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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Laws of Motion
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A circket ball of mass 150 g is moving with a velocity of 12 m/s and is hit by a bat so that the ball is turned back with a velocity of $20 m/s$ .Force on the ball acts for 0.01s, the average force exerted by the bat on the ball

\[(a)\;48 \;N\quad (b)\;480\;N \quad (c)\;40\;N\quad (d)\;400\;N\]
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1 Answer

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$Force=\large\frac{Finace\; momentum-Initial\; momentum}{time}$
$\qquad=\large\frac{\Large\frac{150}{1000} \times 12 -\Large\frac{150}{1000} (-20)}{0.01}$
$\qquad=\large\frac{4.8}{.01}$
$\qquad=480 \;N$
answered Jul 12, 2013 by meena.p
edited May 27, 2014 by lmohan717
 

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