Question

A fielder takes a catch of a fast moving ball. The moment the ball touches his hands, its velocity is 20 m.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. If the mass of the ball is 250 g, find the retarding force applied by the fielder

Answer 1

GiveN :

• Initial velocity (u) = 20 m/s
• Final velocity (v) = 0 m/s
• Time interval (t) = 0.5 s
• Mass of ball (m) = 250 g = 0.25 kg

To FinD :

• Retarding Force applied

Formulae Used :

• v = u + at
• F = ma

where,

• v is final velocity.
• u is initial velocity.
• a is acceleration.
• t is time interval.
• m is mass of ball.
• F is force applied.

SolutioN :

Use 1st equation of motion :

⇒v = u + at

⇒0 = 20 + 0.5 * a

⇒-20 = 0.5a

⇒a = -20/0.5

⇒a = - 40

∴ Acceleration of ball is - 40 m/s²

__________________________

Now, use formula for calculating Force :

⇒F = ma

⇒F = 0.25 * -40

⇒F = -10

∴ Retarding force is -10 N

_______________________

$\: \bigstar \: \underline{\underline{\textsf{\textbf{Note :}}}}$ Acceleration and Force have magnitude -40 m/s² and -10 N respectively. Negative sign shows that they are in opposite directions of motion.