Question

In cricket match a bowler throws a ball of 0.5 kg with a speed of 20 m/s. When a batsman swings the bat, the ball strikes with the bat normal toit and returns in opposite direction with speed of 30 m/s. If the time of the contact of the ball with the bat is 0.1 s, then the force acting on the bat is ............. N.
(A) 250
(B) 25
(C) 50
(D) 125

Answer 1

Answer:

(A) 250

Explanation:

Given,

Mass of ball= 0.5 kg

let's assume the velocity of ball towards batsman is positive then the velocity of ball away from batsman will be negative.Hence,

speed of ball to thrown by bowler= 20 m/s

speed of ball return back after hitting the bat = -30 m/s

$acceleration\ of\ ball\ =\ \dfrac{change\ in\ velocity}{time}\\\\acceleration\ of\ ball\ =\ \dfrac{20-(-30)}{0.1}}\\\\acceleration\ of\ ball\ =\ \dfrac{50}{0.1}=500\ ms^{-2}$

Hence Force acting on bat= mass of ball x acceleration

                                           =500 x 0.5 N

                                            =250 N

Hence the force acting on bat will be 250 N.