Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
0 votes

A body of mass 2 kg is moving along x-direction with velocity 2m/s. If a force of 4N is acting on it along y direction for 1 second, then what will be the final velocity

\[(a)\;0 \quad (b)\; 8 m/s \quad (c)\;2 m/s \quad(d)\;2 \sqrt 2 m/s \]

Can you answer this question?

1 Answer

0 votes
$F=\large\frac{dp}{dt}=\frac{m(v_2 -v_1)}{t}$
$4 \hat j =2(v_2-2 \hat i)$
$2v_2=4 j+4i$
$|v_2|=|2 j+2 i|$
$\qquad=2 \sqrt 2$
answered Jul 8, 2013 by meena.p
edited May 27, 2014 by lmohan717

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App