Dynamics

We start from the statement of the second law — force is proportional to the rate of change of momentum — and reach the familiar working formula F = ma.

Step 1 — write the law as an equation. "Proportional to" means "equal to, apart from a constant". Call that constant k:

Step 2 — put in the definition of momentum, p = mv:

Step 3 — take the mass outside the differentiation. This step needs the mass to be constant; a constant multiplier can always be moved outside a derivative:

Step 4 — recognise the derivative. By definition dv/dt is the acceleration a:

Step 5 — fix the constant k by choosing the unit of force. We define one newton as the force that gives a mass of 1 kg an acceleration of 1 m s⁻². Putting F = 1, m = 1, a = 1 into Step 4:

Step 6 — substitute k = 1:

Result. \vec{F} = m\vec{a}, with the acceleration in the same direction as the force.

What it means. The same force gives a small mass a big acceleration and a big mass a small one — mass is the measure of inertia.

Condition used. Step 3 assumed the mass does not change. For a rocket burning fuel or a raindrop collecting water, go back to F = dp/dt and differentiate the product properly. The law also holds only in an inertial (non-accelerating) frame.

We start from Newton's second law in its momentum form and reach the statement that the impulse of a force equals the change in momentum it produces.

Step 1 — the second law. Force is the rate of change of momentum:

Step 2 — multiply both sides by dt. This just moves dt to the other side, so that we have "force times a tiny time" on the left:

Step 3 — add up all the tiny bits. The contact lasts from t_1 to t_2, during which the momentum goes from \vec{p}_1 to \vec{p}_2. Adding up tiny pieces is what integration does, so we integrate both sides between matching limits:

Step 4 — do the right-hand integral. Integrating d\vec{p} simply gives \vec{p}, so we evaluate it at the two limits:

Step 5 — name the left-hand side. By definition, that integral is the impulse \vec{J}:

Step 6 — put in p = mv. With initial velocity \vec{u} and final velocity \vec{v}:

Step 7 — the constant-force version. If we replace the real changing force by its average value F_{av} acting for the whole time \Delta t = t_2 - t_1, the integral becomes a simple product:

Result.

What it means. The area under a force–time graph is the momentum handed over. For a fixed change of momentum, a longer contact time means a smaller force — which is the whole idea behind airbags, crumple zones and a cricketer drawing his hands back while catching.

Condition used. The mass is constant, and \vec{F} is the net force on the body. Impulse is a vector, so in a rebound problem you must give u and v opposite signs.

We start from Newton's third law applied to two colliding bodies and reach the statement that their total momentum is the same before and after the collision.

Setting up. Two smooth spheres A and B of masses m_1 and m_2 move along the same straight line with velocities u_1 and u_2, where u_1 > u_2 so that A catches up with B. They touch for a short time t and then separate with velocities v_1 and v_2. No outside force (no friction, no push) acts on the pair.

Step 1 — name the two contact forces. While they touch, B pushes A with a force F_{12} and A pushes B with a force F_{21}.

Step 2 — apply the third law. Action and reaction are equal in size and opposite in direction:

Step 3 — write each force as a rate of change of momentum. By the second law, the force on a body equals its change in momentum divided by the time. For A the momentum changes from m_1u_1 to m_1v_1:

and for B, from m_2u_2 to m_2v_2:

Step 4 — substitute both into Step 2:

Step 5 — multiply both sides by t. The contact time is the same for both bodies — they touch and separate together — so it cancels:

Step 6 — open the bracket on the right:

Step 7 — collect the "before" terms on one side and the "after" terms on the other. Add m_1u_1 to both sides and add m_2v_2 to both sides:

Result.

What it means. Whatever violence happens during the contact — however the bodies squash, heat up or break — the total momentum of the pair is unchanged. Momentum is only passed from one body to the other.

Conditions used. (i) No net external force acts on the pair during the collision; this is why the third-law pair could cancel. (ii) The contact time is the same for both bodies. (iii) Momentum is a vector: in one dimension you must give leftward velocities a minus sign, and in two dimensions you conserve the x and y components separately. Note that kinetic energy is not assumed to be conserved — this proof works for inelastic collisions too.

