question_answer

A body of mass 2 kg has an initial velocity of 3 meters per second along OE and it is subjected to a force of 4 N in a direction perpendicular to OE. The distance of the body from O after 4 seconds will be                         [CPMT 1976]

A)                         12 m               

B)                         20 m            

C)                         8 m          

D)                           48 m

Correct Answer: B

Solution :

                            Displacement of body in 4 sec along OE             \[{{s}_{x}}={{v}_{x}}t=3\times 4=12\ m\]                  Force along OF (perpendicular to OE) = 4 N             \[\therefore \]  \[{{a}_{y}}=\frac{F}{m}=\frac{4}{2}=2\ m/{{s}^{2}}\]             Displacement of body in 4 sec along OF             Þ \[{{s}_{y}}={{u}_{y}}t+\frac{1}{2}{{a}_{y}}{{t}^{2}}\]\[=\frac{1}{2}\times 2\times {{(4)}^{2}}=16\ m\]   [As \[{{u}_{y}}=0\]]                 \[\therefore \]Net displacement \[s=\sqrt{s_{x}^{2}+s_{y}^{2}}\ =\sqrt{{{(12)}^{2}}+{{(16)}^{2}}}=20\ m\]