Laws of Motion · Lesson 2 of 14

Momentum and Newton’s Second Law

“Connect force with momentum change and apply the vector equation of motion with consistent units and systems.”

Learning Objectives

• Define linear momentum as a vector. • Derive F = ma from the momentum form for fixed mass. • Apply the law separately to perpendicular components. • Distinguish net force, individual forces and the acceleration they produce.

Mass and velocity both matter

A loaded trolley is harder to bring to the same speed in the same time than an empty one. A faster-moving object of the same mass also requires a greater momentum change to stop. Linear momentum combines these two features: p = mv. It has the direction of velocity, SI unit kg m/s, and dimensions [M L T⁻¹]. A stationary body has zero momentum in that frame, even though it still has mass and inertia.

Linear momentumLaTeX

p = m v

The value depends on the reference frame through velocity. Adding momenta requires a common frame.

A force acts not merely according to how much momentum changes, but how quickly it changes. Bringing the same ball to rest over a longer time requires a smaller average resultant force. A direction change also changes momentum: a stone moving steadily around a circle has constant momentum magnitude but a continuously changing momentum vector.

The same constant force applied for the same duration gives the same momentum change, regardless of mass. If both bodies start from rest, their final momenta are equal while the lighter body has greater final speed. This is another way to see why force is connected to momentum rate rather than speed alone.

The general rate statement and the fixed-mass form

Newton’s second lawLaTeX

\sum F_{ext} = \frac{d p}{d t}

This momentum-rate statement applies to a particle or a fixed collection of particles in an inertial frame. Open systems with mass flowing in or out require momentum-flow accounting.

Constant massLaTeX

\frac{d p}{d t} = \frac{d ( m v )}{d t} = m \frac{d v}{d t} = m a ⟹ \sum F_{ext} = m a

The SI force unit is chosen so the proportionality constant is 1. One newton is 1 kg m/s²; force dimensions are [M L T⁻²].

The left side is the vector sum of real external interactions on the selected body. The quantity ma is the result of that sum, not one more force to add beside weight, tension or friction. For a system of several particles, total external force determines acceleration of its centre of mass. Internal forces cancel in pairs in the system total, though they can accelerate its individual parts.

Resolve forces, then calculate acceleration

Component equationsLaTeX

\sum F_{x} = m a_{x}, \sum F_{y} = m a_{y}, \sum F_{z} = m a_{z}

The mass is the same in each direction. Acceleration points along net force, not necessarily along velocity.

A vertical force changes the vertical velocity component. If the horizontal force is zero, horizontal velocity remains constant. This is why gravity changes the vertical motion of a projectile without slowing its horizontal component in the ideal model. Force can point along velocity, opposite velocity, perpendicular to it, or at another angle. The relative directions determine whether speed, direction or both change.

Two perpendicular forces

Problem
A 5 kg body experiences an 8 N force along +x and a 6 N force along +y. Find acceleration.

1.The resultant force is 8 i-hat + 6 j-hat N and has magnitude √(64 + 36) = 10 N.
2.Acceleration is 1.6 i-hat + 1.2 j-hat m/s², with magnitude 10/5 = 2 m/s².
3.Direction is arctan(6/8) = 36.87° above the 8 N force. Adding 8 and 6 as scalar magnitudes would give an incorrect resultant.

Force from a measured change in speed

Problem
A constant force changes the speed of a 3 kg body from 2.0 to 3.5 m/s in 25 s without changing direction. Find the force.

1.Choose the direction of motion positive. Acceleration is (3.5 − 2.0)/25 = 0.060 m/s².
2.Net force is 3 × 0.060 = 0.180 N in the direction of motion.
3.If other forces are present, 0.180 N is their resultant, not necessarily the magnitude of any one applied force.

Stopping a moving body

Problem
A 20 kg body initially moves at 15 m/s. A constant 50 N force acts opposite its motion until it stops. Find stopping time.

1.Acceleration is −50/20 = −2.5 m/s² when the initial direction is positive.
2.From 0 = 15 − 2.5t, stopping time is 6 s.
3.The phrase “until it stops” ends the model there. If the same fixed-direction force continued after zero velocity, it would accelerate the body in the opposite direction.

Infer force by differentiating position

The coordinate direction matters

Problem
A particle has y = ut + ½gt² with y measured positive downward and g constant. Find net force.

1.Differentiation gives v_y = u + gt and a_y = g.
2.The net force is +mg along the downward y-axis.
3.The positive sign comes from the coordinate convention. It does not mean gravity has reversed direction. Position equations must be interpreted with their stated axes.

Present force sets present acceleration

If the applied force changes, acceleration changes immediately in the ideal model. Velocity generally remains continuous unless an idealised impulsive interaction is introduced. A body’s earlier acceleration affects its present velocity and position, but it does not create an additional force that it carries with it. This distinction becomes especially useful when an object leaves an accelerating vehicle or a string is cut.

Quick check

Which statement correctly describes F in F = ma?

Key Takeaways

• Momentum is mass times velocity and is a vector. • For constant mass, net external force equals ma. • The force unit is newton, equal to kg m/s². • Resolve force components before calculating acceleration. • Acceleration is determined by current net force; ma is not an extra force.

Previous · Lesson 1

Inertia and Newton’s First Law

Next · Lesson 3

Impulse and Collisions