Question:

A man throws a ball of mass $ 3.0\, kg $ with a speed of $ 5.0 \,m/s $ . His hand is in contact with the ball for $ 0.2 \,s $ . If he throws $ 4 $ balls in $ 2 $ seconds, the average force exerted by him in $ 1 $ second is

Updated On: Jun 14, 2022
  • 15 N
  • 30 N
  • 150 N
  • 75 N
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The Correct Option is B

Solution and Explanation

Given, mass of the ball, $m=3kg$, $\Delta V=5\, m/s$
and the time taken the man to throw a ball,
$\Delta t=\frac{2}{4}=0.5\,s$
$\therefore$ Change in momentum of the ball, $\Delta=m \cdot \Delta v$
$=3 \times 5 = 15\, N-s$
Since, we know,
Force = Rate of change in linear momentum
$F=\frac{\Delta p}{\Delta t}=\frac{15}{0.5}$ (Putting values)
$=\frac{15}{5}\times10=30\,N$
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