Newton's laws and momentum

AS · 12 min

Kinematics describes motion; dynamics explains it. Newton's three laws connect force to the change in motion it causes, and momentum is the quantity that makes the connection precise. This note covers mass and weight, $F = ma$ , momentum, force as rate of change of momentum, and how to state and apply all three laws. Definitions here are examined word for word in Paper 2, and $F = ma$ problems with lifts, slopes and connected objects appear every session.

Mass and inertia

Definition

Mass is the property of an object that resists change in its motion.

A loaded supermarket trolley is harder to start moving and harder to stop than an empty one, because it has more mass. This resistance to change in motion is called inertia. Mass is a scalar, measured in kilograms, and it is the same wherever the object is.

Newton's first law

Definition

Newton's first law: an object remains at rest or continues to move with constant velocity unless acted on by a resultant (external) force.

"Constant velocity" means constant speed and constant direction. So the first law says that a resultant force is needed to change speed or direction, not to keep something moving. A puck sliding on ice slows down only because of friction; with no resultant force it would continue forever.

The converse is equally useful: if an object is at rest or moving at constant velocity, the resultant force on it is zero. A car cruising at a steady $30\ \text{m s}^{-1}$ on a straight road has its driving force exactly balanced by air resistance and friction.

Momentum

Definition

Linear momentum is the product of mass and velocity.

Key result

p = mv

Unit: $\text{kg m s}^{-1}$ , which is the same as $\text{N s}$ . Momentum is a vector, in the direction of the velocity.

Newton's second law

Definition

Force is the rate of change of momentum.

Newton's second law: the resultant force acting on an object is equal to the rate of change of its momentum, and the change in momentum takes place in the direction of the resultant force.

Key result

F = \frac{\Delta p}{\Delta t}

For an object of constant mass, $\Delta p = m\Delta v$ , so

F = m\frac{\Delta v}{\Delta t} = ma

The acceleration is always in the same direction as the resultant force.

This is also how the newton is defined: $1\ \text{N}$ is the resultant force that gives a mass of $1\ \text{kg}$ an acceleration of $1\ \text{m s}^{-2}$ . So $1\ \text{N} = 1\ \text{kg m s}^{-2}$ .

$F = \Delta p/\Delta t$ is more general than $F = ma$ . It also handles situations where mass is changing, such as a jet of water, a rocket expelling gas, or sand falling onto a conveyor belt.

Watch out

In $F = ma$ , $F$ is the resultant force: the vector sum of all the forces on the object. If a $3000\ \text{N}$ driving force acts against $600\ \text{N}$ of resistance, use $2400\ \text{N}$ , not $3000\ \text{N}$ . Draw a free-body diagram first and add the forces with signs.

Resultant force and acceleration

A car of mass $1200\ \text{kg}$ has a driving force of $3000\ \text{N}$ and experiences a total resistive force of $600\ \text{N}$ . Calculate its acceleration.

Solution

F = 3000 - 600 = 2400\ \text{N} \text{ forwards}

a = \frac{F}{m} = \frac{2400}{1200} = 2.0\ \text{m s}^{-2} \text{ forwards}

Force from a change in momentum

A ball of mass $0.15\ \text{kg}$ travelling at $20\ \text{m s}^{-1}$ is hit straight back along its path at $30\ \text{m s}^{-1}$ . The bat is in contact with the ball for $0.010\ \text{s}$ . Calculate the average force exerted on the ball.

Solution

Take the ball's final direction as positive. Initial momentum $= 0.15 \times (-20) = -3.0\ \text{N s}$ ; final momentum $= 0.15 \times 30 = 4.5\ \text{N s}$ .

\Delta p = 4.5 - (-3.0) = 7.5\ \text{N s}

F = \frac{\Delta p}{\Delta t} = \frac{7.5}{0.010} = 750\ \text{N}

in the direction of the ball's final motion.

Force–time graphs

Rearranging $F = \Delta p/\Delta t$ gives $\Delta p = F\Delta t$ . For a force that varies, the change in momentum is the area under the force–time graph. (The product $F\Delta t$ is sometimes called the impulse; the name is not required, but the idea is useful.)