We start from F = ma written separately for each hanging mass, and reach formulas for the acceleration of the system and the tension in the string.

Setting up. Two masses m_1 and m_2, with m_1 > m_2, hang from the two ends of a light inextensible string passing over a smooth pulley. Two facts follow from those words, and they are worth marks:

• Inextensible string: the string cannot stretch, so when m_1 goes down by 1 cm, m_2 goes up by 1 cm. Both masses therefore have the same acceleration a.
• Light string over a smooth pulley: no force is needed to accelerate the string itself and none is lost to friction, so the tension T is the same everywhere in the string.

Since m_1 > m_2, common sense says m_1 goes down and m_2 goes up.

Step 1 — free-body equation for m_1. Two forces act on it: its weight m_1g downwards and the tension T upwards. It accelerates downwards, so take downwards as positive for this mass. Net force = mass \times acceleration:

Step 2 — free-body equation for m_2. The same two forces act, but this mass accelerates upwards, so take upwards as positive for it:

Step 3 — add the two equations. We add them because T appears with a minus sign in one and a plus sign in the other, so it will disappear:

Step 4 — cancel T and collect terms. On the left -T + T = 0; on the right take out the common factor a:

Step 5 — take out g on the left:

Step 6 — divide both sides by (m_1 + m_2) to get a alone:

Step 7 — now find T. Go back to Step 2 and make T the subject by adding m_2g to both sides:

Step 8 — substitute the a we just found:

Step 9 — put the bracket over one common denominator (m_1+m_2):

Step 10 — expand the top line:

Step 11 — the two m_2g terms cancel, leaving 2m_1g:

Step 12 — write it tidily:

Result.

What it means. The unbalanced force is only the difference in weights (m_1-m_2)g, but it has to move the total mass (m_1+m_2) — so the machine slows gravity down, which is exactly why Atwood used it to measure g. If m_1 = m_2 then a = 0 and T = m_1g, as expected. Also note m_2g < T < m_1g: the tension always lies between the two weights.

Assumptions used. Light inextensible string, frictionless and massless pulley, no air resistance. If the pulley had mass, part of the torque would go into spinning it and a would be smaller.

We start from a block that is just about to slide down a rough slope, and reach the result \mu_s = \tan\theta — which also proves that the angle of repose and the angle of friction are the same angle.

Setting up. A block of mass m rests on a rough plane whose tilt \theta is slowly increased. At the angle of repose the block is on the very point of sliding: it is still in equilibrium, but the friction has reached its largest possible value, f_{limiting} = \mu_s N.

Choose axes along the slope and perpendicular to the slope — not horizontal and vertical. Then the only force needing to be resolved is the weight mg, which acts vertically downwards.

Step 1 — resolve the weight. The angle between the weight and the perpendicular to the slope is also \theta (simple geometry of the tilted plane). So the weight splits into a part pressing into the slope and a part pulling down the slope:

Step 2 — balance perpendicular to the slope. The block does not sink into the plane or lift off it, so the normal reaction N exactly balances the pressing component:

Step 3 — balance along the slope. The block is not yet moving, so the friction up the slope exactly balances the component pulling it down:

Step 4 — divide Step 3 by Step 2. Dividing one true equation by another is allowed as long as we are not dividing by zero, and N \ne 0 here:

Step 5 — cancel mg, which appears on top and bottom:

Step 6 — use \sin\theta/\cos\theta = \tan\theta:

Step 7 — identify the left-hand side. By definition, limiting friction divided by normal reaction is the coefficient of static friction:

Step 8 — compare with the angle of friction. The angle of friction \lambda is defined by \mu_s = \tan\lambda. Two tangents are equal, so for angles between 0^{\circ} and 90^{\circ} the angles themselves are equal:

Result.

What it means. mg cancelled in Step 5, so the angle of repose does not depend on the mass of the block — only on what the two surfaces are made of. This is why a heap of dry sand always settles to the same slope, whether the heap is small or huge, and it gives a one-line laboratory method for \mu_s: tilt until it slides, then measure \tan\theta = h/b.

Conditions used. The block is on the point of sliding (so friction is at its limiting value — not before, not after), the plane is straight, and no other force such as a push or a string tension acts on the block.