Question

A player caught a cricket ball of mass 200 g which was moving at velocity of 20 ms-1. If catching process was completed in 0.1 s, the force exerted by the ball on the hands of player is
Options:
20 N
40 N
30 N
8ON​

Answer 1

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

Answer 2

Answer:

Given :-

• Mass (M) = 200 g
• Initial velocity (U) = 20 m/s
• Time taken (T) = 0.1 sec

To Find :-

Force exerted by ball

Solution :-

As we know that

$\huge \bf \: F = M \huge( \frac{V - U}{T} )$

Here,

• F is force
• M is mass
• U is initial velocity
• V is final velocity
• T is time

200 g = 200/1000=0.2 kg

$\sf \: f = 0.2 ( \dfrac{0 - 20}{0.1} )$

$\sf \: f = 0.2( \dfrac{ - 20}{0.1} )$

$\sf \: f = 1 \times - 20$

$\huge \mathfrak \green{ F = 20 \: newton}$

Hence :-

Option A is correct