Newton's Second Law — Question 5

Question 5

A force can be perpendicular to velocity at an instant

Use $\sum\vec F=m\vec a$ for constant mass in an inertial frame. At one instant a 2 kg particle has velocity $(3,4)$ m/s and net force $(8,-6)$ N.

1. Find acceleration and the instantaneous rate of change of speed.

2. Is the particle unaccelerated?

3. If this force stays constant, find velocity one second later and compare speeds.

Question 5 - Solution

A force can be perpendicular to velocity at an instant

(a) Project acceleration onto velocity.

$$\vec a=(8,-6)/2=(4,-3),\quad\boxed{\dot v=\frac{\vec v\cdot\vec a}{|\vec v|}=\frac{12-12}{5}=0}$$ $\vec a=(4,-3)$ m/s$^2$. Since $\vec v\cdot\vec a=12-12=0$, $d|\vec v|/dt=(\vec v\cdot\vec a)/|\vec v|=0$ at that instant.

(b) Keep the perpendicular component.

$$|\vec a|=\sqrt{4^2+(-3)^2}=5\,\mathrm{m/s^2}\ne0$$ Acceleration has magnitude 5 m/s$^2$ and changes direction of motion at that instant. Zero instantaneous speed derivative does not mean zero vector acceleration.

(c) Do not extend an instantaneous orthogonality claim.

$$\vec v(1)=(3,4)+(4,-3)=(7,1),\quad|\vec v(1)|=\sqrt{50}=7.071\,\mathrm{m/s}$$ One second later $\vec v=(7,1)$ m/s, with speed $\sqrt{50}\approx7.071$ m/s, larger than the initial 5 m/s. A fixed force does not remain perpendicular to the changing velocity.