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)

30 N

Answer 2

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)
$=\frac{15}{5}\times10=30\,N$