Home / IBDP Physics 2025 SL&HL: 2.3 Newton’s second law in terms of momentum Study Notes

IBDP Physics 2025 SL&HL: 2.3 Newton’s second law in terms of momentum Study Notes

Learning Objectives

Students should understand
• Newton’s three laws of motion
• forces as interactions between bodies
• that forces acting on a body can be represented in a free-body diagram
• that free-body diagrams can be analysed to find the resultant force on a system
• the nature and use of the following contact forces

◦ normal force FN is the component of the contact force acting perpendicular to the surface that counteracts the body
◦ surface frictional force Ff acting in a direction parallel to the plane of contact between a body and a surface, on a stationary body as given by Ff ≤ μsFN or a body in motion as given by Ff = μdFN where μs and μd are the coefficients of static and dynamic friction respectively
◦ tension
◦ elastic restoring force FH following Hooke’s law as given by FH = –kx where k is the spring constant
◦ viscous drag force Fd acting on a small sphere opposing its motion through a fluid as given by Fd = 6πηrv where η is the fluid viscosity, r is the radius of the sphere and v is the velocity of the sphere through the fluid
◦ buoyancy Fb acting on a body due to the displacement of the fluid as given by Fb = ρVg where V is the volume of fluid displaced

• the nature and use of the following field forces

◦ gravitational force Fg is the weight of the body and calculated is given by Fg = mg
◦ electric force Fe
◦ magnetic force Fm

• that linear momentum as given by p = mv remains constant unless the system is acted upon by a resultant external force
• that a resultant external force applied to a system constitutes an impulse J as given by J = FΔt where F is the average resultant force and Δt is the time of contact
• that the applied external impulse equals the change in momentum of the system
that Newton’s second law in the form F = ma assumes mass is constant where as F = Δp Δt allows for situations where mass is changing
• the elastic and inelastic collisions of two bodies
• explosions
• energy considerations in elastic collisions, inelastic collisions, and explosions
• that bodies moving along a circular trajectory at a constant speed experience an acceleration that is directed radially towards the centre of the circle—known as a centripetal acceleration as given by
       \(a=\frac{v^2}{r}=\omega^2r=\frac{4\pi^2r}{T^2}\)
• that circular motion is caused by a centripetal force acting perpendicular to the velocity
• that a centripetal force causes the body to change direction even if its magnitude of velocity may remain constant
• that the motion along a circular trajectory can be described in terms of the angular velocity ω which is related to the linear speed v by the equation as given by

    \(v=\frac{2\pi r}{T}=\omega r\)

NETON’S SECOND LAW OF MOTION

The rate of change of momentum of a body is directly proportional to the external force applied on it and the change takes place in the direction of force applied.
i.e.,   
This is the equation of motion of constant mass system. For variable mass system such as rocket propulsion
And,
The SI unit of force is newton. (One newton force is that much force which produces an acceleration of 1ms–2 in a body of mass 1 kg.
The CGS unit of force is dyne. (1N = 105 dyne)
The gravitational unit of force is kg-wt (kg-f) or g-wt (g-f)
1 kg-wt (kg-f) = 9.8 N, 1 g-wt (g-f) = 980 dyne

LAW OF CONSERVATION OF LINEAR MOMENTUM

A system is said to be isolated, when no external force acts on it. For such isolated system, the linear momentum is constant i.e., conserved.
The linear momentum is defined as
 …..(1)
where is the velocity of the body, whose mass is m. The direction of is same as the direction of the velocity of the body. It is a vector quantity.  From Newton’s second law,
 …..(2)
i.e., time rate of change in momentum of the body is equal to total external force applied on the body.
If or = constant   …..(3)
This is called law of conservation of  momentum.
Now let us consider a rigid body consisting of a large number of particles moving with different velocities, then total linear momentum of the rigid body is equal to the summation of individual linear momentum of all particles
i.e.,
or  
where are individual linear momentum of first, second and nth particle respectively.
If this rigid body is isolated i.e., no external force is applied on it, then constant (from Newton’s second law).
Further we know that internal forces (such as intermolecular forces etc.) also act inside the body, but these can only change individual linear momentum of the particles
(i.e., p1, p2………), but their total momentum remains constant.
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