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

• A. 15 N
• B. 30 N
• C. 150 N
• D. 75 N

Answer 1

Answer: (B)

Given, mass of the ball, $m=3\,kg$, $\Delta V=5\, m/s$
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)
$=\frac{15}{5}\times10=30\,N$