Laws Of Motion Question 243
Question: 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]
Options:
A) 12 m
B) 20 m
C) 8 m
D) 48 m
Show Answer
Answer:
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 therefore $ 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 $
BETA
