If you have a point mass moving with some velocity 'v' and then a force is applied perpendicular to 'v', physics/mathematics claims no change in the magnitude of 'v' , just a change in direction. But that doesn't make intuitive sense to me because wouldn't a force cause an instantaneous change in the point mass velocity perpendicular to 'v'?

Therefore, wouldn't there be instantaneous velocity vectors perpendicular to each other and a resultant vector with a greater magnitude than 'v'?

  • $\begingroup$ The force as it turns, decreases the original component of velocity and increases the component the force is acting $\endgroup$ May 9, 2023 at 13:12
  • $\begingroup$ But the force is perpendicular to the original velocity so how can it cause a decrease? $\endgroup$
    – Dubious
    May 9, 2023 at 13:14
  • $\begingroup$ Yeah, velocity in the initial direction would not change. $\endgroup$ May 9, 2023 at 13:15
  • $\begingroup$ thinking in terms of infinistesimal amount of time, for the force at zero and the force at the next instant dt, one increases the component of velocity in the direction of the force, and at time dt, the force has now changed to perfectly cancel out the same addition of force in the original direction $\endgroup$ May 9, 2023 at 13:17
  • $\begingroup$ Ohh so do you mean that force is always perpendicular to direction of object? $\endgroup$ May 9, 2023 at 13:24

1 Answer 1


Since the velocity vector of an object is always tangent to the path it takes by definition, changes in the magnitude of the vector involve acceleration components also tangent to the velocity vector and hence the path.

Corollary to this, acceleration components not tangent to the velocity vector do not affect the magnitude of the velocity vector, but its direction.

Mathematically if we trace an object along its path, and designate $\boldsymbol{\hat t}$ the tangent vector and $\boldsymbol{\hat n}$ the normal vector (perpendicular) then we can always decompose the velocity vector as

$$ \boldsymbol{v} = v\; \boldsymbol{\hat{t}} \tag{1} $$

where the scalar $v$ is the speed (velocity magnitude).

Not the change in velocity is acceleration and it always decomposes along the two directions as

$$ \boldsymbol{a} = \dot{v}\; \boldsymbol{\hat{t}} + \frac{v^2}{\rho} \;\boldsymbol{\hat{n}} \tag{2} $$

where $\dot{v}$ designates change in speed (linear acceleration) and $\rho$ is the radius of curvature. This radius is an indication of how the direction of velocity changes along the path. It is purely geometric quantity because it removes the dependency of speed (and hence time) from the change in direction.

The smaller the radius, the higher the arc of movement the more rapid the change in direction is. If the radius of curvature is infinite then the second term goes to zero and the body is moving is a straight line but changing speed.

  • 1
    $\begingroup$ Hi John - many thanks and it does make sense the way you have answered my question. I think I may have an incorrect perception of F=ma in its non-vector form . When a net force is applied to a point mass, I have always equated this to an acceleration and therefore a rate of change in the magnitude of its velocity. $\endgroup$
    – Dubious
    May 9, 2023 at 14:00
  • $\begingroup$ @Dubious - yes, I forgot to mention this fact, but since mass is scalar, the acceleration vector is always parallel to the force vector. $\endgroup$ May 9, 2023 at 14:27

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