(0, 0) -- (0.025, 400) (0.025, 400) -- (0.05, 0) fill 0 0.025 y = 16000 x fill 0.025 0.05 y = 400 - 16000 (x - 0.025)

Kicking a ball

The graph shows how the force on a ball of mass $0.20\ \text{kg}$ varies during a kick. The ball is initially at rest. Find its speed as it leaves the foot.

Solution

Area under the graph (a triangle):

\Delta p = \tfrac{1}{2} \times 0.050 \times 400 = 10\ \text{N s}

v = \frac{\Delta p}{m} = \frac{10}{0.20} = 50\ \text{m s}^{-1}

A changing-mass system

A helicopter hovers by pushing air downwards. Its rotor accelerates $120\ \text{kg}$ of air each second from rest to $8.0\ \text{m s}^{-1}$ downwards. Find the force on the air, and the maximum mass of helicopter this could support.

Solution

Rate of change of momentum of the air:

F = \frac{\Delta(mv)}{\Delta t} = \frac{\Delta m}{\Delta t}\,v = 120 \times 8.0 = 960\ \text{N} \text{ downwards}

By Newton's third law the air pushes up on the rotor with $960\ \text{N}$ . Hovering needs this to equal the weight:

m = \frac{960}{9.81} = 97.9\ \text{kg}

(A small helicopter; real ones move far more air.)

Weight

Definition

Weight is the effect of a gravitational field on a mass: the gravitational force acting on it. The weight of an object is the product of its mass and the acceleration of free fall.

Key result

W = mg

with $g = 9.81\ \text{m s}^{-2}$ near the Earth's surface.

Weight is a force, measured in newtons, and acts vertically downwards through the centre of gravity. Mass is a property of the object; weight depends on where it is. A $70\ \text{kg}$ astronaut weighs about $690\ \text{N}$ on Earth and about $110\ \text{N}$ on the Moon, but has a mass of $70\ \text{kg}$ in both places.

$W = mg$ is $F = ma$ applied to free fall: if weight is the only force, the acceleration is $g$ .

Newton's third law

Definition

Newton's third law: if body A exerts a force on body B, then body B exerts a force on body A that is equal in magnitude and opposite in direction.

The two forces of a third-law pair:

act on different bodies;
are of the same type (both gravitational, both contact, both electrostatic...);
are equal in size and opposite in direction, along the same line;
act for the same length of time.

Because they act on different bodies, a third-law pair can never cancel or be added to find a resultant on one object.

Force on the book (lying on a table)	Third-law partner
the Earth pulls the book down (weight, gravitational)	the book pulls the Earth up (gravitational)
the table pushes the book up (normal contact force)	the book pushes the table down (contact)

Watch out

The weight of the book and the normal contact force on it are equal and opposite, but they are not a third-law pair: they act on the same body and are different types of force. They are equal only because the book is in equilibrium (Newton's first law). If the table were in an accelerating lift, they would differ, while each third-law pair would still be exactly equal.

Applying the laws

Method

Solving a dynamics problem

Draw a free-body diagram for each object separately, showing every force acting on that object (weight, normal contact force, tension, friction, drag, thrust).
Choose a positive direction, usually the direction of acceleration.
For each object write resultant force $= ma$ in that direction.
Solve the equations (simultaneously if there are several objects linked by a string).
Check signs and whether the answer is reasonable.

Person in a lift

A person of mass $60\ \text{kg}$ stands on bathroom scales in a lift. Find the reading on the scales (the normal contact force) when the lift (a) accelerates upwards at $1.5\ \text{m s}^{-2}$ , (b) moves at constant velocity, (c) accelerates downwards at $1.5\ \text{m s}^{-2}$ .

Solution

Forces on the person: weight $mg = 588.6\ \text{N}$ down, normal force $R$ up.

(a) Upwards positive: $R - mg = ma$ , so $R = m(g + a) = 60(9.81 + 1.5) = 679\ \text{N}$ .

(b) $a = 0$ : $R = mg = 589\ \text{N}$ .

