Question

In List - I condition on initial velocity $\overrightarrow{u}$, force $\overrightarrow{F}$ and acceleration $\overrightarrow{a}$ of a particle is given.
Resultant motion is described in List-II.
Match List-I with List-II
$\begin{array}{cccc}& \text{List-I}& & \text{List-II}\\ \text{(I)}& \overrightarrow{u}\times \overrightarrow{F}=0\text{and}& \text{(P)}& \text{Path will be}\\ & \overrightarrow{F}=\text{constant}& & \text{circular path}\\ \text{(II)}& \overrightarrow{u}.\overrightarrow{F}=0\text{and}& \text{(Q)}& \text{Speed will increase}\\ & \overrightarrow{F}=\text{constant}& & \\ \text{(III)}& \overrightarrow{v}.\overrightarrow{F}=0\text{all the time}& \text{(R)}& \text{Path will be}\\ & \text{and}|\overrightarrow{F}|=\text{constant}& & \text{straight line}\\ & \text{and the particle}& & \\ & \text{always remains in}& & \\ & \text{one plane}\left(\overrightarrow{v}\text{is instantaneous velocity}\right)& & \\ \text{(IV)}& \overrightarrow{u}=2\hat{i}-3\hat{j}\text{and}& \text{(S)}& \text{Path will be}\\ & \text{acceleration at all time}& & \text{parabolic}\\ & \overrightarrow{a}=6\hat{i}-9\hat{j}& & \end{array}$

• A. $I\to R;II\to Q,S;III\to P;IV\to Q,R$
• B. $I\to Q;II\to S;III\to P,R;IV\to Q,R$
• C. $I\to R;II\to Q;III\to R;IV\to Q,R$
• D. $I\to R,S;II\to Q,S;III\to P;IV\to Q,R$

Answer 1

The correct option is A $I\to R;II\to Q,S;III\to P;IV\to Q,R$

(I) $\overrightarrow{F}=\text{constant}$ and $\overrightarrow{u}\times \overrightarrow{F}=0$
Therefore, initial velocity is either in direction of constant force or opposite to it. Hence, the particle will move in straight line and speed may increase or decrease.

(II) $\overrightarrow{u}.\overrightarrow{F}=0$ and $\overrightarrow{F}=\text{constant}$
Initial velocity is perpendicular to constant force, hence the path will be parabolic with speed of particle increasing.

(III) $\overrightarrow{u}.\overrightarrow{F}=0$ means instantaneous velocity is always perpendicular to force. Hence, the speed will
remain constant. And also $|\overrightarrow{F}|=\text{constant}$. Since, the particle moves in one plane, the resulting
motion has to be circular.

(IV) $\overrightarrow{u}=2\hat{i}-3\hat{j}$ and $\overrightarrow{a}=6\hat{i}-9\hat{j}$. Hence, initial velocity is in same direction of constant acceleration, therefore particle moves in straight line with increasing speed.