Question

A player caught a cricket ball of mass 200g which came to his hand with a speed of 20m/s. If the ball was stopped in 0.1s, find the force exerted by ball on the hands of player?

Answer 1

Answer:

• Mass = 200 g = 0.2 kg
• Initial velocity = 20 m/s
• Time = 0.1 sec
• The Force Applied = ?

$\displaystyle\underline{\bigstar\:\textsf{According to the given Question :}}$

• Here we shall first find the acceleration of the body

$\displaystyle\sf :\implies Acceleration = \dfrac{Final-Initial}{Time}$

$\displaystyle\sf :\implies Acceleration = \dfrac{0-20}{0.1}$

$\displaystyle\sf :\implies Acceleration = \dfrac{-20}{0.1}$

$\displaystyle\sf :\implies \underline{\boxed{\sf Acceleration = 200 \ m/s^2}}$

• So then we know that as per the second law of motion force is given by the product of mass and acceleration
• F = ma
• SI unit of force is Newton
• SI unit of mass is Kg
• SI unit of Acceleration is m/s²

$\displaystyle\sf \dashrightarrow Force = Mass\times Acceleration$

$\displaystyle\sf \dashrightarrow Force = 0.2 \times 200$

$\displaystyle\sf \dashrightarrow Force = 40 \ N$

$\displaystyle\therefore\:\underline{\textsf{ The Force applied by the player is \textbf{ 40 N }}}$

Answer 2

Answer:40N

a=200m/s^2