6. 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.
Answers
Answer:
Retarding force applied by fielder to stop the ball is 10 N .
Explanation:
- initial velocity of ball, u = 20 ms⁻¹
- final velocity of ball, v = 0
(since ball comes to rest)
- time taken for stopping ball, t = 0.5 s
- mass of ball, m = 250 g = 0.25 kg
↪ we need to find the retarding force (Let F ) applied by the fielder to stop the ball
Since ,
▸Force = mass × acceleration
and
▸acceleration = ( change in velocity ) / time taken
therefore,
→ F = m a
→ F = m × ( v - u ) / t
putting known values
→ F = 0.25 × ( 0 - 20 ) / 0.5
→ F = 0.25 × ( - 40 )
→ F = - 10 N
Hence,
Retarding force applied by fielder on the ball to stop it is 10 Newton .
(negative sign with magnitude of force shows that Force is applied in the direction opposite to the motion of ball).
Answer:
Retarding force applied by fielder to stop the ball is 10 N .
Given:
➛Initial velocity of ball, u = 20 m/s
➛Final velocity of ball, v = 0
➛Time taken for stopping ball, t = 0.5 s
➛Mass of ball, m = 250 g = 0.25 kg
To Find:
Retarding force applied by fielder to stop the ball
Solution:
➛Initial velocity of ball, u = 20 m/s.
➛Final velocity of ball, v = 0 .
➛Time taken for stopping ball, t = 0.5 s.
➛Mass of ball, m = 250 g = 0.25 kg.
We know,
➾Force = mass × acceleration
and
➾acceleration = ( change in velocity ) /
time taken
therefore,
➪ F = m a
➪ F = m × ( v - u ) / t
➪F = 0.25 × ( 0 - 20 ) / 0.5
➪ F = 0.25 × ( - 40 )
➪F = - 10 N. ( -ve sign means force is in opposite side of motion)
Therefore,
Retarding force applied by fielder on the ball to stop it is 10 N.