Question

A particle of mass 1 kg has a velocity of 2 m/s. A constant force of 2 N acts on the particle for 1 s in a direction perpendicular to its initial velocity. Find the velocity and displacement of the particle at the end of 1 second.

Answer 1

Let the initial speed of the particle be in x direction.

$\therefore$ $u_x=2\hat{i}$ $m/s$

Given : $M=1kg$ $t=1$ s $F_y=2\hat{j}$ N

x direction : $a_x=0$ $⟹$ $V_x=u_x=2\hat{i}$ $m/s$

Also $S_x=u_{x}t=2\times 1=2\hat{i}$ m

y direction : $a_y=\frac{F_y}{M}=2\hat{j}$ $m/s^2$

$\therefore$ $V_y=u_y+a_{y}t=0+2\times 1=2\hat{j}$ $m/s$

Also $S_y=u_{y}t+\frac{1}{2}{a}_y{t}^2=0+\frac{1}{2}\times 2\times 1^2=1\hat{i}$ m

Total displacement of the particle $S=2\hat{i}+1\hat{j}$ m

$⟹$ $|S|=\sqrt{2^2+1^2}=\sqrt{5}$ m

Velocity of the particle $V=2\hat{i}+2\hat{j}$

$⟹$ $|V|=\sqrt{2^2+2^2}=2\sqrt{2}$ $m/s$