Question

A player caught a cricket ball of mass 150gm moving at a rate of 20m/s. If the catching process be completed in 0.1s, the force of the blow exerted by the ball on the hands of the player is

Answer 1

u=20m/s
v=0
Mass=0.15kg
t=1s
a=(v-u)/t
=(0-20)/0.1
= -20/0.1= -200m/s²(- sign denotes negative acceleration)
F=ma
=0.15x200
=30newton

Answer 2

Given :

Mass of the cricket ball = 200 g

Velocity of the cricket ball = 20 m/s

Time = 0.1 s

To Find :

The force exerted by the ball on the hands of player

Solution :

The relation between Force , Mas and acceleration is given by ,

$\\ \star \: {\boxed{\purple{\sf{F = ma}}}}$

• Since a = $\sf{\dfrac{v-u}{t}}$

$\\ : \implies \sf \:F = m \bigg( \dfrac{v - u}{t} \bigg)$

Where ,

F is force

m is mass

u is initial velocity

v is final velocity

We have ,

m = 200 g = 200/1000 = 0.2 kg

u = 20 m/s

v = 0 [since it came to rest]

t = 0.1 sec

Substituting the values ,

$\\ : \implies \sf \: F = 0.2 \bigg( \dfrac{0 - 20}{0.1} \bigg)$

$\\ : \implies \sf \:F = 0.2\bigg(\dfrac{-20}{0.1}\bigg)$

$\\ : \implies{\underline{\boxed{\pink{\mathfrak{F = - 40 \: N }}}}} \: \bigstar$

Hence ,

The Force exerted by the ball on players hands is 40 N