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Home  >>  JEEMAIN and NEET  >>  Physics  >>  Class11  >>  Laws of Motion

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

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

1 Answer

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

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