A player catches a ball of mass 150 gm moving at a rate of 20 m/s. If the process of catching is to be completed in 0.1 sec. What is the force exerted by the ball on the hands of the player ?
Answer & explanation
Correct answer: option 2
\(\text{Change in momentum : } \Delta p = F.\Delta t\)
\(F = \frac{\Delta p}{\Delta t}\)
\(F = \frac{m \Delta v}{0.15 \text{ sec}}\)
\(F = \frac{0.15 \text{ kg x} 20 \text{ m/s}}{0.15 \text{ sec}}\)
\(F = 30 \text{ N}\)