(c) Downwards acceleration means $a = -1.5\ \text{m s}^{-2}$ with up positive: $R = m(g - 1.5) = 60 \times 8.31 = 499\ \text{N}$ .

The scales read the normal force, so the person feels heavier when accelerating up and lighter when accelerating down. A lift moving down but slowing has an upward acceleration and gives the larger reading.

Connected objects over a pulley

A trolley of mass $2.0\ \text{kg}$ on a smooth horizontal bench is joined by a light inextensible string, passing over a frictionless pulley at the edge of the bench, to a mass of $0.50\ \text{kg}$ hanging freely. Calculate the acceleration of the system and the tension in the string.

Solution

Both objects have the same size of acceleration $a$ (the string is inextensible) and the same tension $T$ throughout (light string, frictionless pulley).

Trolley (horizontal, towards the pulley): $T = 2.0a$ .

Hanging mass (downwards): $0.50g - T = 0.50a$ .

Add the equations to eliminate $T$ :

0.50 \times 9.81 = 2.50a \quad\Rightarrow\quad a = 1.96\ \text{m s}^{-2}

T = 2.0 \times 1.96 = 3.92\ \text{N}

The tension is less than the hanging weight ( $4.9\ \text{N}$ ), as it must be for the hanging mass to accelerate downwards.

Sliding down a slope

A block of mass $5.0\ \text{kg}$ slides down a slope inclined at $30^\circ$ to the horizontal. A constant frictional force of $12\ \text{N}$ acts up the slope. Find the acceleration of the block.

Solution

Resolve along the slope (down the slope positive). The component of weight down the slope is $mg\sin 30^\circ = 5.0 \times 9.81 \times 0.5 = 24.5\ \text{N}$ .

F = 24.5 - 12 = 12.5\ \text{N}

a = \frac{12.5}{5.0} = 2.5\ \text{m s}^{-2} \text{ down the slope}

The normal contact force and the perpendicular component of weight ( $mg\cos 30^\circ$ ) balance, so they do not appear in the equation along the slope.

Exam tip

Learn the three laws and the definitions of momentum and force exactly. For the second law, "force is proportional to the rate of change of momentum" or "resultant force equals rate of change of momentum" both score; " $F = ma$ " alone does not.
"State Newton's third law" needs: equal magnitude, opposite direction, acting on different bodies. Adding "of the same type" is a good safeguard.
"Explain, using Newton's laws, why..." questions want the law named and applied: for example, "the resultant force is zero (first law), so the velocity is constant".
Most errors in lift and pulley questions come from leaving out a force or using the wrong resultant. Write "resultant force $=$ ..." as a separate line; it is often a mark on its own.

Summary

Mass resists change in motion; weight $W = mg$ is the effect of a gravitational field on a mass.
First law: no resultant force means constant velocity (or rest).
Momentum $p = mv$ , a vector, unit $\text{N s}$ .
Second law: force $=$ rate of change of momentum, $F = \Delta p/\Delta t$ ; for constant mass $F = ma$ , with $a$ in the direction of the resultant force.
Area under a force–time graph $=$ change in momentum.
Third law: equal and opposite forces on different bodies, same type; they never cancel each other.
Always draw a free-body diagram and use the resultant force.

