Question

A particle is acted upon by a force of constant magnitude which is always perpendicular to the velocity of the particle, the motion of the particle takes place in a plane. It follows that:

• A. Its velocity is constant
• B. Its acceleration is constant
• C. Its kinetic energy is constant
• D. It moves in a straight line

Answer 1

Answer: (A)

Its velocity is constant

Let the F be the force, Mass of the particle =m , velocity= \vec{F}.d\vec{r}$

We know that :

$dW=\overrightarrow{F}.d\overrightarrow{r}$

$⟹\frac{dW}{dt}=\overrightarrow{F}.\frac{d\overrightarrow{r}}{dt}=\overrightarrow{F}.\overrightarrow{v}$

$⟹W={\int }_{t_1}^{t_2}\overrightarrow{F}.\overrightarrow{v}dt\dots \left(1\right)$

Again we have

$\overrightarrow{F}=m\overrightarrow{a}=m\frac{d\overrightarrow{v}}{dt}$

$⟹W={\int }_{t_1}^{t_2}m\frac{d\overrightarrow{v}}{dt}.\overrightarrow{v}dt$

$W={\int }_{v_1}^{v_2}m\overrightarrow{v}dv$

Form both equations we get

${\int }_{t_1}^{t_2}\overrightarrow{F}.\overrightarrow{v}dt={\int }_{v_1}^{v_2}m\overrightarrow{v}d\overrightarrow{v}$

Since F is perpendicular to v which is given $\overrightarrow{F}.\overrightarrow{v}=0$

${\int }_{v_1}^{v_2}m\overrightarrow{v}d\overrightarrow{v}=0$

Change in kinetic energy is zero

This means that KInetic energy remains constant.