question_answer

A cricket ball of mass 0.5 kg strikes a cricket bat normally with a velocity of 20 ms 1 and rebounds with velocity of 10 ms-1. The impulse of the force exerted by the ball on the bat is

A)  15 Ns                                   

B)  25 Ns

C)  30 Ns                                   

D)  10 Ns

Correct Answer: A

Solution :

                 During collision of ball with the wall horizontal momentum changes (vertical momentum remains constant) \[\therefore \] \[F=\frac{change\text{ }in\text{ }horizontal\text{ }momentum}{time\text{ }of\text{ }contact}\] \[=\frac{m(v+u)\cos \theta }{t}\] \[=0.5(20+10)\cos \theta =15Ns\]