Practice

Question

State Newton's first law and explain why a resultant force is not needed to keep an object moving.
Define linear momentum and state its SI unit in base units.
A $1500\ \text{kg}$ car brakes from $20\ \text{m s}^{-1}$ to rest in $4.0\ \text{s}$ . Calculate the average braking force.
A $0.060\ \text{kg}$ tennis ball moving at $25\ \text{m s}^{-1}$ hits a wall at right angles and rebounds at $20\ \text{m s}^{-1}$ . Contact lasts $4.0\ \text{ms}$ . Calculate the change in momentum and the average force on the wall.
Identify the Newton's third law partner of each force: (a) the weight of an apple hanging on a tree; (b) the push of a swimmer's hands on the water; (c) the tension in a rope pulling a sledge, acting on the sledge.
A $0.40\ \text{kg}$ mass hangs from a string. Find the tension in the string when the mass (a) is at rest, (b) is accelerated upwards at $2.0\ \text{m s}^{-2}$ , (c) is lowered with an acceleration of $2.0\ \text{m s}^{-2}$ downwards.
Water leaves a hose at $15\ \text{m s}^{-1}$ at a rate of $0.80\ \text{kg s}^{-1}$ and hits a wall horizontally, stopping without rebounding. Calculate the force on the wall.
Masses of $3.0\ \text{kg}$ and $2.0\ \text{kg}$ hang on either side of a frictionless pulley, joined by a light inextensible string. Find their acceleration and the tension.
A $70\ \text{kg}$ skier is pulled up a $20^\circ$ slope at constant speed by a tow rope parallel to the slope. Friction is $50\ \text{N}$ . Find the tension in the rope. What would the tension need to be for an acceleration of $0.50\ \text{m s}^{-2}$ up the slope?
A rocket of mass $2.0 \times 10^{4}\ \text{kg}$ on the launch pad ejects gas downwards at $2500\ \text{m s}^{-1}$ relative to the rocket. (a) What rate of gas ejection, in $\text{kg s}^{-1}$ , is needed just to lift off? (b) The rocket ejects gas at $100\ \text{kg s}^{-1}$ . Calculate its initial acceleration.

Answers

An object stays at rest or moves with constant velocity unless a resultant force acts on it. A resultant force changes the velocity; with no resultant force the velocity simply stays the same, so motion continues without any force. (In everyday life friction is a force that has to be balanced.)
Momentum is the product of mass and velocity; unit $\text{kg m s}^{-1}$ .
$F = m\Delta v/\Delta t = 1500 \times 20/4.0 = 7500\ \text{N}$ (opposite to the motion).
$\Delta p = 0.060 \times (20 - (-25)) = 2.7\ \text{N s}$ away from the wall. Force on ball $= 2.7/0.0040 = 675\ \text{N}$ away from the wall; by the third law the force on the wall is $675\ \text{N}$ (about $680\ \text{N}$ ) into the wall.
(a) The apple's gravitational pull on the Earth, upwards. (b) The push of the water on the swimmer's hands (forwards). (c) The pull of the sledge on the rope, backwards.
(a) $T = 0.40 \times 9.81 = 3.92\ \text{N}$ . (b) $T = 0.40(9.81 + 2.0) = 4.72\ \text{N}$ . (c) $T = 0.40(9.81 - 2.0) = 3.12\ \text{N}$ .
$F = (\Delta m/\Delta t)\,v = 0.80 \times 15 = 12\ \text{N}$ .
$(3.0 - 2.0)g = 5.0a$ , so $a = 9.81/5.0 = 1.96\ \text{m s}^{-2}$ . For the $2.0\ \text{kg}$ mass: $T - 2.0g = 2.0a$ , so $T = 2.0(9.81 + 1.96) = 23.5\ \text{N}$ .
Constant speed means zero resultant: $T = mg\sin 20^\circ + 50 = 70(9.81)(0.342) + 50 = 234.9 + 50 = 285\ \text{N}$ . With $a = 0.50$ : $T = 285 + 70 \times 0.50 = 320\ \text{N}$ .
(a) The thrust is the rate of change of momentum of the gas, $(\Delta m/\Delta t) \times 2500$ . Lift-off needs thrust $\ge$ weight $= 2.0 \times 10^{4} \times 9.81 = 1.962 \times 10^{5}\ \text{N}$ , so $\Delta m/\Delta t = 1.962 \times 10^{5}/2500 = 78.5\ \text{kg s}^{-1}$ . (b) Thrust $= 100 \times 2500 = 2.5 \times 10^{5}\ \text{N}$ ; resultant $= 2.5 \times 10^{5} - 1.962 \times 10^{5} = 5.38 \times 10^{4}\ \text{N}$ ; $a = 5.38 \times 10^{4}/2.0 \times 10^{4} = 2.7\ \text{m s}^{-2}$ .

Mass and inertia

Newton's first law

Momentum

Newton's second law

Force–time graphs

Weight

Newton's third law

Applying the laws

Summary

Practice

Console