any force acting for any time, however short, changes its momentum, and that change is the impulse. 📌 Do you know? (book): hair acts like a crumple zone on your skull: a 5 N force might fracture a bare skull, but with skin and hair covering it, about 50 N would be needed — the padding stretches the impact time, so the force drops.

An isolated system is one on which no outside force acts — gas molecules sealed in a vessel are the book's example, because nothing outside can push them. Inside such a system, objects can push each other around all they like, but the total momentum cannot change. The proof is short and beautiful. When two balls collide, the first feels force F for time t and the second feels the reaction F′ = −F for the same time. Add their changes of momentum and the forces cancel: the first ball's gain is exactly the second ball's loss, so (m₁v₁ + m₂v₂) = (m₁v₁′ + m₂v₂′). Total before equals total after — that is the law of conservation of momentum: the total linear momentum of an isolated system remains constant.

One warning before you use it: momentum is a vector, so signs are part of the answer, not decoration. Pick a positive direction first and stick to it; a velocity the other way goes in with a minus. Example 2.8 shows why. Two balls, 2.0 kg and 3.0 kg, move towards each other at 6.0 and 4 m s⁻¹. Take the 2 kg ball's direction as positive. Momentum before = 2 × 6 + 3 × (−4) = 12 − 12 = 0. After the collision the bigger ball moves at 3.0 m s⁻¹ — the book's solution takes that as negative, in the 3 kg ball's original sense — so momentum after = 2 × v₁′ + 3 × (−3) = 0, giving v₁′ = 4.5 m s⁻¹. If your sign convention differs, say so in your solution; the physics is the same, only the labels flip.

📌 Do you know? (book): when a moving car stops suddenly, passengers keep moving forward towards the windscreen. Seatbelts change the forces of motion and stop the passengers from moving on, which greatly reduces the chance of injury. 📌 Do you know? (book): a motorcyclist's safety helmet is padded so as to extend the time of any collision, and so cut the force — the same impulse idea, worn on your head.

A collision is called elastic when no kinetic energy is lost in it. That is a strong claim — momentum and total energy are conserved in every collision, but kinetic energy is only conserved in elastic ones. Drop a tennis ball and it does not bounce back to its starting height, because some kinetic energy turns into heat and sound as the ball squashes; that is an inelastic collision. Drop a hard ball on a marble floor and it comes back very nearly to the same height, losing almost nothing: close to elastic, though nothing in real life is perfectly so.

For two smooth balls in one dimension, conservation of momentum gives m₁(v₁ − v₁′) = m₂(v₂′ − v₂), and conservation of kinetic energy gives a matching equation. Divide one by the other and something lovely drops out: v₁ − v₂ = v₂′ − v₁′. In plain language, the magnitude of the relative velocity of approach equals the magnitude of the relative velocity of separation. Solving the two equations together gives the final velocities: v₁′ = ((m₁ − m₂)/(m₁ + m₂))v₁ + (2m₂/(m₁ + m₂))v₂, and v₂′ = (2m₁/(m₁ + m₂))v₁ + ((m₂ − m₁)/(m₁ + m₂))v₂.

The special cases are what examiners actually ask. (i) Equal masses, m₁ = m₂: the balls simply swap velocities, v₁′ = v₂ and v₂′ = v₁. (ii) Equal masses with the target at rest: the moving ball stops dead and the target takes off with the first ball's velocity — the billiard-ball trick you have seen on a pool table. (iii) A light body hitting a massive body at rest: it bounces back with almost the same speed, v₁′ ≈ −v₁, while the massive body barely moves — the squash player's ball off the wall. (iv) A massive body hitting a light body at rest: the massive body carries on almost unchanged, v₁′ ≈ v₁, and the light one flies off at about twice the incident speed, v₂′ ≈ 2v₁.

Example 2.9: a 70 g ball at 9 m s⁻¹ hits a 140 g ball at rest, perfectly elastically. v₁′ = ((70 − 140)/(70 + 140)) × 9 = (−70/210) × 9 = −3 m s⁻¹ — the minus means it rebounds backwards. v₂′ = (2 × 70/(70 + 140)) × 9 = (140/210) × 9 = 6 m s⁻¹ forwards. Grams cancel against grams, so no conversion was needed here, but check that in every other numerical.

Most real collisions are inelastic: kinetic energy is lost, even though momentum is still conserved. The extreme case, a perfectly inelastic collision, is when the two bodies stick together afterwards and move off with one common velocity. In one dimension that velocity is easy: total momentum before, m₁v₁ + m₂v₂, must equal (m₁ + m₂)v_f after, so v_f = (m₁v₁ + m₂v₂)/(m₁ + m₂). If the target is at rest, that becomes v_f = (m₁/(m₁ + m₂))v₁ — the moving body's speed is cut down by exactly the mass ratio.

Two dimensions only add one step: resolve momentum along each axis separately. For an elastic collision in two dimensions, momentum along x gives m₁v₁ + m₂v₂ = m₁v₁′cos θ₁ + m₂v₂′cos θ₂, along y it gives 0 = m₁v₁′sin θ₁ − m₂v₂′sin θ₂, and kinetic energy conservation adds a third equation. For a perfectly inelastic collision in two dimensions, the stuck-together mass M = m₁ + m₂ moves off at an angle φ, with m₁v₁ + m₂v₂cos θ = Mv_fcos φ along x and m₂v₂sin θ = Mv_fsin φ along y. The kinetic energy lost is simply ΔK.E. = (K.E.)_i − (K.E.)_f, where (K.E.)_i = ½m₁v₁² + ½m₂v₂² and (K.E.)_f = ½Mv_f² — and since kinetic energy is a scalar, you never need to split velocities into components for it.

The book's everyday examples make the lost energy concrete. In a karate chop, the hand does not bounce back off the bricks; the energy goes into breaking them, plus heat and sound. In a car crash, the crumpling bodywork absorbs the energy so the passengers do not have to — that deformation is the point, not an accident. And when a bat hits a ball, some energy is lost deforming the ball and some goes into heat and sound, so even a lively-looking hit is, strictly, inelastic.

A rocket is conservation of momentum wearing an engine. It carries its own fuel, burns it, and throws the hot gas out of the back at enormous speed. The gas gains momentum backwards; the rocket must gain an equal momentum forwards, because there is no outside force to change the total. That is the whole secret — a rocket does not push against the air or the ground, which is why it works perfectly well in the vacuum of space, where there is nothing to push against at all.

The book puts numbers to the idea. A typical rocket burns about 10000 kg s⁻¹ of fuel and throws the burnt gases out at over 4000 m s⁻¹. More than 80% of a rocket's launch mass is fuel alone, which is why rockets are built in stages: when one stage has done its job it is dropped, so the remaining stages do not have to carry its dead weight any further. If m is the mass of gas ejected per second at velocity v relative to the rocket, the momentum change per second is mv, and that equals the thrust on the rocket. The rocket's acceleration is therefore a = mv / M, where M is the rocket's mass. And as fuel burns away, M falls, so the acceleration grows even if the engine's output never changes.

One detail from the book's side note is worth keeping: in the combustion chamber, fuel and oxygen mix and burn, the hot gases rush out at very high velocity, and the gain in momentum of the gases equals the gain in momentum of the rocket. Gas and rocket push each other apart in opposite directions — Newton's third law, doing the lifting.