Question

A fielder takes a catch of a fast-moving ball. The moment the ball touches his hand, its velocity is 20m.s-1. The fielder after holding the ball, pulls his hands in the backward direction, i.e., in the direction of motion of the ball and finally stops it after 0.5 s. In the mass of the ball is 250g, find the retarding force applied by the fielder.

Answer 1

Given:

Initial velocity of ball before touching the hand of fielder,u= 20 m/s

Final velocity of ball,v= 0 m/s

(Since, ball stops finally)

Time taken by fielder to stop the ball,t= 0.5 s

Mass of ball, m= 250g= 0.25 kg

To Find:

Retarding force applied by the fielder

Solution:

We know that,

• Retarding force is equal to actual force in magnitude but have opposite sign with respect to actual force

• According to first equation of motion for constant acceleration,

$\purple{\underline{\boxed{\bf{v=u+at}}}}$

where,

v is final velocity

u is initial velocity

a is acceleration

t is time taken

• $\pink{\underline{\boxed{\bf{Force=Mass\times Acceleration}}}}$

$\rule{190}{1}$

Let the acceleration of ball after reaching to the hands of fielder be a

So, on applying first equation of motion on ball, we get

$\longrightarrow\rm{v=u+at}$

$\longrightarrow\rm{0=20+a(0.5)}$

$\longrightarrow\rm{-20=0.5a}$

$\longrightarrow\rm{0.5a=-20}$

$\longrightarrow\rm{a=\dfrac{-20}{0.5}}$

$\longrightarrow\rm{a=-40\ m/s^{2}}$

Here, negative sign shows that direction of acceleration is opposite to direction of motion

$\rule{190}{1}$

Let the retarding force applied by fielder be F

So,

$\longrightarrow\rm{F=ma}$

$\longrightarrow\rm{F=0.25\times(-40)}$

$\longrightarrow\rm\green{F=-10\ N}$

Hence, the retarding force applied by fielder is -10 N.

Answer 2

$\huge\sf\pink{Answer}$

☞ Force = -10 N

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$\huge\sf\blue{Given}$

✭ Initial Velocity (u) = 20 m/s

✭ Final Velocity (v) = 0 m/s

✭ Time (t) = 0.5 s

✭ Mass (m) = 250 g = 0.25 kg

$\rule{110}1$

$\huge\sf\gray{To \:Find}$

◈ Force Applied?

$\rule{110}1$

$\huge\sf\purple{Steps}$

We know that,

$\underline{\boxed{\sf F = MA}}$

So to complete this formula we need the value of a,so it can be found with the help of,

$\underline{\boxed{\sf v=u+at}}$

Substituting the given values,

➝ $\sf 0 = 20 + a\times 0.5$

➝ $\sf -20 = 0.5a$

➝ $\sf\dfrac{-20}{0.5} = a$

➝ $\sf\red{ a = -40 m/s}$

So on substituting the Value of a in the first formula,

➢ $\sf F = 0.25 \times -40$

➢ $\sf\orange{F = -10 N}$

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